AC Circuit Analysis

Two ways of connecting a 3-phase system

1. 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
2. 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.

Take the supply voltage as reference: V = 200∠0° V, f = 50 Hz.

ZC = 3 + j2.5 Ω, ZA = 2 + j1.5 Ω, ZB = 5 − j3.5 Ω.

Step 1: Parallel impedance ZAB

ZA × ZB = (2 + j1.5)(5 − j3.5) = 15.25 + j0.5
ZA + ZB = 7 − j2
ZAB = (15.25 + j0.5)/(7 − j2)
    = 1.995 + j0.642 Ω  = 2.096∠17.82° Ω

Step 2: Total impedance and total current

Z  = ZC + ZAB = (3 + j2.5) + (1.995 + j0.642)
   = 4.995 + j3.142 Ω = 5.901∠32.17° Ω
IC = V/Z = 200∠0° / 5.901∠32.17°
   = 33.89∠−32.17° A  (= 28.69 − j18.04 A)

Step 3: Branch currents

VAB = IC × ZAB = 33.89∠−32.17° × 2.096∠17.82°
    = 71.03∠−14.34° V
IA  = VAB/ZA = 71.03∠−14.34° / 2.5∠36.87°
    = 28.41∠−51.21° A
IB  = VAB/ZB = 71.03∠−14.34° / 6.103∠−34.99°
    = 11.64∠20.65° A
Check: IA + IB = (17.80 − j22.15) + (10.89 + j4.10)
               = 28.69 − j18.04 A = IC  ✓

Step 4: Power factor and powers

pf = cos 32.17° = 0.8465 (lagging)
S  = V × I = 200 × 33.89    = 6778.5 VA
P  = V I cos φ              = 5738.1 W
Q  = V I sin φ              = 3608.7 VAR (inductive)
Check: P = I² R = 33.89² × 4.995 = 5738 W ✓

Answer:

Significance of power factor

Power factor (cos φ = P/S) shows what fraction of the apparent power does useful work. For a given power at a given voltage, I = P/(V cos φ), so a low power factor means a larger current. This causes higher copper losses (I²R), more voltage drop, poorer efficiency, and needs bigger cables, transformers and generators (rated in kVA). Utilities therefore charge penalties for low pf and consumers improve it with capacitors.

Calculation

V = 200∠0° V. Series part Z1 = 3 + j6 Ω; branches Z2 = 4 + j8 Ω, Z3 = 5 − j8 Ω.

Parallel combination:

Z2 × Z3 = (4 + j8)(5 − j8) = 20 − j32 + j40 + 64
        = 84 + j8
Z2 + Z3 = 9 + j0
Zp = (84 + j8)/9 = 9.333 + j0.889 Ω = 9.376∠5.44° Ω

Total impedance:

Z = Z1 + Zp = (3 + j6) + (9.333 + j0.889)
  = 12.333 + j6.889 Ω
|Z| = √(12.333² + 6.889²) = 14.127 Ω
φ  = tan⁻¹(6.889/12.333) = 29.19°
Z  = 14.127∠29.19° Ω

Circuit current:

I = V/Z = 200∠0° / 14.127∠29.19°
  = 14.157∠−29.19° A

Branch currents (for completeness): Vp = I × Zp = 132.73∠−23.75° V, I2 = Vp/Z2 = 14.84∠−87.18° A, I3 = Vp/Z3 = 14.07∠34.25° A.

Power factor:

pf = cos 29.19° = R/|Z| = 12.333/14.127 = 0.873 (lagging)

Power consumed:

P = V I cos φ = 200 × 14.157 × 0.873 = 2472 W
Check: P = I² R = 14.157² × 12.333 = 2472 W ✓

Answer: Z = 12.33 + j6.89 Ω = 14.13∠29.19° Ω, I = 14.16∠−29.19° A, pf = 0.873 lagging, P ≈ 2472 W (Q ≈ 1381 VAR, S ≈ 2831 VA).

A three-phase system has three windings (phases R, Y, B) whose voltages are equal in magnitude and 120° apart. The six ends of the windings can be connected in two ways.

1. Star (Y) connection

The similar ends (say the finishing ends) of the three windings are joined to a common point called the star or neutral point (N). The other three ends are connected to the three lines. A fourth wire may be taken from N (3-phase 4-wire system).

            R
            o  I_L
            |
           (Ph R)
            |
            N o------ neutral
          /   \
     (Ph B)   (Ph Y)
        /       \
       o         o
       B         Y

• Line voltage V_RY = V_R − V_Y (phasor difference).
• V_L = √3 V_ph, line voltage leads the phase voltage by 30°.
• I_L = I_ph (each line is in series with one phase).
• Gives two voltages: 400 V line-to-line and 230 V line-to-neutral (Nepal's LV supply), so it is used for distribution.

2. Delta (Δ) or mesh connection

The finishing end of one winding is joined to the starting end of the next, forming a closed triangle. Lines are taken from the three junctions. There is no neutral.

            R
            o
           / \
     Ph RB/   \Ph RY
         /     \
        o-------o
        B Ph YB  Y

• V_L = V_ph (each phase is directly between two lines).
• Line current I_R = I_RY − I_BR (phasor difference).
• I_L = √3 I_ph, line current lags the phase current by 30°.
• Used for motors and transmission where no neutral is needed.

Comparison

Power in both: P = √3 V_L I_L cos φ = 3 V_ph I_ph cos φ.