Electrical Engg (EEE) - MCQ Practice Questions

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 $i_{ds}$.

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.