- 2081 Baishakh (new course) · 2+4 marks
Three-phase connections and series RC phasor
Short answer
Star (Y): V_L = √3 V_ph, I_L = I_ph. Delta (Δ): V_L = V_ph, I_L = √3 I_ph. In a series RC circuit, common current I is reference. V_R = IR in phase, V_C = I X_C lags 90°; total V = √(V_R²+V_C²) = I√(R²+X_C²). Impedance Z = R − jX_C, current leads voltage by φ = tan⁻¹(X_C/R).
What are the two ways of connecting a 3-phase system and mention their relation between phase and line quantities. Derive the expression for series connected RC with phasor diagram.
Two ways of connecting a 3-phase system
- Star (Y) connection: one end of each of the three phase windings is joined to a common point called the neutral (N); the other three ends go to the lines.
- V_L = √3 V_ph (line voltage leads the phase voltage by 30°)
- I_L = I_ph
- Delta (Δ) connection: the three windings are joined end to end in a closed loop; the lines are taken from the three junctions.
- V_L = V_ph
- I_L = √3 I_ph (line current lags the phase current by 30°)
In both cases the total power is P = √3 V_L I_L cos φ.
Series RC circuit
A resistance R and capacitance C are in series across v = Vm sin ωt. The same current I flows through both, so I is taken as the reference phasor.
o---[ R ]---| C |---o
V_R V_C
<-------- V -------->
- Voltage across R: V_R = I R, in phase with I.
- Voltage across C: V_C = I X_C, lagging I by 90°, where X_C = 1/(ωC) = 1/(2πfC).
Phasor sum:
V = V_R − j V_C = I (R − j X_C)
|V| = √(V_R² + V_C²) = I √(R² + X_C²)
Z = V/I = R − j X_C
|Z| = √(R² + X_C²)
φ = tan⁻¹(X_C/R) (current leads voltage)
Phasor diagram:
I, V_R
o-------------> (reference)
|\
| \ φ
| \
V_C \ V
| \
v v
The voltage V lags the current I by φ (i.e. current leads voltage).
If v = Vm sin ωt, then i = Im sin(ωt + φ) with Im = Vm/|Z|.
Power: power factor cos φ = R/|Z| (leading). Average power P = V I cos φ = I²R; the capacitor takes no average power, only reactive power Q = I² X_C.
✓ Answer checked by Claude Opus 5.5 (high effort)
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