Fundamentals of Electrical and Electronics Circuits

Average value is the mean of all instantaneous values of an alternating quantity over one complete cycle (or over a half cycle, because the full-cycle mean of a symmetrical wave is zero).

$$I_{av}=\frac{1}{T}\int_0^T i\,dt$$

For a sine wave, taken over a half cycle: $I_{av}=\frac{2}{\pi}I_m = 0.637\,I_m$.

RMS (effective) value is the value of direct current that produces the same heating effect in a resistor as the alternating current does in the same time. It is the square root of the mean of the squares of the instantaneous values over one cycle.

$$I_{rms}=\sqrt{\frac{1}{T}\int_0^T i^2\,dt}$$

For a sine wave: $I_{rms}=\frac{I_m}{\sqrt2}=0.707\,I_m$.

The ratio $I_{rms}/I_{av}=1.11$ is called the form factor of a sine wave. Ammeters and voltmeters on AC supply read RMS values (for example, 230 V supply has a peak of $230\sqrt2 = 325$ V).

Assumptions (from the figure): the 3 A current source is in the top branch of the right mesh in series with the 6 Ω resistor, and it forces 3 A to flow leftwards there. Mesh currents $I_1$ (left) and $I_2$ (right) are both taken clockwise. The 12 V source aids $I_1$ and the 6 V source opposes $I_2$.

Step 1: Right mesh (current source in the branch). Clockwise $I_2$ flows to the right along the top branch, but the source forces 3 A to the left, so

$$I_2=-3\ \text{A}$$

A current source in a mesh needs no KVL equation for that mesh; its current fixes the mesh current directly.

Step 2: KVL in the left mesh.

$$\begin{aligned} 12 &= 5I_1 + 2(I_1-I_2)\\ 12 &= 7I_1 - 2I_2\\ 7I_1 &= 12 + 2(-3) = 6\\ I_1 &= \frac{6}{7} = 0.857\ \text{A} \end{aligned}$$

Step 3: Current in the 2 Ω shared branch.

$$I_{2\Omega}=I_1-I_2 = 0.857-(-3)=3.857\ \text{A}$$

Answer: $I_1 = 0.857$ A (clockwise), $I_2 = -3$ A (i.e. 3 A anticlockwise, fixed by the source); current in the 2 Ω resistor = 3.857 A.

Step 1: Bring both quantities to the same form.

$$i=20\sqrt2\cos(314t-45^\circ)=20\sqrt2\sin(314t+45^\circ)$$ $$v=400\sqrt2\sin(314t+40^\circ)$$

RMS values: $V=400$ V, $I=20$ A. Phase angle of current is $+45^\circ$ and of voltage $+40^\circ$, so the current leads the voltage by

$$\phi = 45^\circ-40^\circ = 5^\circ$$

A leading current means the circuit is a series R–C circuit.

Step 2: Impedance and parameters.

$$\begin{aligned} Z &= \frac{V}{I}=\frac{400}{20}=20\ \Omega\\ R &= Z\cos\phi = 20\cos5^\circ = 19.92\ \Omega\\ X_C &= Z\sin\phi = 20\sin5^\circ = 1.743\ \Omega\\ C &= \frac{1}{\omega X_C}=\frac{1}{314\times1.743}=1827\ \mu\text{F} \end{aligned}$$

Step 3: Voltage drops.

$$\begin{aligned} V_R &= IR = 20\times19.92 = 398.5\ \text{V}\\ V_C &= IX_C = 20\times1.743 = 34.86\ \text{V} \end{aligned}$$

Check: $\sqrt{398.5^2+34.86^2}=400$ V.

Step 4: Power factor and power.

$$\cos\phi=\cos5^\circ=0.996\ \text{(leading)}$$ $$P = VI\cos\phi = 400\times20\times0.996 = 7970\ \text{W}\ (=I^2R)$$

Answer: $R=19.92\ \Omega$, $C=1827\ \mu$F ($X_C=1.743\ \Omega$); $V_R=398.5$ V, $V_C=34.86$ V; pf = 0.996 leading; $P=7.97$ kW.

Given: $W_1=2100$ W, $W_2=1550$ W, $V_L=400$ V, star connection.

(i) Per-phase average power. Total power is the sum of the readings:

$$P = W_1+W_2 = 3650\ \text{W}, \qquad P_{ph}=\frac{3650}{3}=1216.67\ \text{W}$$

(ii) Per-phase reactive power. Total reactive power is $\sqrt3\,(W_1-W_2)$:

$$Q=\sqrt3(2100-1550)=952.63\ \text{VAR}, \qquad Q_{ph}=\frac{952.63}{3}=317.54\ \text{VAR}$$

(iii) Power factor.

$$\tan\phi=\frac{Q}{P}=\frac{952.63}{3650}=0.261\ \Rightarrow\ \phi=14.63^\circ$$ $$\cos\phi=0.9676\ \text{(lagging, for an inductive load)}$$

(iv) Phase impedance. Line current from $P=\sqrt3 V_L I_L\cos\phi$:

$$I_L=\frac{3650}{\sqrt3\times400\times0.9676}=5.445\ \text{A}=I_{ph}$$ $$V_{ph}=\frac{400}{\sqrt3}=230.94\ \text{V}, \qquad Z_{ph}=\frac{230.94}{5.445}=42.41\ \Omega$$

Then $R_{ph}=Z\cos\phi=41.04\ \Omega$ and $X_{ph}=Z\sin\phi=10.71\ \Omega$, so $Z_{ph}=41.04+j10.71\ \Omega$.

Answer: (i) 1216.67 W, (ii) 317.54 VAR, (iii) 0.9676 lagging, (iv) 42.41 Ω ($41.04+j10.71\ \Omega$).

Circuit reading (from the figure). Node C is joined to the bottom rail, so take C as the reference (0 V). The 2 A source and the 12 Ω resistor $R_1$ are both between node D and the rail. Between D and B there are two paths in parallel: the 6 Ω bridge $R_4$ and the series path D–A–B ($8+4=12\ \Omega$; node A has no other connection, so no separate equation is needed). The 20 Ω resistor joins D to the reference and the 10 Ω resistor joins B to the reference.

Unknowns: $V_D$ and $V_B$.

KCL at node D (2 A flows in, the rest leaves through the branches):

$$2=\frac{V_D}{12}+\frac{V_D}{20}+\frac{V_D-V_B}{6}+\frac{V_D-V_B}{12}$$

KCL at node B:

$$\frac{V_B-V_D}{6}+\frac{V_B-V_D}{12}+\frac{V_B}{10}=0$$

Collect terms (conductances in siemens):

$$\begin{aligned} 0.3833\,V_D-0.25\,V_B&=2\\ -0.25\,V_D+0.35\,V_B&=0 \end{aligned}$$

From the second equation, $V_B=0.7143\,V_D$. Substituting in the first:

$$0.3833V_D-0.1786V_D=2 \Rightarrow V_D=9.767\ \text{V},\quad V_B=6.977\ \text{V}$$

Current through the 6 Ω resistor:

$$I_{6\Omega}=\frac{V_D-V_B}{6}=\frac{9.767-6.977}{6}=0.465\ \text{A}\ (D\to B)$$

Answer: $I_{6\Omega}=0.465$ A, from D to B.

An alternating voltage is generated by rotating a coil in a uniform magnetic field (or rotating the field past a fixed coil). By Faraday's law, an emf is induced in the coil whenever the flux linked with it changes.

Principle. Consider a rectangular coil of $N$ turns rotating at constant angular speed $\omega$ between the poles of a magnet. Let the flux through the coil be maximum ($\Phi_m$) when the coil plane is perpendicular to the field. After time $t$ the coil has turned through $\theta=\omega t$ from this position:

$$\Phi=\Phi_m\cos\omega t$$

The induced emf is

$$e=-N\frac{d\Phi}{dt}=N\omega\Phi_m\sin\omega t=E_m\sin\omega t$$

        N  ___________  S
          |  coil     |
     ---> |   (o)     | ---> slip rings
          |___________|      and brushes
         rotates at w         -> load

Working.

1. When the coil plane is perpendicular to the field ($\theta=0^\circ$), the conductors move parallel to the flux, so emf = 0.
2. At $\theta=90^\circ$ the conductors cut the flux at right angles, so emf is maximum, $E_m=N\omega\Phi_m$.
3. At $180^\circ$ the emf is zero again; between $180^\circ$ and $360^\circ$ it is negative, as the conductors under each pole have exchanged places.
4. One revolution of a two-pole machine gives one complete cycle, so frequency $f=\frac{PN_s}{120}$ for $P$ poles at $N_s$ rpm (50 Hz in Nepal).

The coil ends are connected to slip rings, and carbon brushes bring the sinusoidal voltage to the external load. In practice, the armature is fixed and the field rotates, as in an alternator.

Given: $R=5\ \Omega$, $L=30$ mH, $V=230$ V, $f=50$ Hz.

Inductive reactance:

$$X_L=2\pi fL=2\pi\times50\times0.03=9.425\ \Omega$$

Impedance:

$$Z=\sqrt{R^2+X_L^2}=\sqrt{5^2+9.425^2}=10.669\ \Omega$$

Current:

$$I=\frac{V}{Z}=\frac{230}{10.669}=21.56\ \text{A}$$

Power factor:

$$\cos\phi=\frac{R}{Z}=\frac{5}{10.669}=0.469\ \text{(lagging)}$$

Power consumed:

$$P=I^2R=(21.56)^2\times5=2323.7\ \text{W}$$

(Check: $VI\cos\phi=230\times21.56\times0.4686=2323.7$ W.)

Answer: $I=21.56$ A, pf $=0.469$ lagging, $P=2.32$ kW.

A three-phase supply has three voltages of equal magnitude, displaced by $120^\circ$ from each other. Its main advantages over single phase are:

1. More power from the same machine. A three-phase machine of the same frame size gives about 1.5 times the output of a single-phase one, so it is smaller, lighter and cheaper per kW.
2. Less conductor material. For the same power and voltage, a three-phase line needs only 75% of the copper of a single-phase line (3 wires carry 3 phases, as against 2 wires for one).
3. Constant power. In a balanced three-phase load, the instantaneous power is steady, so there is less vibration in motors and smoother torque.
4. Self-starting motors. A rotating magnetic field is produced naturally, so three-phase induction motors start without extra devices.
5. Better efficiency and power factor of three-phase motors and generators.
6. Two voltages available. Line voltage ($\sqrt3\,V_{ph}$) and phase voltage can both be used, from a four-wire system.
7. Smaller rectifier ripple for DC conversion, and easier power transmission with lower line losses.

Given: each coil $R=20\ \Omega$, $L=0.08$ H, delta connection, $V_L=415$ V, $f=50$ Hz. In delta, $V_{ph}=V_L=415$ V.

Coil impedance:

$$X_L=2\pi\times50\times0.08=25.13\ \Omega, \qquad Z_{ph}=\sqrt{20^2+25.13^2}=32.12\ \Omega$$

(i) Phase current:

$$I_{ph}=\frac{415}{32.12}=12.92\ \text{A}$$

(ii) Line current:

$$I_L=\sqrt3\,I_{ph}=\sqrt3\times12.92=22.38\ \text{A}$$

(iv) Power factor:

$$\cos\phi=\frac{R}{Z}=\frac{20}{32.12}=0.623\ \text{lagging}\ (\phi=51.5^\circ)$$

(iii) Total power:

$$P=3I_{ph}^2R=3\times(12.92)^2\times20=10016\ \text{W}$$

(Check: $\sqrt3\times415\times22.38\times0.6227=10016$ W.)

Answer: (i) 12.92 A, (ii) 22.38 A, (iii) 10.02 kW, (iv) 0.623 lagging.

Kirchhoff's laws

KCL (current law): the algebraic sum of the currents meeting at a node is zero. Equivalently, current entering a node equals current leaving it: $\sum I=0$. It follows from conservation of charge.

KVL (voltage law): the algebraic sum of all voltages (sources and drops) around any closed loop is zero: $\sum V=0$. It follows from conservation of energy.

Voltage and current sources

An ideal voltage source keeps a fixed terminal voltage whatever the load current; its internal resistance is zero. A practical voltage source is an ideal source $E$ in series with an internal resistance $r$, so $V=E-Ir$ and the terminal voltage falls with load.

An ideal current source supplies a fixed current whatever the load voltage; its internal resistance is infinite. A practical current source is an ideal source $I_s$ in parallel with an internal resistance $r$.

Source conversion. A practical voltage source $E$ with series $r$ is equivalent to a current source $I_s=E/r$ with the same $r$ in parallel, which simplifies network analysis.

Assumptions (from the figure). Point c is the reference node (0 V). The 25 V source (+ at top) is connected directly from node a to c, so $V_a=25$ V. Node b connects to a through 10 Ω, to c through 4 Ω, and to the 50 V source (+ at top) through the 2 Ω resistor. The left branch (2 V source with 5 Ω) is across the fixed node a, so it only changes the current of the 25 V source and does not affect $V_b$.

KCL at node b (currents leaving b equal zero):

$$\frac{V_b-25}{10}+\frac{V_b}{4}+\frac{V_b-50}{2}=0$$

Multiply by 20:

$$2V_b-50+5V_b+10V_b-500=0$$ $$17V_b=550 \Rightarrow V_b=32.35\ \text{V}$$

Current in branch a–b (from a to b):

$$I_{ab}=\frac{V_a-V_b}{10}=\frac{25-32.35}{10}=-0.735\ \text{A}$$

The negative sign means 0.735 A actually flows from b to a.

Check: current in the 4 Ω resistor is $32.35/4=8.09$ A (b to c). Current from the 50 V source through 2 Ω into b is $(50-32.35)/2=8.82$ A. Current from b to a is 0.735 A. KCL at b: $8.82 = 8.09+0.735$ ✓.

Answer: $V_b=32.35$ V and $I_{ab}=0.735$ A, flowing from b to a.

Given: $V=200\angle53.8^\circ$ V, $Z_1=18+j24=30\angle53.13^\circ\ \Omega$, $Z_2=15-j30=33.54\angle-63.43^\circ\ \Omega$.

Branch currents:

$$I_1=\frac{V}{Z_1}=\frac{200\angle53.8^\circ}{30\angle53.13^\circ}=6.667\angle0.67^\circ\ \text{A}$$ $$I_2=\frac{V}{Z_2}=\frac{200\angle53.8^\circ}{33.54\angle-63.43^\circ}=5.963\angle117.23^\circ\ \text{A}$$

Total current (phasor sum):

$$I=I_1+I_2=(6.667+j0.078)+(-2.728+j5.302)=3.939+j5.380$$

In polar form, $I=6.667\angle53.8^\circ$ A (the total current is in phase with the source angle, so the net circuit is purely resistive at this voltage angle: the reactive currents cancel).

Real power:

$$P=I_1^2R_1+I_2^2R_2=(6.667)^2\times18+(5.963)^2\times15=800+533.3=1333.3\ \text{W}$$

Check with $P=VI\cos\phi=200\times6.667\times\cos0^\circ=1333.3$ W, where $\phi=0$ since $I$ and $V$ have the same angle.

Answer: $I=6.667\angle53.8^\circ$ A, $I_1=6.667\angle0.67^\circ$ A, $I_2=5.963\angle117.23^\circ$ A, $P=1333.3$ W (unity power factor).