Nepal Engineering Council · Electrical Engineering · Chapter 1
Fundamental of Electrical Engineering
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180 questions in 6 syllabus topics.
1.1 Basic circuit concept
32 questions · AExE0101
1. Ohm's law states that, provided temperature and other physical conditions remain constant, the current through a conductor is
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Think of V = IR with R fixed.
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Answer: D. directly proportional to the voltage across it
Ohm's law: V = IR with R constant, so I is directly proportional to V.
2. An electric heater draws 5 A from a 230 V supply. Its resistance is
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Rearrange V = IR.
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Answer: D. 46 Ω
R = V/I = 230/5 = 46 Ω.
3. Ohm's law is NOT applicable to which of the following?
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Look for the element whose V-I graph is not a straight line.
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Answer: A. A semiconductor diode
A diode has a non-linear V-I characteristic, so V/I is not constant; the others are linear (ohmic) at constant temperature.
4. One kilowatt-hour (kWh) of electrical energy is equal to
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Convert both kilo and hour into SI units.
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Answer: C. 3.6 × 10⁶ J
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
5. A 1.5 kW water heater is used for 4 hours every day. The energy consumed in a 30-day month is
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Energy = power × total operating time.
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Answer: A. 180 kWh
Energy = 1.5 kW × 4 h × 30 = 180 kWh.
6. The hot resistance of a 100 W, 230 V incandescent lamp is
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Use P = V²/R.
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Answer: D. 529 Ω
R = V²/P = 230²/100 = 529 Ω.
7. Three equal resistors connected in series across a constant-voltage supply dissipate a total power P. If they are reconnected in parallel across the same supply, the total power becomes
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Compare the two equivalent resistances; P = V²/R.
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Answer: A. 9P
Series: R_eq = 3R; parallel: R_eq = R/3. Power V²/R_eq increases by a factor (3R)/(R/3) = 9.
8. A wire is drawn out so that its length is doubled while its volume stays the same. Its resistance becomes
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Volume constant means area changes too.
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Answer: A. 4 times the original
Doubling length at constant volume halves the area; R = ρl/A becomes ρ(2l)/(A/2) = 4R.
9. Which of the following materials has a negative temperature coefficient of resistance?
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Pure metals behave alike; one material here is a non-metal.
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Answer: A. Carbon
The resistance of carbon decreases as temperature rises (negative coefficient); pure metals have a positive coefficient.
10. Which of the following metals has the highest electrical conductivity at room temperature?
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Copper is used for cost reasons, not because it is the best.
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Answer: B. Silver
Silver has the lowest resistivity (about 1.6 × 10⁻⁸ Ω·m) of all metals, slightly better than copper.
11. Which of the following is an insulating material?
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Three of these are used to carry current or produce heat/light.
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Answer: C. Mica
Mica has very high resistivity and dielectric strength; nichrome and tungsten are metallic conductors and graphite also conducts.
12. A good insulating material should have
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It must both block current and resist breakdown.
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Answer: C. high resistivity and high dielectric strength
An insulator must block current flow (high resistivity) and withstand high electric stress without breakdown (high dielectric strength).
13. A copper coil has a resistance of 10 Ω at 20 °C. Taking α₂₀ = 0.00393 /°C, its resistance at 70 °C is about
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Use the temperature rise above 20 °C, not 70 °C itself.
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Answer: D. 11.97 Ω
R₇₀ = R₂₀[1 + α₂₀(70 − 20)] = 10(1 + 0.00393 × 50) = 11.97 Ω.
14. In a series circuit, the quantity that is the same through every element is
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There is only one path for charge to flow.
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Answer: A. current
Series elements form a single path, so the same current flows through each.
15. A 4 Ω resistor is connected in series with a parallel combination of 6 Ω and 3 Ω across a 12 V source. The source current is
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Reduce the parallel pair first.
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Answer: D. 2 A
6 ∥ 3 = 2 Ω; total = 4 + 2 = 6 Ω; I = 12/6 = 2 A.
16. A current of 10 A divides between two parallel resistors of 2 Ω and 8 Ω. The current through the 2 Ω resistor is
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The smaller resistor takes the larger share.
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Answer: C. 8 A
I₂Ω = I × 8/(2 + 8) = 8 A; the smaller resistance carries the larger current.
17. A 24 V supply is applied across 4 kΩ and 8 kΩ resistors in series. The voltage across the 8 kΩ resistor is
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Voltage divides in proportion to resistance.
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Answer: B. 16 V
Voltage divider: V = 24 × 8/(4 + 8) = 16 V.
18. A balanced delta network has 30 Ω in each arm. The equivalent star network has, in each arm,
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The star equivalent of a balanced delta is smaller.
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Answer: B. 10 Ω
For balanced networks R_Y = R_Δ/3 = 30/3 = 10 Ω.
19. A star network has R_A = 5 Ω, R_B = 10 Ω and R_C = 20 Ω. In the equivalent delta, the resistance connected between terminals A and B is
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Sum of the two adjacent arms plus their product over the third.
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Answer: A. 17.5 Ω
R_AB = R_A + R_B + R_A·R_B/R_C = 5 + 10 + 50/20 = 17.5 Ω.
20. A delta network has R_AB = 10 Ω, R_BC = 20 Ω and R_CA = 30 Ω. The star-equivalent resistance connected to terminal A is
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Use the two delta arms that meet at terminal A.
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Answer: B. 5 Ω
R_A = R_AB·R_CA/(R_AB + R_BC + R_CA) = 10 × 30/60 = 5 Ω.
21. Kirchhoff's current law is based on the principle of conservation of
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What cannot pile up at a junction?
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Answer: C. charge
Charge cannot accumulate at a node, so the sum of currents entering equals the sum leaving.
22. Kirchhoff's voltage law is a consequence of the principle of conservation of
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Voltage is energy per unit charge.
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Answer: D. energy
Around a closed loop the net work done on a unit charge is zero, so the algebraic sum of voltages is zero.
23. At a node, currents of 5 A and 3 A enter and a current of 2 A leaves through a third branch. The current leaving through the fourth branch is
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Total in = total out.
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Answer: D. 6 A
KCL: 5 + 3 = 2 + I, so I = 6 A.
24. A closed loop contains a 20 V source and three resistors. The voltage drops across two of the resistors are 8 V and 5 V. The drop across the third resistor is
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Sum of drops equals the source emf.
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Answer: B. 7 V
KVL: 20 = 8 + 5 + V, so V = 7 V.
25. A linear circuit is one in which
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It is about the parameters, not the layout.
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Answer: C. the circuit parameters (R, L, C) remain constant irrespective of the voltage or current
Linearity means the element parameters do not change with voltage or current, so the V-I relations are linear (superposition holds).
26. An iron-cored coil operated well into magnetic saturation is an example of
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Does its inductance stay fixed as current rises?
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Answer: A. a non-linear element
In saturation the inductance changes with current, so the element is non-linear (it is still passive and bilateral).
27. Which of the following is a unilateral element?
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Reverse the current direction and see which behaves differently.
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Answer: B. Diode
A diode conducts very differently in the two directions; R, L and C behave the same for either direction of current.
28. A network that contains at least one source of energy is called
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Sources supply energy.
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Answer: C. an active network
Active networks contain energy sources (voltage or current sources); passive networks contain only R, L, C.
29. Which of the following is a passive element?
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Which one cannot supply net energy?
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Answer: D. Inductor
An inductor only stores and returns energy; it cannot deliver net energy, unlike sources.
30. The internal resistance of an ideal voltage source is
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Its terminal voltage never drops with load.
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Answer: C. zero
An ideal voltage source keeps its terminal voltage constant for any current, which requires zero internal resistance.
31. A 12 V battery with internal resistance 0.5 Ω supplies a 5.5 Ω load. The terminal voltage of the battery is
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Subtract the internal drop from the emf.
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Answer: B. 11 V
I = 12/(0.5 + 5.5) = 2 A; V_t = 12 − 2 × 0.5 = 11 V.
32. A current of 2 A flows through a 10 Ω resistor for 5 minutes. The heat energy produced is
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Convert minutes to seconds.
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Answer: B. 12 kJ
P = I²R = 40 W; W = 40 × 300 s = 12 000 J = 12 kJ.
1.2 Network theorems
30 questions · AExE0102
33. The superposition theorem can be applied only to circuits that are
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Which property allows responses to be added?
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Answer: C. linear
Superposition depends on linearity (additivity and homogeneity); it holds for AC or DC sources.
34. While applying the superposition theorem, an ideal independent voltage source that is not being considered is replaced by
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What has zero voltage across it for any current?
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Answer: D. a short circuit
Setting an ideal voltage source to zero makes its terminal voltage zero, i.e. a short circuit (its internal resistance, if any, stays).
35. While applying the superposition theorem, an ideal independent current source that is not being considered is replaced by
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What carries zero current for any voltage?
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Answer: A. an open circuit
Setting a current source to zero means no current can flow through it, i.e. an open circuit.
36. The superposition theorem cannot be used directly to calculate
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Which quantity is not linear in the source values?
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Answer: B. the power dissipated in a resistor
Power is proportional to I² (non-linear), so powers due to individual sources cannot simply be added.
37. During superposition, dependent (controlled) sources in the circuit are
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Only sources that act on their own are switched off.
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Answer: D. left in the circuit as they are
Dependent sources are controlled by circuit variables; they remain active while each independent source acts alone.
38. A 12 V source in series with 4 Ω feeds node A. A 4 Ω resistor connects node A to the reference, and an ideal 2 A current source injects current into node A. By superposition, the voltage of node A is
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Find each source's contribution, then add.
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Answer: D. 10 V
12 V alone: 12 × 4/8 = 6 V. 2 A alone: 2 × (4 ∥ 4) = 4 V. Total = 6 + 4 = 10 V.
39. In a 5 Ω resistor, source X acting alone produces 3 A and source Y acting alone produces 1 A in the opposite direction. The actual power dissipated in the resistor is
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Add currents with sign first, then compute power.
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Answer: C. 20 W
Net current = 3 − 1 = 2 A, so P = 2² × 5 = 20 W. Adding the individual powers (45 + 5 = 50 W) is wrong.
40. Source E₁ = 10 V with series 5 Ω and source E₂ = 20 V with series 10 Ω are both connected to a common node, which is joined to the reference by R₃ = 10 Ω (both sources referenced to the same node). The current in R₃ is
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Each source's contribution is found with the other shorted.
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Answer: B. 1 A
E₁ alone gives 5 V across R₃ and E₂ alone gives 5 V, so V = 10 V and I = 10/10 = 1 A.
41. The Thevenin equivalent of a linear two-terminal network consists of
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Norton uses the other arrangement.
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Answer: C. a voltage source in series with an impedance
Thevenin: V_th in series with R_th (Z_th); the parallel current-source form is Norton's.
42. The Thevenin voltage of a network is the
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The load is removed first.
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Answer: A. open-circuit voltage across the load terminals
V_th is measured or calculated with the load removed (terminals open).
43. When a network contains only independent sources and resistors, the Thevenin resistance is found by
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Switch off the sources, then look into the terminals.
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Answer: D. replacing every source by its internal resistance and finding the resistance seen from the load terminals
With sources deactivated (ideal V-sources shorted, I-sources opened), R_th is the resistance looking into the open terminals.
44. When a network contains dependent sources, the Thevenin resistance is best obtained by
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Dependent sources must stay in the circuit.
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Answer: C. applying a test source at the terminals (or using V_oc/I_sc) with dependent sources kept active
Dependent sources cannot be switched off; R_th = V_test/I_test or V_oc/I_sc.
45. Thevenin's theorem is especially useful when
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Think about changing the load repeatedly.
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Answer: D. the current in one branch must be found for several different values of its load
The rest of the circuit is reduced once to V_th and R_th; each new load then needs only I = V_th/(R_th + R_L).
46. A 24 V source in series with 6 Ω is connected across a 3 Ω resistor, and the output terminals are taken across the 3 Ω resistor. The Thevenin equivalent is
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Open-circuit voltage by divider; resistance with the source shorted.
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Answer: B. 8 V, 2 Ω
V_th = 24 × 3/9 = 8 V; R_th = 6 ∥ 3 = 2 Ω.
47. A network has V_th = 8 V and R_th = 2 Ω. The current in a 6 Ω load connected to it is
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Include R_th in series with the load.
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Answer: B. 1 A
I = V_th/(R_th + R_L) = 8/(2 + 6) = 1 A.
48. A network gives an open-circuit voltage of 20 V and a short-circuit current of 4 A at its terminals. Its Thevenin resistance is
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Ratio of open-circuit voltage to short-circuit current.
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Answer: A. 5 Ω
R_th = V_oc/I_sc = 20/4 = 5 Ω.
49. The Norton current of a network is the
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The terminals are shorted, not opened.
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Answer: A. short-circuit current through the load terminals
I_N is the current that flows when the load terminals are shorted.
50. The Norton equivalent of a linear network consists of
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It is the dual of the Thevenin form.
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Answer: A. a current source in parallel with a resistance
Norton: I_N in parallel with R_N.
51. For the same linear network, the Norton resistance R_N and the Thevenin resistance R_th are related as
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Both are found the same way.
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Answer: C. R_N = R_th
Both are the resistance seen into the terminals with sources deactivated, so they are equal.
52. A Thevenin equivalent of 10 V in series with 5 Ω is converted to its Norton equivalent. The Norton source and resistance are
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I_N = V_th/R_th.
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Answer: A. 2 A in parallel with 5 Ω
I_N = V_th/R_th = 10/5 = 2 A; R_N = R_th = 5 Ω.
53. A Norton equivalent of 3 A in parallel with 4 Ω is converted to a Thevenin equivalent. The Thevenin voltage is
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Multiply the source current by the parallel resistance.
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Answer: C. 12 V
V_th = I_N × R_N = 3 × 4 = 12 V.
54. A 30 V source in series with 10 Ω is connected across a 15 Ω resistor, and the output terminals are taken across the 15 Ω resistor. The Norton equivalent is
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Shorting the output also shorts the 15 Ω resistor.
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Answer: B. 3 A in parallel with 6 Ω
Shorting the terminals bypasses 15 Ω, so I_N = 30/10 = 3 A; R_N = 10 ∥ 15 = 6 Ω.
55. In a DC network, maximum power is transferred to a load resistance R_L when
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Differentiate P with respect to R_L.
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Answer: B. R_L equals the Thevenin resistance of the network
Maximising P = V_th²R_L/(R_th + R_L)² gives R_L = R_th.
56. Under the condition of maximum power transfer from a source with internal resistance, the efficiency of power transfer is
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The same current flows through two equal resistances.
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Answer: B. 50 %
With R_L = R_th, equal power is lost in R_th and delivered to R_L, so efficiency is 50 %.
57. A network has V_th = 20 V and R_th = 5 Ω. The maximum power it can deliver to a resistive load is
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Set R_L = R_th and find I²R_L.
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Answer: A. 20 W
P_max = V_th²/(4R_th) = 400/20 = 20 W.
58. A 12 V source with internal resistance 2 Ω supplies a variable load resistor. The load current at the condition of maximum power transfer is
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First find the load resistance for maximum power.
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Answer: A. 3 A
At maximum power R_L = 2 Ω, so I = 12/(2 + 2) = 3 A.
59. An ideal 5 A current source is connected in parallel with a 4 Ω resistor. The maximum power this combination can deliver to a resistive load is
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Convert to a Thevenin equivalent first.
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Answer: C. 25 W
Thevenin: V_th = 5 × 4 = 20 V, R_th = 4 Ω; P_max = 20²/(4 × 4) = 25 W.
60. An AC source has internal impedance (3 + j4) Ω. For maximum power transfer to a load whose resistance and reactance can both be varied, the load impedance should be
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The reactances should cancel each other.
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Answer: D. (3 − j4) Ω
Maximum power transfer in AC requires Z_L = Z_s* (complex conjugate), cancelling the reactance.
61. An AC source has internal impedance (6 + j8) Ω. If the load is restricted to a pure resistance, maximum power is transferred when the load resistance is
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Use the magnitude of the source impedance.
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Answer: D. 10 Ω
For a purely resistive load, R_L = |Z_s| = √(6² + 8²) = 10 Ω.
62. Power transmission and distribution systems are NOT operated at the maximum power transfer condition mainly because
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Consider where the other half of the power goes.
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Answer: B. efficiency would be only 50 %
At maximum power transfer half the power is wasted in the source/line resistance; power systems need high efficiency.
1.3 Alternating current fundamentals
30 questions · AElE0103
63. The generation of alternating emf in an alternator is based on
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Moving conductors in a magnetic field.
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Answer: D. Faraday's law of electromagnetic induction
A conductor moving relative to a magnetic field has an emf induced in it proportional to the rate of change of flux linkage.
64. For a rectangular coil rotating at constant speed in a uniform magnetic field, the induced emf is maximum when the plane of the coil is
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Maximum emf occurs when flux linkage is changing fastest, not when it is largest.
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Answer: C. parallel to the magnetic field lines
When the coil plane is parallel to the field, the flux linkage is zero but changing fastest (the sides cut flux at right angles), so the emf is maximum.
65. A coil of 100 turns, each of area 0.02 m², rotates at 50 revolutions per second in a uniform field of 0.5 T. The peak value of the induced emf is about
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Use E_m = NBAω with ω = 2πn.
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Answer: B. 314 V
E_m = NBAω = 100 × 0.5 × 0.02 × (2π × 50) ≈ 314 V.
66. An 8-pole alternator runs at 750 rpm. The frequency of the generated emf is
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Use f = PN/120.
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Answer: C. 50 Hz
f = PN/120 = 8 × 750/120 = 50 Hz.
67. In a 6-pole machine, one complete mechanical revolution corresponds to
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Each pair of poles produces one cycle.
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Answer: B. 1080 electrical degrees
Electrical degrees = (P/2) × mechanical degrees = 3 × 360° = 1080°.
68. The time period of a 50 Hz alternating voltage is
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Period is the reciprocal of frequency.
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Answer: D. 20 ms
T = 1/f = 1/50 s = 20 ms.
69. The angular frequency of a 50 Hz supply is
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One cycle is 2π radians.
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Answer: B. 314.16 rad/s
ω = 2πf = 2π × 50 = 314.16 rad/s.
70. An alternating voltage is given by v = 325 sin 314t volts. Its rms value and frequency are
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Divide the peak by √2 and ω by 2π.
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Answer: B. 229.8 V and 50 Hz
V_rms = 325/√2 ≈ 229.8 V; f = ω/2π = 314/2π ≈ 50 Hz.
71. The instantaneous value at t = 0 of the voltage v = 100 sin(628t + 30°) V is
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Put t = 0; only the phase angle remains.
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Answer: A. 50 V
At t = 0, v = 100 sin 30° = 50 V.
72. A current i = 10 sin(100πt) A starts from zero at t = 0. The time taken for it to first reach 5 A is about
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Find the angle whose sine is 0.5, in radians.
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Answer: C. 1.67 ms
sin(100πt) = 0.5 gives 100πt = π/6, so t = 1/600 s ≈ 1.67 ms.
73. If v = V_m sin(ωt + 20°) and i = I_m sin(ωt − 40°), then the current
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Compare the two phase angles.
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Answer: A. lags the voltage by 60°
Phase difference = 20° − (−40°) = 60°, with i behind v, so i lags by 60°.
74. The average value of a sinusoidal current of peak value I_m, taken over one half-cycle, is
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Integrate over half a cycle only.
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Answer: C. 0.637 I_m
I_avg = 2I_m/π ≈ 0.637 I_m over a half-cycle (over a full cycle the average is zero).
75. The average value of a pure sinusoidal voltage taken over one complete cycle is
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Compare the positive and negative halves.
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Answer: A. zero
The positive and negative half-cycles are equal and cancel over a full cycle.
76. The rms value of a sinusoidal current of peak value 20 A is
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Divide by √2.
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Answer: B. 14.14 A
I_rms = I_m/√2 = 20/1.414 = 14.14 A.
77. The rms value of an alternating current is defined as the steady (DC) current which
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It is also called the effective or heating value.
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Answer: A. produces the same heat in a given resistance in the same time
RMS (effective) value is defined by equal heating effect, I²Rt.
78. The form factor of a sinusoidal waveform is
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Form factor = rms ÷ average.
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Answer: D. 1.11
Form factor = rms/average = 0.707 V_m/0.637 V_m = 1.11.
79. The peak (crest) factor of a sinusoidal waveform is
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Peak factor = peak ÷ rms.
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Answer: D. 1.414
Peak factor = V_m/V_rms = √2 = 1.414.
80. The form factor of a symmetrical square wave is
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Its magnitude is constant throughout.
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Answer: D. 1.0
For a square wave rms = average (half-cycle) = V_m, so form factor = 1.
81. The peak factor of a symmetrical triangular waveform is
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Recall the rms of a triangular wave.
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Answer: A. 1.732
For a triangular wave V_rms = V_m/√3, so peak factor = √3 = 1.732.
82. A symmetrical triangular voltage waveform has a peak value of 12 V. Its rms value is
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A triangular wave's rms is not V_m/√2.
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Answer: B. 6.93 V
V_rms = V_m/√3 = 12/1.732 = 6.93 V.
83. The average value (over a full cycle) of a half-wave rectified sine wave of peak 100 V is
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Only one half-cycle is present out of each full cycle.
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Answer: A. 31.8 V
V_avg = V_m/π = 100/π ≈ 31.8 V, because one half-cycle is missing.
84. The rms value of a half-wave rectified sine wave of peak value V_m is
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Mean square is half that of a full sine.
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Answer: A. 0.5 V_m
V_rms = √(V_m²/4) = V_m/2.
85. The peak-to-peak value of a 230 V (rms) sinusoidal supply voltage is about
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Peak is √2 × rms; peak-to-peak is twice the peak.
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Answer: D. 650.5 V
V_pp = 2√2 × 230 ≈ 650.5 V.
86. A sinusoidal current has an average (half-cycle) value of 9 A. Its rms value is about
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Use the form factor.
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Answer: B. 10 A
I_rms = form factor × I_avg = 1.11 × 9 ≈ 10 A.
87. A current i = 4 + 3√2 sin ωt A flows through a resistor. The rms value of this current is
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Combine the DC part and the AC rms part as squares.
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Answer: C. 5 A
I_rms = √(I_DC² + I_ac,rms²) = √(4² + 3²) = 5 A.
88. A sinusoidal current of 10 A rms flows through a 5 Ω resistor. The average power dissipated is
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Use rms, not peak, for power.
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Answer: C. 500 W
P = I_rms²R = 10² × 5 = 500 W.
89. Two currents i₁ = 10 sin ωt A and i₂ = 10 sin(ωt + 90°) A are added. The resultant is
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Add them as phasors, not as numbers.
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Answer: B. 14.14 sin(ωt + 45°) A
Phasor sum = 10∠0° + 10∠90° = 14.14∠45°.
90. AC voltmeters and ammeters used on power circuits are normally calibrated to read
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230 V supply is quoted in which value?
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Answer: C. rms values
Ratings and readings on power systems are in rms (effective) values, e.g. 230 V means 230 V rms.
91. A sinusoidal waveform is preferred for AC power systems mainly because
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Think of what inductors and capacitors do to a waveform.
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Answer: A. its derivative and integral are also sinusoids of the same frequency
Because L and C involve derivatives/integrals, a sinusoidal supply produces sinusoidal voltages and currents everywhere in a linear circuit.
92. In phasor diagrams, phasors are conventionally assumed to rotate
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Leading angles are measured in the positive (counter-clockwise) direction.
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Answer: D. anticlockwise at angular velocity ω
By convention, phasors rotate counter-clockwise at ω rad/s; angles measured anticlockwise are leading.
1.4 Electric circuit responses
30 questions · AElE0104
93. The time constant of a series R-L circuit is
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Check which combination has units of seconds.
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Answer: A. L/R
τ = L/R seconds (henry/ohm = second).
94. A 10 kΩ resistor is connected in series with a 100 µF capacitor. The time constant of the circuit is
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Multiply R in ohms by C in farads.
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Answer: D. 1 s
τ = RC = 10 × 10³ × 100 × 10⁻⁶ = 1 s.
95. When an uncharged capacitor is charged through a resistor from a DC source, after one time constant its voltage reaches about
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Evaluate 1 − e⁻¹.
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Answer: A. 63.2 % of the final value
v = V(1 − e⁻¹) = 0.632 V at t = τ.
96. In a decaying R-L circuit, the current after one time constant falls to about
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Evaluate e⁻¹.
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Answer: C. 36.8 % of its initial value
i = I₀e⁻¹ = 0.368 I₀ at t = τ.
97. A transient in a first-order R-L or R-C circuit is considered practically complete after about
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Find when e^(−t/τ) is below 1 %.
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Answer: A. 5 time constants
After 5τ the response is within about 0.7 % of its final value (1 − e⁻⁵ ≈ 0.993).
98. At the instant of switching (t = 0⁺), an initially uncharged capacitor behaves as
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What is its voltage just after switching?
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Answer: B. a short circuit
Capacitor voltage cannot change instantly; at t = 0⁺ it is still 0 V, like a short circuit.
99. At the instant of switching (t = 0⁺), an inductor carrying no initial current behaves as
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What is its current just after switching?
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Answer: C. an open circuit
Inductor current cannot change instantly; at t = 0⁺ it is still zero, like an open circuit.
100. In a DC circuit that has reached steady state, an ideal inductor and an ideal capacitor behave respectively as
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Use v = L di/dt and i = C dv/dt with nothing changing.
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Answer: C. a short circuit and an open circuit
With constant current, v_L = L di/dt = 0 (short); with constant voltage, i_C = C dv/dt = 0 (open).
101. Which of the following cannot change instantaneously in an electric circuit?
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Which change would require infinite current?
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Answer: D. Voltage across a capacitor
An instant change in capacitor voltage would need infinite current (i = C dv/dt); capacitor current and inductor voltage can jump.
102. A 10 V DC source is switched on to a series circuit of R = 5 Ω and L = 0.5 H. The current 0.1 s after switching is about
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0.1 s is exactly one time constant here.
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Answer: B. 1.26 A
τ = L/R = 0.1 s; i = (10/5)(1 − e^(−t/τ)) = 2(1 − e⁻¹) ≈ 1.26 A.
103. For the same circuit (10 V, R = 5 Ω, L = 0.5 H), the initial rate of rise of current just after switching on is
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Initially there is no drop across R.
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Answer: C. 20 A/s
At t = 0⁺ the current is zero, so the full 10 V appears across L: di/dt = V/L = 10/0.5 = 20 A/s.
104. A capacitor charged to 100 V discharges through a resistor; the time constant is 2 s. The capacitor voltage 4 s after the start of discharge is about
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4 s is two time constants.
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Answer: B. 13.5 V
v = 100e^(−4/2) = 100e⁻² ≈ 13.5 V.
105. A 12 V source charges an initially uncharged capacitor through a resistor; τ = 1 ms. The capacitor voltage at t = 2 ms is about
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2 ms is two time constants of charging.
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Answer: D. 10.38 V
v = 12(1 − e⁻²) ≈ 12 × 0.865 = 10.38 V.
106. The initial charging current of an uncharged capacitor connected through a 20 kΩ resistor to a 100 V DC supply is
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The capacitor voltage is still zero at the start.
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Answer: B. 5 mA
At t = 0⁺ the capacitor acts as a short, so i = V/R = 100/20 000 = 5 mA.
107. In an R-C charging circuit with RC = 10 ms, the time taken for the capacitor voltage to reach half its final value is about
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Solve e^(−t/τ) = 0.5.
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Answer: D. 6.93 ms
1 − e^(−t/τ) = 0.5 gives t = τ ln 2 = 0.693 × 10 ms = 6.93 ms.
108. The energy stored in a 2 H inductor carrying a steady current of 3 A is
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Use ½LI².
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Answer: D. 9 J
W = ½LI² = ½ × 2 × 9 = 9 J.
109. The energy stored in a 50 µF capacitor charged to 200 V is
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Use ½CV² with C in farads.
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Answer: B. 1 J
W = ½CV² = ½ × 50 × 10⁻⁶ × 200² = 1 J.
110. Doubling the resistance in a series R-L circuit (inductance unchanged) makes the time constant
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R is in the denominator.
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Answer: C. half as large
τ = L/R, so doubling R halves τ.
111. A coil of inductance 200 mH and resistance 40 Ω is switched onto a DC supply. The current practically reaches its steady value after about
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Find τ first, then use the 5τ rule.
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Answer: D. 25 ms
τ = L/R = 0.2/40 = 5 ms; steady state ≈ 5τ = 25 ms.
112. When the current in a highly inductive circuit is suddenly interrupted by opening a switch, a high voltage appears across the switch because
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Think of Faraday's law for self-inductance.
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Answer: C. the induced emf L di/dt is very large
A very rapid fall in current makes di/dt large, so the self-induced emf L di/dt can be many times the supply voltage.
113. A series R-L-C circuit is critically damped when
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Set the discriminant of the characteristic equation to zero.
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Answer: B. R = 2√(L/C)
Critical damping occurs when the roots of s² + (R/L)s + 1/(LC) = 0 are equal: (R/2L)² = 1/(LC), i.e. R = 2√(L/C).
114. A series R-L-C circuit has L = 1 H and C = 1 µF. The value of R for critical damping is
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Use R = 2√(L/C).
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Answer: A. 2 kΩ
R = 2√(L/C) = 2√(1/10⁻⁶) = 2 × 1000 = 2000 Ω.
115. A series R-L-C circuit has R = 100 Ω, L = 0.1 H and C = 10 µF. Its transient response to a step input is
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Compare R with 2√(L/C).
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Answer: A. underdamped (oscillatory, decaying)
ζ = (R/2)√(C/L) = 50 × √(10⁻⁴) = 0.5 < 1, so the response is underdamped. (Critical R would be 200 Ω.)
116. If the resistance of a series R-L-C circuit is zero, its natural response to a step input is
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Nothing dissipates energy.
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Answer: C. a sustained oscillation of constant amplitude
With R = 0 there is no damping (ζ = 0), so energy keeps exchanging between L and C indefinitely.
117. The complete response of a linear circuit to a suddenly applied source is the sum of
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One part dies out, the other remains.
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Answer: A. the natural (transient) response and the forced (steady-state) response
Total response = transient (natural, decays to zero) + steady-state (forced by the source).
118. A series R-L circuit with R = 3 Ω and X_L = 4 Ω is connected to a 100 V AC supply. In steady state, the current is
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Find |Z| and tan φ = X_L/R.
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Answer: B. 20 A lagging the voltage by 53.1°
Z = √(3² + 4²) = 5 Ω, I = 100/5 = 20 A, φ = tan⁻¹(4/3) = 53.1° lagging.
119. In a series R-C circuit supplied with sinusoidal voltage, the steady-state current
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Mixed resistive and capacitive behaviour.
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Answer: D. leads the voltage by an angle between 0° and 90°
The capacitive reactance makes the current lead; the resistance keeps the angle below 90°.
120. In a purely capacitive AC circuit, the current
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Differentiate a sine.
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Answer: A. leads the voltage by 90°
i = C dv/dt; for v = V_m sin ωt, i = ωCV_m cos ωt, which leads v by 90°.
121. The reactance of a 0.1 H inductor at 50 Hz is
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X_L = 2πfL.
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Answer: B. 31.4 Ω
X_L = 2πfL = 2π × 50 × 0.1 = 31.4 Ω.
122. The reactance of a 100 µF capacitor at 50 Hz is
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X_C = 1/(2πfC), with C in farads.
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Answer: A. 31.83 Ω
X_C = 1/(2πfC) = 1/(2π × 50 × 100 × 10⁻⁶) = 31.83 Ω.
1.5 AC series and parallel circuits
29 questions · AElE0105
123. The units of active, reactive and apparent power are respectively
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Reactive power has its own special unit.
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Answer: C. W, VAR and VA
Active power P in watts, reactive power Q in volt-ampere reactive, apparent power S in volt-amperes.
124. Power factor of an AC circuit is defined as the ratio of
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It equals cos φ in the power triangle.
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Answer: A. active power to apparent power
pf = P/S = cos φ for sinusoidal quantities.
125. A single-phase load draws 10 A from a 230 V supply at a power factor of 0.8 lagging. The active power consumed is
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P = VI cos φ.
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Answer: A. 1840 W
P = VI cos φ = 230 × 10 × 0.8 = 1840 W.
126. A load takes 50 kVA and 40 kW. Its reactive power and power factor are
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Use the power triangle S² = P² + Q².
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Answer: D. 30 kVAR and 0.8
Q = √(S² − P²) = √(2500 − 1600) = 30 kVAR; pf = 40/50 = 0.8.
127. The power factor of a purely inductive circuit is
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What is cos 90°?
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Answer: A. zero, lagging
In a pure inductor the current lags by 90°, so cos 90° = 0 (lagging).
128. The average power consumed by a pure inductor connected to a sinusoidal supply is
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Energy is stored and returned every cycle.
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Answer: B. zero
Energy stored in the magnetic field in one quarter-cycle is returned in the next, so the average power is zero (I²X_L is reactive power).
129. A series circuit with R = 6 Ω and X_L = 8 Ω is connected to a 100 V AC supply. The active power consumed is
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Only the resistance consumes active power.
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Answer: D. 600 W
Z = 10 Ω, I = 10 A, P = I²R = 100 × 6 = 600 W.
130. A series circuit with R = 20 Ω and X_L = 15 Ω carries a current of 5 A. The reactive power drawn is
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Reactive power is associated with the reactance.
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Answer: C. 375 VAR
Q = I²X_L = 25 × 15 = 375 VAR. (P = 500 W and S = 625 VA.)
131. The impedance of a load is (8 + j6) Ω. Its power factor is
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pf = R/|Z|; the sign of j tells lag or lead.
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Answer: C. 0.8 lagging
|Z| = 10 Ω; pf = R/|Z| = 8/10 = 0.8; positive (inductive) reactance means lagging.
132. A load has V = 100∠0° V and I = 5∠−30° A. The active and reactive power absorbed by the load are
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Use S = VI* (conjugate of the current).
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Answer: D. 433 W and 250 VAR (inductive)
S = VI* = 100 × 5∠30° = 433 + j250; current lagging, so the load absorbs inductive VAR.
133. Which of the following loads operates at a leading power factor?
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Leading current comes from capacitance.
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Answer: D. A series R-C circuit
Capacitive loads draw current that leads the voltage.
134. The power factor of an inductive load is usually improved by
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You need a source of leading reactive power.
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Answer: C. connecting capacitors in parallel with the load
Shunt capacitors supply leading VAR that cancel part of the load's lagging VAR without changing the load voltage.
135. For the same active power at the same voltage, a low power factor results in
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Look at I = P/(V cos φ).
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Answer: D. larger current and therefore higher I²R losses
I = P/(V cos φ); a low cos φ increases current, copper losses, voltage drop and the required conductor and equipment rating.
136. A 100 kW load operates at 0.6 power factor lagging. The capacitor kVAR needed to raise the power factor to 0.9 lagging is about
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Find the reactive power before and after with active power fixed.
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Answer: D. 84.9 kVAR
Q_C = P(tan φ₁ − tan φ₂) = 100(1.333 − 0.484) ≈ 84.9 kVAR.
137. At series resonance in an R-L-C circuit, the impedance is
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The reactances cancel.
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Answer: A. minimum and equal to R
X_L = X_C, so they cancel and Z = R, its minimum; current is maximum and pf is unity.
138. A series R-L-C circuit has L = 10 mH and C = 10 µF. Its resonant frequency is about
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Don't forget the 2π.
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Answer: B. 503 Hz
f₀ = 1/(2π√(LC)) = 1/(2π√(10⁻⁷)) ≈ 503 Hz. (3162 is ω₀ in rad/s.)
139. A series R-L-C circuit has R = 10 Ω, L = 0.1 H and C = 10 µF. Its quality factor at resonance is
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Q = (1/R)√(L/C) for a series circuit.
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Answer: A. 10
Q = (1/R)√(L/C) = (1/10)√(0.1/10⁻⁵) = (1/10)(100) = 10.
140. A series resonant circuit with quality factor 10 is supplied with 10 V at its resonant frequency. The voltage across the capacitor is
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Series resonance magnifies voltage by Q.
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Answer: B. 100 V
At resonance V_C = Q × V = 10 × 10 = 100 V (voltage magnification).
141. A series resonant circuit has a resonant frequency of 1 kHz and a quality factor of 20. Its bandwidth is
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Bandwidth = f₀/Q.
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Answer: B. 50 Hz
BW = f₀/Q = 1000/20 = 50 Hz.
142. At frequencies below series resonance, a series R-L-C circuit behaves as
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At low frequency, which reactance is larger?
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Answer: A. a capacitive circuit with leading current
Below f₀, X_C = 1/(ωC) exceeds X_L = ωL, so the net reactance is capacitive.
143. At the half-power (cut-off) frequencies of a series resonant circuit, the current is
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Half the power, not half the current.
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Answer: A. 0.707 times its value at resonance
Power ∝ I², so half power occurs at I = I₀/√2 = 0.707 I₀.
144. Increasing the quality factor of a resonant circuit (same resonant frequency)
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Bandwidth and Q are inversely related.
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Answer: C. reduces the bandwidth and improves selectivity
BW = f₀/Q, so a higher Q gives a narrower, more selective response.
145. If the capacitance of a series resonant circuit is made four times larger (inductance unchanged), the resonant frequency becomes
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f₀ varies with the square root of C.
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Answer: B. half of the original
f₀ ∝ 1/√C, so 4C gives f₀/√4 = f₀/2.
146. At parallel resonance of an ideal L-C circuit, the impedance and the supply current are respectively
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Branch currents are equal and opposite.
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Answer: C. maximum and minimum
The branch currents in L and C cancel, so the line current is minimum and the impedance maximum.
147. A parallel resonant circuit is also known as
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The series circuit accepts; the parallel one ...
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Answer: D. a rejector circuit
It offers maximum impedance (rejects current) at resonance; the series resonant circuit is the acceptor circuit.
148. A coil with R = 10 Ω and L = 0.1 H is connected in parallel with a 10 µF capacitor. The dynamic impedance of the circuit at resonance is
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Use Z_D = L/(CR).
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Answer: B. 1000 Ω
Z_D = L/(CR) = 0.1/(10 × 10⁻⁶ × 10) = 1000 Ω.
149. Compared with an ideal L-C tank of the same L and C, the resonant frequency of a parallel circuit whose coil has resistance R is
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The R²/L² term is subtracted.
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Answer: B. slightly lower
f₀ = (1/2π)√(1/(LC) − R²/L²), which is less than 1/(2π√(LC)).
150. A 10 Ω resistor and an inductor of reactance 10 Ω are connected in parallel across a 100 V AC supply. The supply current is
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Branch currents add as phasors.
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Answer: C. 14.14 A
I_R = 10 A (in phase), I_L = 10 A (lagging 90°); I = √(10² + 10²) = 14.14 A.
151. A series circuit has R = 30 Ω, X_L = 70 Ω and X_C = 30 Ω. The magnitude of its impedance is
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The two reactances oppose each other.
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Answer: A. 50 Ω
Z = √(R² + (X_L − X_C)²) = √(900 + 1600) = 50 Ω.
1.6 Three phase systems
29 questions · AElE0106
152. In a balanced star-connected system, the line and phase quantities are related as
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In star, each line is in series with one phase.
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Answer: A. V_L = √3 V_ph and I_L = I_ph
In star, each line carries its phase current, and the line voltage is the phasor difference of two phase voltages, √3 times larger.
153. In a balanced delta-connected system, the line and phase quantities are related as
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In delta, each phase is across two lines.
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Answer: A. V_L = V_ph and I_L = √3 I_ph
In delta, each phase is directly across two lines, and each line current is the difference of two phase currents.
154. In a balanced three-phase supply, the three phase voltages are displaced from one another by
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360° shared among three phases.
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Answer: A. 120 electrical degrees
The three windings are spaced 120° electrical apart, so the voltages are 120° apart.
155. In a balanced delta-connected load (positive sequence), each line current
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Draw I_ab − I_ca as phasors.
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Answer: C. lags its corresponding phase current by 30°
I_a = I_ab − I_ca = √3 I_ab∠−30°, so the line current lags the phase current by 30°.
156. The instantaneous sum of the three phase voltages of a balanced three-phase supply is
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Add three equal phasors spaced 120°.
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Answer: A. zero
Three equal phasors 120° apart add to zero at every instant.
157. A 400 V (line), three-phase, star-connected supply has a phase voltage of about
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Divide by √3.
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Answer: A. 230.9 V
V_ph = V_L/√3 = 400/1.732 = 230.9 V.
158. An 11 kV, star-connected generator has a phase voltage of about
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Rated voltage of a three-phase machine is the line voltage.
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Answer: C. 6.35 kV
V_ph = 11/√3 ≈ 6.35 kV.
159. Three 10 Ω resistors are connected in delta across a 400 V, three-phase supply. The phase and line currents are
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In delta the full line voltage is across each resistor.
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Answer: D. 40 A and 69.3 A
I_ph = 400/10 = 40 A; I_L = √3 × 40 = 69.3 A.
160. A balanced delta-connected load has a phase current of 10 A. The line current is
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Multiply by √3.
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Answer: A. 17.32 A
I_L = √3 I_ph = 1.732 × 10 = 17.32 A.
161. Three identical resistors draw a total power P when connected in star across a three-phase supply. If they are reconnected in delta across the same supply, the total power becomes
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Compare the voltage across each resistor.
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Answer: C. 3P
In delta each resistor gets √3 times the voltage, so power per resistor rises by (√3)² = 3.
162. A balanced three-phase load draws 20 A per line from a 415 V supply at 0.85 power factor. The total active power is about
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Use √3 V_L I_L cos φ.
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Answer: B. 12.22 kW
P = √3 V_L I_L cos φ = 1.732 × 415 × 20 × 0.85 ≈ 12.22 kW.
163. A 20 kW, three-phase balanced load operates from a 400 V supply at 0.8 power factor. The line current is about
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Rearrange P = √3 V_L I_L cos φ.
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Answer: B. 36.1 A
I_L = P/(√3 V_L cos φ) = 20 000/(1.732 × 400 × 0.8) ≈ 36.1 A.
164. A balanced star-connected load of (8 + j6) Ω per phase is supplied from a 400 V, three-phase supply. The total active power is
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Find phase current first, then 3I²R.
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Answer: C. 12.8 kW
V_ph = 230.9 V, I = 230.9/10 = 23.09 A, P = 3 × 23.09² × 8 = 12.8 kW.
165. A balanced three-phase load takes 30 kW at a power factor of 0.6 lagging. Its apparent power is
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S = P/pf.
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Answer: D. 50 kVA
S = P/cos φ = 30/0.6 = 50 kVA (Q = 40 kVAR).
166. The total reactive power of a balanced three-phase load is given by
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Same form as active power, but with the other trig function.
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Answer: B. √3 V_L I_L sin φ
Q = 3 V_ph I_ph sin φ = √3 V_L I_L sin φ for both star and delta.
167. The power factor of a balanced three-phase load is the cosine of the angle between
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It is the angle of the load impedance per phase.
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Answer: D. the phase voltage and the phase current
φ is the impedance angle of each phase, i.e. between phase voltage and phase current; line quantities differ by an extra 30° in some cases.
168. The total instantaneous power delivered to a balanced three-phase load is
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Add the three pulsating components.
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Answer: B. constant
The double-frequency pulsating terms of the three phases cancel, leaving a constant total power; this gives smooth torque in motors.
169. In the two-wattmeter method of measuring three-phase power, if one wattmeter reads zero the load power factor is
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Find φ that makes cos(30° + φ) = 0.
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Answer: D. 0.5
W₁ = V_L I_L cos(30° − φ) and W₂ = V_L I_L cos(30° + φ); W₂ = 0 when φ = 60°, so pf = 0.5.
170. In a two-wattmeter test on a balanced three-phase load, the readings are 8 kW and 4 kW. The load power factor is about
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Use tan φ = √3(W₁ − W₂)/(W₁ + W₂).
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Answer: B. 0.866
tan φ = √3(W₁ − W₂)/(W₁ + W₂) = √3 × 4/12 = 0.577, so φ = 30° and pf = 0.866.
171. In a balanced star-connected load, the current in the neutral conductor is
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Add three equal currents 120° apart.
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Answer: D. zero
The three balanced line currents are equal and 120° apart, so their phasor sum is zero.
172. The main purpose of the neutral wire in a three-phase four-wire distribution system is to
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Loads on a distribution system are rarely equal.
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Answer: A. carry unbalanced current and keep the phase voltages nearly equal
The neutral returns the unbalance current, holding each load at its own phase voltage even when loads differ.
173. A three-phase four-wire system (with a neutral) can be obtained only from
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Where does the neutral come from?
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Answer: C. a star-connected source
Only a star connection has a common star point from which the neutral can be taken.
174. A four-wire star system supplies unity-power-factor phase currents of 10 A, 10 A and 5 A (balanced voltages, positive sequence). The neutral current is
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Treat the currents as phasors 120° apart.
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Answer: B. 5 A
The phasor sum 10∠0° + 10∠−120° + 5∠120° = −5∠120°, magnitude 5 A.
175. Unbalanced loading of a three-phase four-wire distribution system results in
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The line currents no longer cancel.
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Answer: A. current flowing in the neutral conductor
When the phase currents are unequal their phasor sum is not zero, and that sum flows in the neutral.
176. If the neutral of a star-connected, three-phase, four-wire lighting load breaks while the loads are unbalanced, then
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The star point of the load is no longer held at neutral potential.
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Answer: B. the phase voltages become unequal (neutral shift), so some lamps get over-voltage and others under-voltage
Without the neutral, the star point floats; the phase with the highest impedance (least load) gets a higher voltage and the most heavily loaded phase a lower voltage.
177. Interchanging any two supply lines of a three-phase induction motor
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Think about phase sequence.
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Answer: D. reverses its direction of rotation
Swapping two lines reverses the phase sequence, which reverses the rotating magnetic field.
178. Compared with a single-phase system transmitting the same power at the same voltage and losses, a three-phase system
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One reason three-phase is used in grids.
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Answer: C. requires less conductor material
Three-phase transmission needs only about 75 % of the copper of single-phase for the same power, voltage and losses.
179. In a two-wattmeter test on a balanced three-phase load, the readings are 8 kW and 4 kW. The total reactive power of the load is about
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Reactive power uses the difference of the readings.
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Answer: D. 6.93 kVAR
Q = √3(W₁ − W₂) = 1.732 × 4 = 6.93 kVAR.
180. In a balanced three-phase star connection (positive sequence), the line voltage V_RY
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Subtract V_Y from V_R as phasors.
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Answer: C. leads the phase voltage V_R by 30°
V_RY = V_R − V_Y = √3 V_R∠30°.