Electrical Engg (EEE) — MCQ Practice Questions & Answers

Question 1

A separately excited DC motor runs at 1000 rpm at rated load with an armature voltage of 200 V and armature current of 20 A. If the armature resistance is 1\,\Omega , what is the back EMF of the motor?

Accepted Answer

Correct answer: A. 180\,	ext{V} — The back EMF is given by: E_b = V_a - I_a R_a Substituting the values: \begin{aligned} E_b &= 200 - (20 	imes 1) \ &= 200 - 20 \ &= \boxed{180\,	ext{V}} \end{aligned}

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B. 200\,	ext{V}

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C. 220\,	ext{V}

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D. 160\,	ext{V}

Question 2

In a Variable Frequency Drive (VFD) controlling an induction motor, the V/f ratio is kept constant primarily to:

Accepted Answer

Correct answer: A. Maintain constant air-gap flux in the motor — 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.

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B. Minimize copper losses in the stator winding

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C. Reduce harmonic distortion in the supply current

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D. Improve the power factor of the motor

Question 3

A DC shunt motor has an armature resistance of 0.5\,\Omega and a field resistance of 200\,\Omega . It is connected to a 220\,	ext{V} supply. If the armature current at full load is 40\,	ext{A} , the developed mechanical power is:

Accepted Answer

Correct answer: A. 8000\,	ext{W} — The back EMF is: E_b = V - I_a R_a = 220 - (40 	imes 0.5) = 220 - 20 = 200\,	ext{V} The developed mechanical power is: \begin{aligned} P_{	ext{mech}} &= E_b 	imes I_a \ &= 200 	imes 40 \ &= \boxed{8000\,	ext{W}} \end{aligned}

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B. 8800\,	ext{W}

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C. 7200\,	ext{W}

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D. 8400\,	ext{W}

Question 4

In a three-phase induction motor drive, the slip at maximum torque is denoted by s_m . If the rotor resistance per phase is R_2 and the rotor leakage reactance per phase at standstill is X_2 , then s_m is given by:

Accepted Answer

Correct answer: A. s_m = \dfrac{R_2}{X_2} — 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: \boxed{s_m = \frac{R_2}{X_2}} 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.

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B. s_m = \dfrac{X_2}{R_2}

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C. s_m = \dfrac{R_2^2}{X_2}

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D. s_m = \dfrac{R_2}{X_2^2}

Question 5

Which of the following methods of speed control of a DC shunt motor gives speed below the base (rated) speed?

Accepted Answer

Correct answer: A. Armature voltage control — The speed of a DC motor is given by N \propto \dfrac{V_a - I_a R_a}{\phi} . 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 \phi ) gives speeds **above** base speed at constant power. Hence armature voltage control is used for sub-base speed operation.

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B. Field flux weakening

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C. Increasing the supply frequency

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D. Reducing the armature resistance

Question 6

In regenerative braking of a DC separately excited motor, the motor acts as a:

Accepted Answer

Correct answer: A. Generator, returning energy back to the supply — 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.

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B. Generator, dissipating energy in braking resistors

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C. Motor, absorbing energy from the supply to brake

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D. Transformer, transferring energy between phases

Question 7

A squirrel cage induction motor is to be started using a star-delta starter. Compared to direct-on-line (DOL) starting, the starting torque with star-delta starting is:

Accepted Answer

Correct answer: A. \dfrac{1}{3} of the DOL starting torque — 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: T_{	ext{star}} = \left(\frac{1}{\sqrt{3}}ight)^2 T_{	ext{DOL}} = \frac{1}{3}\,T_{	ext{DOL}} Similarly, the starting current drawn from the supply is also reduced to \dfrac{1}{3} of the DOL starting current. Hence \boxed{T_{	ext{star}} = \dfrac{1}{3}\,T_{	ext{DOL}}} .

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B. \dfrac{1}{\sqrt{3}} of the DOL starting torque

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C. \sqrt{3} times the DOL starting torque

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D. Equal to the DOL starting torque

Question 8

In a chopper-controlled DC drive, a DC motor is supplied from a 200 V source through a chopper with a duty cycle of 0.6. Assuming continuous current conduction, the average output voltage applied to the motor armature is:

Accepted Answer

Correct answer: A. 120\,	ext{V} — For a step-down (buck) chopper, the average output voltage is: V_o = \delta 	imes V_s where \delta is the duty cycle and V_s is the source voltage. \begin{aligned} V_o &= 0.6 	imes 200 \ &= \boxed{120\,	ext{V}} \end{aligned}

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B. 133\,	ext{V}

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C. 80\,	ext{V}

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D. 150\,	ext{V}

Electrical Engg (EEE) - MCQ Practice Questions