Electrical Engg (EEE) - MCQ Practice Questions

The back EMF is given by:

Substituting the values:

The air-gap flux in an induction motor is proportional to $V/f$. By keeping the V/f ratio constant as frequency is varied, the magnetic flux in the air gap remains constant, which ensures the motor can develop rated torque throughout the speed range without magnetic saturation or under-excitation. This is the fundamental principle of scalar (V/f) control.

The back EMF is:

The developed mechanical power is:

From the equivalent circuit of an induction motor, differentiating the torque expression with respect to slip and setting it to zero gives the slip at maximum torque:

This shows that increasing rotor resistance (e.g., by external resistance in wound-rotor motors) increases $s_m$, shifting the maximum torque to a higher slip value, which is the principle of rotor resistance speed control.

The speed of a DC motor is given by $N\propto \frac{V_a-I_a{R}_a}{\varphi }$. Reducing the armature voltage $V_a$ below the rated value reduces the back EMF and hence the speed — this gives speeds below base speed at constant torque. Field weakening (reducing $\varphi$) gives speeds above base speed at constant power. Hence armature voltage control is used for sub-base speed operation.

During regenerative braking, the kinetic energy of the load drives the motor above synchronous/no-load speed, causing the back EMF $E_b$ to exceed the supply voltage $V_a$. The machine then acts as a generator, and the current reverses direction, feeding electrical energy back into the DC supply. This is energy-efficient braking, unlike dynamic braking which dissipates energy as heat in resistors.

In star-delta starting, the motor is first connected in star, reducing the voltage across each winding to $V/\sqrt{3}$ of the line voltage. Since torque is proportional to the square of the voltage applied to each winding:

Similarly, the starting current drawn from the supply is also reduced to $\frac{1}{3}$ of the DOL starting current. Hence $\boxed{T_{\text{star}}=\frac{1}{3}{T}_{\text{DOL}}}$.

For a step-down (buck) chopper, the average output voltage is:

where $\delta$ is the duty cycle and $V_s$ is the source voltage.

For fans, blowers, and centrifugal pumps, the torque required varies as the square of the speed:

Consequently, the power consumed varies as the cube of speed: $P\propto N^3$. This is in contrast to constant torque loads (e.g., conveyors, hoists) where $T=\text{constant}$. This characteristic makes VFDs particularly effective for fan/pump drives, offering significant energy savings at reduced speeds.

In Field-Oriented Control (FOC), the stator current vector is decomposed into two orthogonal components in a rotating reference frame aligned with the rotor flux:

This decoupling allows the induction motor to be controlled like a separately excited DC motor, achieving fast dynamic response. The torque is given by $T_e\propto {\psi }_r\cdot i_{qs}$, where ${\psi }_r$ is the rotor flux controlled by $i_{ds}$.

Understanding:

We must find the back-EMF of a separately excited DC motor.

Formula:

The armature voltage equation of a DC motor is:

Step 1: Rearrange for back-EMF

Solving for $E_b$:

Step 2: Substitute values

Answer:

The back-EMF of the motor is 200 V.

Quick Tip:

The voltage drop $I_a{R}_a$ represents copper losses in the armature winding. The back-EMF is always less than the applied voltage during motoring operation.

Understanding:

We must find the initial braking current when the motor is switched to dynamic braking.

Formula:

During dynamic braking the motor acts as a generator. The back-EMF drives current through the series combination of armature resistance and braking resistance:

Step 1: Substitute values

Answer:

The initial braking current is 40 A.

Quick Tip:

The braking resistance $R_b$ is chosen to limit the braking current to a safe value (typically 1.5–2 times rated current). Here $R_a$ is small but must not be neglected.

Understanding:

We must find the air-gap power given the rotor copper loss and slip.

Formula:

The fundamental power balance relationship of an induction motor is:

Step 1: Rearrange for air-gap power

Step 2: Substitute values

Step 3: Verify mechanical power developed

This confirms the rotor copper loss: $20000-19200=800\text{W}$ ✓

Answer:

The air-gap power is 20000 W.

Quick Tip:

Remember: air-gap power splits as $s:(1-s)$ between rotor copper loss and mechanical power developed. This ratio is always tested.

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.