# DRDO MODEL QUESTIONS WITH ANSWERS ELECTRICAL ENGINEERING PRACTICE

**DRDO MODEL QUESTIONS WITH ANSWERS ELECTRICAL ENGINEERING PRACTICE **

Q.1:- The resistance of a strip of copper of rectangular cross-section is 2Ω. A metal of resistivity twice that of copper is coated on its upper surface to a thickness equal to that of copper strip.

The resistance of composite strip will be

(a) 6Ω

(b) 4/3Ω

(c) 3/2Ω

(d) 3/4Ω

**(b) Copper and coated metal strips have resistance of 2 ohms and 4 ohms respectively. These two strips are in parallel. Hence the resistance of the composite strip will be (2×4) / (2+4) =4/3 ohms.**

Q.2:- The electric field lines and equipotential lines

Are parallel to each other.

Are one and the same.

Cut each other orthogonally.

Can be inclined to each other at any angle.

**-(c) Since no electric can exist along any surface, all points of which are at the same potential, electric field lines and equipotential lines are orthogonal to each other.**

Q.3:- For an SCR, di/dt protection is achieved through the use of

R in series with SCR.

L in series with SCR.

RL in series with SCR.

RLC in series with SCR.

**( b) For an SCR, di/dt protection is achieved through the used of L in series with SCR. A snubber circuit connected across an SCR is to suppress dv / dt.**

Q.4:- In a linear system, an input of 5 sin wt produces an output of 10 cos wt. The output corresponding to input 10 cos wt will be equal to

(a) + 5 sin wt

(b) – 5 sin wt

(c) + 20 sin wt

(d) – 20 sin wt.

**( d)** Sin wtCos wt

**Differentiating**** **

** Cos wt2 Sin wt**** **

** An input of 10 Cos wt will cause a **** **

**Response of – 20 Sin wt.**

Q.5:- For the system shown in figure, with a damping ratio § of 0.7 and an undamped natural frequency wn of 4 rad / sec, the values of K and are

K = 4, a = 0.35

K = 8, a = 0.455

K = 16, a = 0.225

K = 64, 0.9

**(c)** M (s) =

=

The changed equation is

s (s+2) + K (1 + as) = 0 or s² + s (2+ak) + k =0

Compare with s²+2 s+ = 0

K = = 4² = 16;

2 = (2+ak)

a= = = 0.225

Q.6:- A single instruction to clear the lower four bits of the accumulator I 8085 assembly language is:

(a) XRI OHF

(b) ANI FOH

(c) XRI FOH

(d) ANI OFH.

**(b) A single instruction to clear the lower four bits of the accumulator in 8085 assembly language is ANI FOH.**

Q.7:- The p-type substrates in a conventional pn-junction isolated integrated circuit should be connected to:

Nowhere, i.e., left floating.

A dc ground potential.

The most positive potential available in the circuit.

The most negative potential available in the circuit.

**(d) it should be connected to the most negative potential available in the circuit.**

Q.8:- The velocity of electromagnetic ratio waves is

3 x 〖10〗^6 m/s

3 x 〖10〗^8 m/s

3 x 〖10〗^10 m/s

3 x 〖10〗^12 m/s

**(b) The velocity of electromagnetic wave is 3 x 10⁸ m / s.**

Q.9:- A certain 8- bit microprocessor executes “SUB A, B” by loading B to the ALU, then loading A to the ALU, Subtracting A from B, and storing the result in B. Which of the following instructions would cause the ZERO flag to be set?

SUB – (X), (X) +

SUB )X), (X) +

SUB – ( X), (X)

all of the above

** (a) Note that
(X) + signifies “locate the operand at location X, and then increment X by 1.”
− (X) signifies “decrement X by 1, and then locate the operand at location old – X minus one.”
Q.10:- which of the following induction motor will have the least shaft diameter?
20 HP, 2880 rpm
20 HP, 1440 rpm
20 HP, 960 rpm
20 HP, 730 rpm**

Consider the instruction:

SUB – (X), (X) +

Observe that at both times the same operand, namely a, is being accessed. Since a – a = 0 the zero flag will be set.

Q.10:- which of the following induction motor will have the least shaft diameter?

(a) 20 HP, 2880 rpm

(b) 20 HP, 1440 rpm

(c) 20 HP, 960 rpm

(d) 20 HP, 730 rpm

** (a)** The horsepower is proportional to the product of rpm (N) and torque (T)

Since the horsepower of all the motors is the same, the motor having highest rpm (2880 in this case) will have least torque. Diameter of the shaft depends on torque, therefore, 20 HP, 2880 rpm motor will have the least diameter of shaft