GATE EE Solved Problems

GATE 2013 Electrical Engineering (EE) Power Systems (2013)

Solved problems

Author: Prof. Mithun Mondal Subject: Power Systems Year: 2013 Total Questions: 2
Section 01

1-Mark Questions

QQuestion 1 1 Mark

A single-phase load is supplied by a single-phase voltage source. If the current flowing from the load to the source is \(10\angle -150°\) A and if the voltage at the load terminals is \(100\angle 60°\) V, then the

AOptions

  1. load absorbs real power and delivers reactive power.
  2. load absorbs real power and absorbs reactive power.
  3. load delivers real power and delivers reactive power.
  4. load delivers real power and absorbs reactive power.

SSolution

Given:

  • Current from load to source: \(\bar{I} = 10\angle -150°\) A
  • Voltage at load terminals: \(\bar{V} = 100\angle 60°\) V

Note: Current direction is from load to source (opposite to conventional direction for a load).

Current into the load: \(\bar{I}_{load} = -\bar{I} = 10\angle 30°\) A

Phase angle:

\[\phi = \angle V - \angle I_{load} = 60° - 30° = 30°\]

Complex power:

\[\bar{S} = \bar{V} \times \bar{I}_{load}^* = 100\angle 60° \times 10\angle -30° = 1000\angle 30°\]
\[P = |\bar{S}|\cos\phi = 1000\cos(30°) = 866 \text{ W}\]
\[Q = |\bar{S}|\sin\phi = 1000\sin(30°) = 500 \text{ VAr}\]

Since we used current into the load:

  • \(P > 0\): Load absorbs real power
  • \(Q > 0\): Load absorbs reactive power (inductive)

Forming \(\bar{V}\bar{I}^* = 1000\angle 210°\) with \(\bar{I}\) as given does not change this conclusion: because \(\bar{I}\) leaves the load, that product is the power the load delivers to the source. Its negative, \(1000\angle 30°\), is again the power the load draws.

\[P = +866 \text{ W}, \qquad Q = +500 \text{ VAr}\]

The current entering the load lags the terminal voltage by 30°, so the load is inductive: it absorbs real power and absorbs reactive power.

Correct answer: B

Section 02

2-Mark Questions

QQuestion 2 2 Mark

For a power system network with \(n\) nodes, \(Z_{33}\) of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is \(1.3\angle -10°\) per unit. If a capacitor having reactance of \(-j3.5\) per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is

AOptions

  1. \(0.325\angle -100°\)
  2. \(0.325\angle 80°\)
  3. \(0.371\angle -100°\)
  4. \(0.433\angle 80°\)

SSolution

Given:

  • \(Z_{33} = j0.5\) pu (driving point impedance at bus 3)
  • \(V_3 = 1.3\angle -10°\) pu (voltage before adding capacitor)
  • Capacitor reactance: \(X_C = -j3.5\) pu

Thevenin equivalent at bus 3:

The Thevenin impedance is \(Z_{th} = Z_{33} = j0.5\) pu

The Thevenin voltage is the open circuit voltage = \(V_3 = 1.3\angle -10°\) pu

After adding capacitor:

The capacitor is connected across the Thevenin equivalent, so it forms a single loop with \(Z_{th}\) driven by the open-circuit voltage \(V_3\). The capacitor current is therefore

\[I_C = \frac{V_3}{Z_{33} + X_C} = \frac{1.3\angle -10°}{j0.5 - j3.5} = \frac{1.3\angle -10°}{-j3}\]
\[= \frac{1.3\angle -10°}{3\angle -90°} = \frac{1.3}{3}\angle(-10° + 90°) = 0.433\angle 80°\]

Correct answer: D