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# ELECTRICAL ENGINEERING - page 13 / 15

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Solving for  we have

=

10, 1000 ohms (rejecting the negative value)

36:- (a) One may arrive at the correct choice by comparing the two sides dimensionally. Since 1 /  has the dimension of velocity squared,

The pertinent choice is  =

37:- (d) The waveform of the load current will be triangular.

38:- (d) The voltage across is = 35 – 5 = 30V. The current in the circuit is

39:- (a) It will settle down at 1+ OP.

40:- (c) Let aRb. Since the relation is symmetric bRb. Since transitivity holds good, aRb and bRa implied aRa. Also bRa implies aRa and bRb. If R has to be an equivalence relation, it has to be reflexive, i.e. for any x belonging to A, xRx should hold good. So, R need not be reflexive and so R need not be and equivalence relation.

For example, let a = (1, 2, 3)

Let R = {(1,2),(2,1),(1,1),(2,2)}

R is both symmetric and transitive but R is not reflexive as (3,3) is not present in R.

41:- (c) Its base-emitter junction increases by about 20 mV.

42:- (b) Critical Frequency, fc = 9

=  = 16.839 MHz.

43:- (b) Arithmetic

44:- (a) “Figure of merit” is used to compare the efficiency of thermo-electric materials.

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