GATE EE Solved Problems

GATE 2010 Electrical Engineering (EE) Power Systems (2010)

Solved problems

Author: Prof. Mithun Mondal Subject: Power Systems Year: 2010 Total Questions: 9
Section 01

1-Mark Questions

QQuestion 1 1 Mark

Power is transferred from system A to system B by an HVDC link as shown in the figure. If the voltages \(V_{AB}\) and \(V_{CD}\) are as indicated in the figure, and \(I > 0\), then

Figure 1.1
Figure 1.1

AOptions

  1. \(V_{AB} < 0, V_{CD} < 0, V_{AB} > V_{CD}\)
  2. \(V_{AB} > 0, V_{CD} > 0, V_{AB} > V_{CD}\)
  3. \(V_{AB} > 0, V_{CD} > 0, V_{AB} < V_{CD}\)
  4. \(V_{AB} > 0, V_{CD} < 0\)

SSolution

HVDC link operation:

Power flow direction: System A \(\rightarrow\) System B

At Rectifier (System A):

  • Converts AC to DC
  • DC voltage \(V_{AB}\) is positive
  • Acts as source of DC power
  • \(V_{AB} > 0\)

At Inverter (System B):

  • Converts DC to AC
  • Acts as load for DC system
  • For power transfer: acts like a voltage source opposing current
  • \(V_{CD} > 0\) but \(V_{CD} < V_{AB}\)

Power transfer condition:

For DC current \(I > 0\) to flow from A to B:

\[V_{AB} > V_{CD}\]

The voltage drop in the line is:

\[V_{AB} - V_{CD} = I \times R_{line}\]

Both voltages are positive, but rectifier voltage must exceed inverter voltage.

Correct answer: B

QQuestion 2 1 Mark

Consider a step voltage wave of magnitude 1 pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant the travelling wave reaches the reactor is

Figure 2.1
Figure 2.1

AOptions

  1. \(-1\) pu
  2. 1 pu
  3. 2 pu
  4. 3 pu

SSolution

Travelling wave on transmission line:

Incident wave: \(V_i = 1\) pu

Termination: Pure reactor (inductive)

Reflection coefficient:

For a reactor (inductive termination):

\[\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}\]

where \(Z_L = j\omega L\) (reactor) and \(Z_0\) is characteristic impedance.

At the instant wave arrives (very high frequency):

\[Z_L = j\omega L \rightarrow j\infty \text{ (for high } \omega\text{)}\]
\[\Gamma = \frac{j\infty - Z_0}{j\infty + Z_0} \approx 1\]

Reflected wave:

\[V_r = \Gamma \times V_i = 1 \times 1 = 1 \text{ pu}\]

Total voltage at reactor:

\[V_{total} = V_i + V_r = 1 + 1 = 2 \text{ pu}\]

The reactor acts like an open circuit to the fast-rising wavefront, so voltage doubles (similar to open-circuit termination).

Correct answer: C

QQuestion 3 1 Mark

Consider two buses connected by an impedance of \((0 + j5)\) \(\Omega\). The bus 1 voltage is \(100\angle 30°\) V, and bus 2 voltage is \(100\angle 0°\) V. The real and reactive power supplied by bus 1, respectively, are

AOptions

  1. 1000 W, 268 VAr
  2. \(-1000\) W, \(-134\) VAr
  3. 276.9 W, \(-56.7\) VAr
  4. \(-276.9\) W, 56.7 VAr

SSolution

Given:

  • \(V_1 = 100\angle 30°\) V
  • \(V_2 = 100\angle 0°\) V
  • \(Z = j5\) \(\Omega\)

Current from bus 1 to bus 2:

\[I_{12} = \frac{V_1 - V_2}{Z} = \frac{100\angle 30° - 100\angle 0°}{j5}\]
\[V_1 - V_2 = 100(\cos 30° + j\sin 30°) - 100\]
\[= 100(0.866 + j0.5 - 1) = 100(-0.134 + j0.5)\]
\[= -13.4 + j50\]
\[I_{12} = \frac{-13.4 + j50}{j5} = \frac{-13.4 + j50}{j5}\]

Multiply by \(-j\):

\[I_{12} = \frac{(-13.4 + j50)(-j)}{5} = \frac{50 + j13.4}{5} = 10 + j2.68 \text{ A}\]

Complex power from bus 1:

\[S_1 = V_1 I_{12}^* = 100\angle 30° \times (10 - j2.68)\]
\[= 100\angle 30° \times 10.35\angle -15°\]
\[= 1035\angle 15°\]
\[P_1 = 1035\cos(15°) = 1000 \text{ W}\]
\[Q_1 = 1035\sin(15°) = 268 \text{ VAr}\]

Alternatively, using direct formula:

\[P_1 = \frac{V_1 V_2}{X}\sin(\delta_1 - \delta_2) = \frac{100 \times 100}{5}\sin(30°) = 2000 \times 0.5 = 1000 \text{ W}\]
\[Q_1 = \frac{V_1(V_1 - V_2\cos\delta)}{X}\]

Correct answer: A

QQuestion 4 1 Mark

A three-phase, 33 kV oil circuit breaker is rated 1200 A, 2000 MVA, 3 s. The symmetrical breaking current is

AOptions

  1. 1200 A
  2. 3600 A
  3. 35 kA
  4. 104.8 kA

SSolution

Given:

  • Voltage rating: 33 kV (line-to-line)
  • Current rating: 1200 A
  • Breaking capacity: 2000 MVA
  • Operating time: 3 s

Symmetrical breaking current:

The breaking capacity relates to the fault level the breaker can interrupt:

\[\text{Breaking capacity} = \sqrt{3} \times V_L \times I_{breaking}\]
\[2000 \times 10^6 = \sqrt{3} \times 33 \times 10^3 \times I_{breaking}\]
\[I_{breaking} = \frac{2000 \times 10^6}{\sqrt{3} \times 33 \times 10^3}\]
\[= \frac{2000 \times 10^3}{57.16} = 35,000 \text{ A} = 35 \text{ kA}\]

The symmetrical breaking current is the RMS value of the fault current that the breaker can interrupt.

Correct answer: C

QQuestion 5 1 Mark

Consider a stator winding of an alternator with an internal high-resistance ground fault. The currents under the fault condition are as shown in the figure. The winding is protected using a differential current scheme with current transformers of ratio 400/5 A as shown. The current through the operating coil is

AOptions

  1. 0.17875 A
  2. 0.2 A
  3. 0.375 A
  4. 60 A

SSolution

Given:

  • CT ratio: 400/5 A
  • Line currents: 220 A and 250 A (from figure)

CT secondary currents:

Current from CT1:

\[I_{CT1} = 220 \times \frac{5}{400} = \frac{1100}{400} = 2.75 \text{ A}\]

Current from CT2:

\[I_{CT2} = 250 \times \frac{5}{400} = \frac{1250}{400} = 3.125 \text{ A}\]

Differential current (through operating coil):

Under normal conditions, these would be equal. During internal fault:

\[I_{op} = |I_{CT2} - I_{CT1}| = |3.125 - 2.75| = 0.375 \text{ A}\]

This differential current indicates internal fault and operates the relay.

Correct answer: C

QQuestion 6 1 Mark

The zero-sequence circuit of the three-phase transformer shown in the figure is

(Star-Delta transformer connection shown)

SSolution

Zero-sequence behavior:

For a Star-Delta transformer:

Primary (Star side):

  • Zero-sequence currents need return path through neutral
  • If neutral is grounded: zero-sequence can flow
  • If neutral is isolated: zero-sequence is blocked

Secondary (Delta side):

  • Zero-sequence currents circulate within delta
  • No zero-sequence current in lines
  • Delta provides path for zero-sequence

Zero-sequence equivalent:

Primary side: Open circuit (if neutral not grounded) or through impedance (if grounded) Secondary side: Open circuit to external system (zero-sequence trapped in delta)

The zero-sequence impedance is effectively infinite from line terminals because:

  • Delta blocks zero-sequence from appearing in lines
  • Star side without ground also blocks zero-sequence

Correct answer: Check figure options for open circuit representation

Section 02

2-Mark Questions

QQuestion 7 2 Mark

Consider a three-phase, 50 Hz, 11 kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figure. The voltage across the two insulators is

Figure 7.1
Figure 7.1

AOptions

  1. \(e_1 = 3.74\) kV, \(e_2 = 2.61\) kV
  2. \(e_1 = 3.46\) kV, \(e_2 = 2.89\) kV
  3. \(e_1 = 6.0\) kV, \(e_2 = 4.23\) kV
  4. \(e_1 = 5.5\) kV, \(e_2 = 5.5\) kV

SSolution

Given:

  • Line voltage: 11 kV (line-to-line)
  • Self capacitance: \(C_s = 5C\) (where \(C\) is shunt capacitance)
  • Two insulators in string

Phase voltage:

\[V_{ph} = \frac{11}{\sqrt{3}} = 6.35 \text{ kV}\]

Voltage distribution:

Let \(e_1\) = voltage across lower insulator (nearest conductor) Let \(e_2\) = voltage across upper insulator

Using current balance at junction:

\[i_1 = i_2 + i_3\]

where: - \(i_1\) = current through \(C_s\) (self capacitance of lower insulator) - \(i_2\) = current through \(C_s\) (self capacitance of upper insulator) - \(i_3\) = current through \(C\) (shunt capacitance to ground from junction)

\[e_1 \omega C_s = e_2 \omega C_s + (e_1 + e_2)\omega C\]
\[5C e_1 = 5C e_2 + C(e_1 + e_2)\]
\[5e_1 = 5e_2 + e_1 + e_2\]
\[4e_1 = 6e_2\]
\[e_1 = 1.5 e_2\]

Also: \(e_1 + e_2 = 6.35\) kV

\[1.5e_2 + e_2 = 6.35\]
\[e_2 = \frac{6.35}{2.5} = 2.54 \text{ kV}\]
\[e_1 = 1.5 \times 2.54 = 3.81 \text{ kV}\]

Closest to option A.

Correct answer: A

QQuestion 8 2 Mark

Consider a three-core, three-phase, 50 Hz, 11 kV cable whose conductors are denoted as R, Y and B in the figure. The inter-phase capacitance (\(C_1\)) between each pair of conductors is 0.2 \(\mu\)F and the capacitance between each line conductor and the sheath is 0.4 \(\mu\)F. The per-phase charging current is

Figure 8.1
Figure 8.1

AOptions

  1. 2.0 A
  2. 2.4 A
  3. 2.7 A
  4. 3.5 A

SSolution

Given:

  • Line voltage: 11 kV (line-to-line)
  • \(C_1 = 0.2\) \(\mu\)F (inter-phase)
  • \(C_2 = 0.4\) \(\mu\)F (line-to-sheath)
  • Frequency: 50 Hz

Effective capacitance per phase:

For three-phase cable, effective capacitance to neutral:

\[C_n = C_2 + 3C_1 = 0.4 + 3(0.2) = 0.4 + 0.6 = 1.0 \text{ } \mu\text{F}\]

Phase voltage:

\[V_{ph} = \frac{11}{\sqrt{3}} = 6.35 \text{ kV}\]

Charging current per phase:

\[I_c = V_{ph} \times \omega C_n\]
\[= 6.35 \times 10^3 \times 2\pi \times 50 \times 1.0 \times 10^{-6}\]
\[= 6350 \times 314.16 \times 10^{-6}\]
\[= 1.995 \approx 2.0 \text{ A}\]

Correct answer: A

QQuestion 9 2 Mark

For the power system shown in the figure below, the specifications of the components are: G1: 25 kV, 100 MVA, X=9%; G2: 25 kV, 100 MVA, X=9%; T1: 25 kV/220 kV, 90 MVA, X=12%; T2: 220 kV/25 kV, 90 MVA, X=12%; Line1: 220 kV, X= 150 ohms. Choose 25 kV as the base voltage at generator G1, and 200 MVA as the MVA base. The impedance diagram is

Figure 9.1
Figure 9.1

SSolution

Base values:

  • \(S_{base} = 200\) MVA (new base)
  • \(V_{base,G1} = 25\) kV

Converting to new base:

Generator G1:

\[X_{G1,new} = 0.09 \times \frac{200}{100} \times \left(\frac{25}{25}\right)^2 = 0.18 \text{ pu}\]

Generator G2: Same rating as G1:

\[X_{G2,new} = 0.18 \text{ pu}\]

Transformer T1:

\[X_{T1,new} = 0.12 \times \frac{200}{90} \times \left(\frac{25}{25}\right)^2 = 0.267 \approx 0.27 \text{ pu}\]

Transformer T2: Same as T1:

\[X_{T2,new} = 0.27 \text{ pu}\]

Transmission Line:

Base impedance at 220 kV:

\[Z_{base} = \frac{V^2_{base}}{S_{base}} = \frac{(220)^2}{200} = 242 \text{ } \Omega\]

Line reactance:

\[X_{line,new} = \frac{150}{242} = 0.62 \text{ pu}\]

Impedance diagram: - G1: j0.18 - T1: j0.27 - Line: j0.62 - T2: j0.27 - G2: j0.18

Correct answer: C