Electrical Engg (EEE) - MCQ Practice Questions

Understanding:

We must determine how torque changes in a DC series motor when armature current is doubled.

Formula:

For a DC series motor the flux is proportional to armature current ($\varphi \propto I_a$), so:

Step 1: Compute new torque

Step 2: Find the ratio

Answer:

The torque increases by a factor of 4.

Quick Tip:

The square-law torque characteristic of a series motor makes it ideal for high-starting-torque applications like traction and cranes. The same property makes it dangerous at no load — speed rises without limit.

Understanding:

We must calculate the percentage slip of a three-phase induction motor.

Formula:

Synchronous speed:

Percentage slip:

Step 1: Calculate synchronous speed

Step 2: Calculate percentage slip

Answer:

The percentage slip of the motor is 4%.

Quick Tip:

For a 4-pole, 50 Hz motor the synchronous speed is always 1500 rpm. Full-load slip of 3–5% is typical for squirrel cage induction motors.

Understanding:

We must identify the effect of reducing stator voltage on the torque-speed characteristic of an induction motor.

Formula:

The maximum (pull-out) torque of an induction motor is given by:

where $V_s$ is the stator voltage and $X_s$ is the equivalent reactance. The synchronous speed depends only on supply frequency $f$ and number of poles $P$:

Step 1: Effect on synchronous speed

Since $f$ and $P$ are unchanged when only voltage is varied, $N_s$ remains constant.

Step 2: Effect on maximum torque

Because $T_{\text{max}}\propto V_s^2$, reducing voltage to half reduces maximum torque to one-quarter of its original value. The torque-speed curve shifts downward while maintaining the same synchronous speed.

Step 3: Effect on slip at maximum torque

The slip at maximum torque $s_m=\frac{R_2}{X_2}$ depends on rotor resistance and reactance, not on stator voltage, so it remains unchanged.

Answer:

Maximum torque varies as the square of stator voltage; synchronous speed is unaffected by voltage changes.

Quick Tip:

Stator voltage control is inefficient for speed control because the excess slip power is wasted as rotor copper loss. It is mainly used for soft-starting or fan/pump loads where the load torque is low at reduced speed.

Understanding:

We must find the average armature current in a chopper-fed DC motor drive.

Formula:

The duty cycle and average output voltage of a step-down chopper are:

Average armature current:

Step 1: Calculate duty cycle

Step 2: Calculate average armature voltage

Step 3: Calculate average armature current

Answer:

The average armature current is 10 A.

Quick Tip:

Always compute the average voltage first using $\delta V_s$, then apply the standard armature circuit equation. The ripple current (due to motor inductance) is a separate calculation.

Understanding:

We must identify which quadrant of a torque-speed plane corresponds to reverse rotation with positive (braking) torque.

Step 1: Define the four quadrants

In the torque-speed ($T$–$N$) plane used for electrical drives:

Step 2: Match the condition

The question specifies reverse rotation ($N<0$) and braking torque in the positive direction ($T>0$). This combination lies in Quadrant IV.

Step 3: Physical interpretation

In Quadrant IV the motor is decelerating from reverse rotation. Energy flows back to the supply (regenerative braking in the reverse direction). A dual-converter or four-quadrant chopper is needed to achieve this.

Answer:

Reverse rotation with positive braking torque corresponds to Quadrant IV.

Quick Tip:

A simple mnemonic: Quadrants I and III are motoring (torque and speed have the same sign); Quadrants II and IV are braking/generating (torque and speed have opposite signs).

Understanding:

We must find the stator voltage when a VFD operates at a reduced frequency under constant $V/f$ control.

Formula:

Constant $V/f$ control maintains:

Therefore:

Step 1: Calculate the $V/f$ ratio at base

Step 2: Find voltage at 30 Hz

Answer:

The stator voltage at 30 Hz is 240 V.

Quick Tip:

Constant $V/f$ keeps the air-gap flux approximately constant, which prevents magnetic saturation at low frequencies and avoids torque reduction. Above base speed, voltage is capped at $V_0$ and flux-weakening begins.