DRDO Electrical Solved Sample Papers with solutions Set-I

DRDO Electrical Solved Sample Papers with solutions Set-I

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

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

(a)    Are parallel to each other.

(b)   Are one and the same.

(c)     Cut each other orthogonally.

(d)   Can be inclined to each other at any angle.

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

(a)    R in series with SCR.

(b)   L in series with SCR.

(a)    RL in series with SCR.

(c)    RLC in series with SCR.

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.

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

(a)    K = 4,     a = 0.35

(b)   K = 8,     a = 0.455

(c)    K = 16,   a = 0.225

(d)   K = 64,   0.9

Solutions 

1(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.

2(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.

3( 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.

4( d) Sin wtCos wt

Differentiating

Cos wt2 Sin wt

An input of 10 Cos wt will cause a

Response of – 20 Sin wt.

5 (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

 

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