GATE 2020 Electrical Engineering (EE) Electrical Machines (2020)

Section 01

0-Mark Questions

QQuestion 1 0 Mark

A single-phase, 4 kVA, 200 V/100 V, 50 Hz transformer with laminated CRGO steel core has rated no-load loss of 450 W. When the high-voltage winding is excited with 160 V, 40 Hz sinusoidal ac supply, the no-load losses are found to be 320 W. When the high-voltage winding of the same transformer is supplied from a 100 V, 25 Hz sinusoidal ac source, the no-load losses will be \_\_\_\_\_\_\_\_\_ W (rounded off to 2 decimal places).

SSolution

Given: At 200 V, 50 Hz: $P_c = 450$ W; At 160 V, 40 Hz: $P_c = 320$ W

Since $V/f$ is constant: $\frac{200}{50} = \frac{160}{40} = \frac{100}{25} = 4$

Core loss equation: $P_c = Af + Bf^2$

Setting up equations: \begin{align*} 450 &= A(50) + B(50)^2 \\ 320 &= A(40) + B(40)^2 \end{align*}

Dividing by frequency: \begin{align*} 9 &= A + 50B \\ 8 &= A + 40B \end{align*}

Subtracting: $1 = 10B \Rightarrow B = 0.1$

Therefore: $A = 9 - 50(0.1) = 4$

At 100 V, 25 Hz:

\[P_c = 4(25) + 0.1(25)^2 = 100 + 62.5 = \boxed{162.50 \text{ W}}\]

QQuestion 2 0 Mark

A three-phase, 50 Hz, 4-pole induction motor runs at no-load with a slip of 1%. With full load, the slip increases to 5%. The percentage speed regulation of the motor (rounded off to 2 decimal places) is \_\_\_\_\_\_\_\_.

SSolution

Given: 4-pole, 50 Hz motor; $s_{nl} = 0.01$, $s_{fl} = 0.05$

Synchronous speed: $N_s = \frac{120f}{P} = \frac{120 \times 50}{4} = 1500$ rpm

No-load speed: $N_0 = N_s(1 - s_{nl}) = 1500(1 - 0.01) = 1485$ rpm

Full-load speed: $N = N_s(1 - s_{fl}) = 1500(1 - 0.05) = 1425$ rpm

Speed regulation:

\[%SR = \frac{N_0 - N}{N} \times 100 = \frac{1485 - 1425}{1425} \times 100 = \boxed{4.21%}\]

QQuestion 3 0 Mark

A three-phase cylindrical rotor synchronous generator has a synchronous reactance $X_s$ and a negligible armature resistance. The magnitude of per phase terminal voltage is $V_A$ and the magnitude of per phase induced emf is $E_A$. Considering the following two statements:

P: For any three-phase balanced leading load connected across the terminals of this synchronous generator, $V_A$ is always more than $E_A$.

Q: For any three-phase balanced lagging load connected across the terminals of this synchronous generator, $V_A$ is always less than $E_A$.

Which of the following options is correct?

AOptions

P is true and Q is true.
P is true and Q is false.
P is false and Q is false. \CorrectChoice P is false and Q is true.

SSolution

For a synchronous generator with negligible armature resistance:

Statement P (Leading load): When the load is leading (capacitive), the armature reaction is magnetizing. The terminal voltage $V_A$ can be less than, equal to, or greater than $E_A$ depending on the power factor and load conditions. Therefore, P is false.

Statement Q (Lagging load): When the load is lagging (inductive), the armature reaction is demagnetizing. The voltage drop across $X_s$ subtracts from $E_A$, making $V_A < E_A$ for all lagging loads. Therefore, Q is true.

Correct answer: D

QQuestion 4 0 Mark

A single 50 Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to be increased by 10 MW, as a result of which, system frequency dropped to 49.75 Hz. Further increase in load in the system resulted in a frequency of 49.25 Hz. At this condition, the power in MW supplied by the generator is \_\_\_\_\_\_\_\_ (rounded off to 2 decimal places).

SSolution

The droop characteristic is linear. Given points: (100 MW, 50 Hz) and (110 MW, 49.75 Hz)

Slope: $\tan\theta = \frac{50 - 49.75}{110 - 100} = \frac{0.25}{10}$

At 49.25 Hz, using similar triangles:

\[\frac{50 - 49.75}{110 - 100} = \frac{49.75 - 49.25}{x - 110}\]

\[\frac{0.25}{10} = \frac{0.5}{x - 110}\]

\[x - 110 = \frac{0.5 \times 10}{0.25} = 20\]

\[x = \boxed{130.00 \text{ MW}}\]

QQuestion 5 0 Mark

A 250 V dc shunt motor has an armature resistance of 0.2 $\Omega$ and a field resistance of 100 $\Omega$. When the motor is operated on no-load at rated voltage, it draws an armature current of 5 A and runs at 1200 rpm. When a load is coupled to the motor, it draws total line current of 50 A at rated voltage, with a 5% reduction in the air-gap flux due to armature reaction. Voltage drop across the brushes can be taken as 1 V per brush under all operating conditions. The speed of the motor, in rpm, under this loaded condition, is closest to:

AOptions

900
1200
1000 \CorrectChoice 1220

SSolution

Given: $V = 250$ V, $R_a = 0.2$ $\Omega$, $R_{sh} = 100$ $\Omega$, Brush drop = 2 V total

Shunt field current: $I_{sh} = \frac{250}{100} = 2.5$ A (constant)

No-load: $I_{a0} = 5 - 2.5 = 2.5$ A

\[E_{b0} = 250 - 2.5(0.2) - 2 = 247.5 \text{ V}\]

On load: $I_{aL} = 50 - 2.5 = 47.5$ A

\[E_{bL} = 250 - 47.5(0.2) - 2 = 238.5 \text{ V}\]

Using speed equation: $\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2}$

\[\frac{N_2}{1200} = \frac{238.5}{247.5} \times \frac{1}{0.95}\]

\[N_2 = 1200 \times \frac{238.5}{247.5 \times 0.95} = \boxed{1217.22 \approx 1220 \text{ rpm}}\]

Correct answer: D

QQuestion 6 0 Mark

The temperature of the coolant oil bath for a transformer is monitored using a thermistor circuit. The thermistor has resistance $R_{thermistor} = 2(1 + \alpha T)$ k$\Omega$, where $T$ is temperature in Ã‚Â°C. The temperature coefficient $\alpha = -(4 \pm 0.25)$%/Ã‚Â°C. Circuit parameters: $R_1 = 1$ k$\Omega$, $R_2 = 1.3$ k$\Omega$, $R_3 = 2.6$ k$\Omega$. The error in the output signal (in V, rounded off to 2 decimal places) at 150Ã‚Â°C is \_\_\_\_\_\_\_\_.

SSolution

Given: $\alpha = -(0.04 \pm 0.0025)$ /Ã‚Â°C, so $\alpha_{max} = -0.0425$ /Ã‚Â°C, $\alpha_{min} = -0.0375$ /Ã‚Â°C

At $T = 150$Ã‚Â°C with $\alpha = -0.04$ /Ã‚Â°C:

\[R_{th} = 2[1 + (-0.04)(150)] = 2[1 - 6] = -10 \text{ k}\Omega\]

Output voltage: $V_0 = \left[\frac{R_1}{R_1 + R_{th}}\right] \times 3 + \left[\frac{R_2}{R_3}\right] \times 0.1$

\[V_0 = \frac{1}{1 - 10} \times 3 + \frac{1.3}{2.6} \times 0.1 = \frac{-3}{9} + 0.05 = 0.5 \text{ V}\]

With $\alpha_{max} = -0.0425$: $R_{th} = -10.75$ k$\Omega$ $\Rightarrow V_0 = 0.46$ V

With $\alpha_{min} = -0.0375$: $R_{th} = -9.25$ k$\Omega$ $\Rightarrow V_0 = 0.54$ V

Error = $|0.5 - 0.46| = |0.5 - 0.54| = \boxed{0.04 \text{ V}}$$

QQuestion 7 0 Mark

A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and subtransient reactance of 0.2 pu. It is operating at $(1+j0)$ pu terminal voltage with an internal emf of $(1+j0.7)$ pu. Following a three-phase solid short circuit fault at the terminal of the generator, the magnitude of the subtransient internal emf (rounded off to 2 decimal places) is \_\_\_\_\_\_\_\_ pu.

SSolution

Given: $X_d = 0.7$ pu, $X_d" = 0.2$ pu, $V_t = 1\angle 0Ã‚Â°$ pu, $E_f = 1 + j0.7$ pu

Pre-fault current:

\[I_0 = \frac{E_f - V_t}{jX_d} = \frac{(1+j0.7) - 1}{j0.7} = \frac{j0.7}{j0.7} = 1 \text{ pu}\]

Subtransient internal emf:

\[E_f" = V_t + jX_d" I_0 = 1 + j(0.2)(1) = 1 + j0.2\]

\[|E_f"| = \sqrt{1^2 + 0.2^2} = \sqrt{1.04} = \boxed{1.02 \text{ pu}}\]

QQuestion 8 0 Mark

A cylindrical rotor synchronous generator with constant real power output and constant terminal voltage is supplying 100 A current to a 0.9 lagging power factor load. An ideal reactor is now connected in parallel with the load, as a result of which the total lagging reactive power requirement of the load is twice the previous value while the real power remains unchanged. The armature current is now \_\_\_\_\_\_\_ A (rounded off to 2 decimal places).

SSolution

Given: $I_{a1} = 100$ A, $\cos\phi_1 = 0.9$ (lagging)

From $\cos\phi_1 = 0.9$: $\tan\phi_1 = \frac{\sqrt{1-0.9^2}}{0.9} = 0.484$

This gives: $\frac{Q_1}{P} = 0.484$

After reactor connection: $Q_2 = 2Q_1$, so $\frac{Q_2}{P} = 2(0.484) = 0.9686$

Therefore: $\tan\phi_2 = 0.9686$, which gives $\cos\phi_2 = 0.7182$

Since real power is constant:

\[I_{a1}\cos\phi_1 = I_{a2}\cos\phi_2\]

\[100 \times 0.9 = I_{a2} \times 0.7182\]

\[I_{a2} = \frac{90}{0.7182} = \boxed{125.29 \text{ A}}\]

QQuestion 9 0 Mark

Consider a permanent magnet dc (PMDC) motor which is initially at rest. At $t = 0$, a dc voltage of 5 V is applied to the motor. Its speed monotonically increases from 0 rad/s to 6.32 rad/s in 0.5 s and finally settles to 10 rad/s. Assuming that the armature inductance of the motor is negligible, the transfer function for the motor is:

AOptions

$\frac{10}{0.5s + 1}$
$\frac{2}{0.5s}$
$\frac{10}{0.5s}$

SSolution

First-order system: $G(s) = \frac{K}{Ts + 1}$

From steady-state response to 5 V input:

\[\text{Steady state speed} = 10 \text{ rad/s} = 5K\]

\[K = 2\]

At 63.2% of steady state (one time constant):

\[0.632 \times 10 = 6.32 \text{ rad/s at } t = T = 0.5 \text{ s}\]

Therefore: $T = 0.5$ s

Transfer function: $G(s) = \boxed{\frac{2}{0.5s + 1}}$

Correct answer: A