Tutorial 03- Solution

ESO-210

Solution1)

Phase impedance = (8+j6) ohm = 10∠36.86° ohm ∴ Power factor, cos = R/Z= 8/10 =0.8 lagging as it is an inductive load.ΦPhase voltage = 230/√3 v = 132.79 volt ∴ Line current = phase current = Vphase/ Zphase = 132.79/10 A = 13.28 A

Total VA = √3 Vline Iline = √3 * 230 * 13.28 = 5290 VA Active Power = VA cos = 5290 * 0.8 = 4232 WΦ Reactive Power = VA sin =5290 * 0.6 = 3174 VARΦ

Solution2) VAB= 400∠0° V, VBC = 400∠-120°V , VCA= 400∠120° V

For wattmeter 1, IXY = IA = VAC/ 10 + VAB /10∠90° = (40∠-90° - 40∠120°) A = 77.274∠-75° A

Potential coil voltage V1V2 of W1 = VAC/2 = -200∠120° volt = 200∠-60° volt

∴ Reading of W1 = V1V2 *IXY * cos∠V1V2 Ixy = 200* 77.274 * cos 15°

= 14.908 kW

For Wattmeter 2, IXY = IB = VCA/ 10 + VCB /20∠-90° = 58.186 ∠130° A

Potential coil voltage V1V2 of W2 = VCA/2 = 200∠120° volt

∴ Reading of W2 = V1V2 *IXY * cos∠V1V2 Ixy = 200* 58.186 * cos (-10°)

= 11.46kW

W1

W2

Solution3)

XC= 1/WC = 398 ohm I= 400/(300 – j398) A = 0.8 ∠53° A

Now writing loop equation, 300 I + VXC + VCA = 0 Or, VXC = - VXC – 300*I = -400∠120° - 300* 0.8 ∠53° = -400∠120° - 240∠53° = 541 ∠-84.1° Volt

Solution4)

a) Applying KVL

Applying KCL at load neutral

By solving equation (1), (2) and (3) we get

b) Converting source as well as load into delta we get

Applying KVL & solving

So the line currents are

Solution5)

Writing loop equations and solving we get,

Solution6)

Let the load is Z∠θ ∴ Wattmeter reading is W = Vbc Ia cos ∠bca = Vl Il cos (90 – θ) = Vl Il sin θ ∴ W = Q/√3 = Reactive power /√3

Or, Q = √3 * W