GATE EE Solved Problems

GATE 2017 Electrical Engineering (EE) Power Systems (2017)

Solved problems

Author: Prof. Mithun Mondal Subject: Power Systems Year: 2017 Total Questions: 7
Section 01

1-Mark Questions

QQuestion 1 1 Mark

The per-unit impedance of a transmission line on its own base is 0.2 pu. What will be the per-unit impedance on a base that is twice the original voltage base and four times the original MVA base? \begin{figure}[H] \centering

Figure 1.1
Figure 1.1

\end{figure}

AOptions

  1. 0.05 pu
  2. 0.1 pu
  3. 0.2 pu
  4. 0.8 pu

SSolution

\(Z_{pu,new} = Z_{pu,old} \times \frac{MVA_{new}}{MVA_{old}} \times \frac{V_{old}^2}{V_{new}^2}\) \(Z_{pu,new} = 0.2 \times \frac{4MVA}{MVA} \times \frac{V^2}{(2V)^2} = 0.2 \times 4 \times \frac{1}{4} = 0.2\) pu Correct answer: C.

QQuestion 2 1 Mark

The power transfer capability of a transmission line is limited by:

AOptions

  1. Thermal limits
  2. Voltage stability limits
  3. Transient stability limits
  4. All of the above

SSolution

Power transfer capability is limited by: 1. Thermal limits (conductor heating) 2. Voltage stability limits (voltage collapse) 3. Transient stability limits (loss of synchronism) All three factors must be considered for safe operation. Correct answer: D.

QQuestion 3 1 Mark

The protection scheme commonly used for transmission lines is: \begin{figure}[H] \centering

Figure 3.1
Figure 3.1

\end{figure}

AOptions

  1. Overcurrent protection
  2. Distance protection
  3. Differential protection
  4. Pilot protection

SSolution

Distance protection is the most commonly used protection scheme for transmission lines. It measures impedance from relay location to fault point and operates when impedance falls within set zones. Correct answer: B.

Section 02

2-Mark Questions

QQuestion 4 2 Mark

A transmission line has a surge impedance of 400Ω and surge impedance loading of 250MW. What is the line voltage?

\begin{figure}[H] \centering

Figure 4.1
Figure 4.1

\end{figure}

AOptions

  1. 316 kV
  2. 400 kV
  3. 500 kV
  4. 632 kV

SSolution

Surge Impedance Loading: \(SIL = \frac{V^2}{Z_c}\) \(250 \times 10^6 = \frac{V^2}{400}\) \(V^2 = 250 \times 10^6 \times 400 = 100 \times 10^9\) \(V = \sqrt{100 \times 10^9} = 316,227V \approx 316kV\) Correct answer: A.

QQuestion 5 2 Mark

In a power system, the fault current for a line-to-ground fault is: \begin{figure}[H] \centering

Figure 5.1
Figure 5.1

\end{figure}

AOptions

  1. \(\frac{3V_f}{Z_1 + Z_2 + Z_0}\)
  2. \(\frac{V_f}{Z_1 + Z_2 + Z_0}\)
  3. \(\frac{\sqrt{3}V_f}{Z_1 + Z_2 + Z_0}\)
  4. \(\frac{3V_f}{Z_1 + Z_2}\)

SSolution

For a single line-to-ground fault, all three sequence networks are connected in series. Fault current: \(I_f = \frac{3V_f}{Z_1 + Z_2 + Z_0}\) where \(V_f\) is the prefault voltage, \(Z_1\), \(Z_2\), \(Z_0\) are positive, negative, and zero sequence impedances. Correct answer: A.

QQuestion 6 2 Mark

A load bus in a power system has a voltage magnitude of 0.95 pu and the voltage phase angle is -5°. If the base voltage is 132 kV, what is the actual voltage?

AOptions

  1. 125.4 kV ∠-5°
  2. 132 kV ∠-5°
  3. 138.6 kV ∠-5°
  4. 139.0 kV ∠-5°

SSolution

Actual voltage = per-unit voltage × base voltage \(V_{actual} = 0.95 × 132kV = 125.4kV\) Phase angle remains the same: -5° Therefore: \(V_{actual} = 125.4kV ∠-5°\) Correct answer: A.

QQuestion 7 2 Mark

In a power system, reactive power compensation is primarily used for:

AOptions

  1. Reducing transmission losses
  2. Voltage regulation
  3. Power factor improvement
  4. All of the above

SSolution

Reactive power compensation serves multiple purposes: 1. Voltage regulation (maintaining voltage levelsreducing reactive power demand) 3. Reducing transmission losses (by reducing current magnitude) Therefore, all options are correct. Correct answer: D.