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Electronics and Communication Engineering - Electronic Devices and Circuits

Exercise : Electronic Devices and Circuits - Section 9
26.
If V = 4 in the figure, the value of Is is given by
6 A
5/2 A
12 A
none
Answer: Option
Explanation:

At node V1

at node V2

2V2 - V1 = 0 V1 = 2V2

V2 = 4 V1 = 8 Is = x 8 - 4 = 6A.

27.
A current of 1 A in the coil of an iron cored electromagnet causes B = 0.5T. If current is 2A, B =
0.5 T
1 T
1 T or more
1 T or less
Answer: Option
Explanation:

If saturation starts B will be less than it.

28.
Two current sources each of rms value 10 A have frequencies 40 Hz and 50 Hz respectively. They are connected in series, the expression for resultant wave is
i = 10 sin 40 t + 10 sin 50 t
i = 14.14 sin 40 t + 14.14 sin 50 t
i = 14.14 sin 80 pt + 14.14 sin 100 pt
i = 20 sin 80 pt + 20 sin 100 pt
Answer: Option
Explanation:

Peak value = 14.14 A; ω1 = 2p x 40, ω2 = 2p x 50

29.
To find the current in a 20 Ω resistance connected in a circuit, Norton's theorem is used. IN = 7.5 A. The current though 20 Ω resistance.
is 7.5 A
is less than 7.5 A
may be 7.5 A or less
may be 7.5 A or more or less
Answer: Option
Explanation:

Norton's equivalent is as shown in figure. Some current will flow through RN. There-fore, current through 20Ω resistance is less than 7.5 A.

30.
While calculating Norton's resistance, all current sources are short circuited.
True
False
Answer: Option
Explanation:

All current sources are open circuited.

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