Electrical Engg (EEE) - MCQ Practice Questions

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 $i_{ds}$.

By Gauss's law, the total electric flux through any closed surface enclosing charge $Q$ is

Here $Q_{\text{enc}}=4\mu \text{C}=4\times 10^{-6}\text{C}$, so

The radius of the surface does not affect the total flux; only the enclosed charge matters.

The intrinsic impedance of free space is

The relationship between peak fields is

From $\nabla \cdot \vec{B}=0$ (Gauss's law for magnetism), applying the divergence theorem to a pillbox surface at the interface gives

The normal component of $\vec{B}$ is always continuous across any interface. Note that $H_n$ is not necessarily continuous when ${\mu }_1\ne {\mu }_2$.

A distortionless (Heaviside) line requires that the attenuation $\alpha$ is frequency-independent and the phase velocity $v_p=\omega /\beta$ is constant (no dispersion). This is achieved when

which rearranges to

Under this condition, $\alpha =\sqrt{RG}$ and $\beta =\omega \sqrt{LC}$, giving a flat frequency response.

Skin depth varies as

Therefore,

So

As frequency increases, skin depth decreases, confining current to a thinner surface layer.

Faraday's law in differential (point) form is one of Maxwell's four equations:

The negative sign is a consequence of Lenz's law — the induced electric field opposes the change in magnetic flux. Option A is missing the negative sign; option C incorrectly uses the divergence operator; option D is Ampere's law with a wrong sign.

The force per unit length between two parallel current-carrying conductors is

Substituting:

Currents in the same direction attract each other (by the right-hand rule, each wire is in the other's magnetic field directed to produce attraction).

At the surface of a perfect conductor ($\sigma \to \infty$, skin depth $\to 0$), the boundary condition requires the tangential electric field to be zero. The reflection coefficient for the electric field is

where ${\eta }_2=0$ for a perfect conductor. Thus

The reflected electric field is equal in magnitude but opposite in phase to the incident field, resulting in a standing wave with a null at the conductor surface.

The capacitance of a parallel-plate capacitor with a dielectric is

Substituting:

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.