square waveform as shown in figure is applied across 1 mH ideal
inductor. The current through the inductor is a . wave of peak
amplitude. -1 1V 0 0.5 t (m sec) 1 [Gate 1987: 2 Marks] 0.5 1.5 1 t
m sec 1V
8. The voltage across the terminals a and b in Figure is 3A 2 2
1 1V ab (a) 0.5 V (b) 3.0 V (c) 3.5 V (d) 4.0 V
[Gate 1998: 1 Mark] 5. The voltage V in Figure is always equal
to
DC 2 2 A V+- 5V (a) 9 V (b) 5 V (c) 1 V (d) None of these
[Gate 1997: 1 Mark]
7. In the circuit shown in the figure the current iD through the
ideal diode (zero cut in voltage and zero forward resistance)
equals
10V DC 4 4 1 2A iD (a) 0 A (b) 4 A (c) 1 A (d) None of these
[Gate 1997: 3 Marks]
The voltage V in Figure is DC DC 10V 5V 3 +- ab (a) 10 V (b) 15
V (c) 5 V (d) None of the these
[Gate 1997: 1 Mark] Ans. (a) 4. The Voltage V in Figure is equal
to DC DC DC 5V 4V 4V + V - 2 (a) 3 V (b) -3 V (c) 5 V (d) None of
these = + + = + Since the voltage of 2A current source is not
known, it is not possible to find the value of voltage V.
[Gate 1997: 1 Mark] Ans. (a) Apply KVL V + 5 4 = 4 V = 4 + 4 5 =
3V
1. A
Ans. The current through the inductor is = . The integration of
a square wave is a triangular wave so the current through the
inductor is a triangular wave of 1 volt peak amplitude. Slope of
triangular wave is 2
2. Two 2H inductance coils are connected in series and are also
magnetically coupled to each other the coefficient of coupling
being 0.1. The total inductance of the combination can be (a) 0.4 H
(b) 3.2 H (c) 4.0 H (d) 4.4 H
M L1 L2 2H 2H Ans. (d)
The equivalent inductance = + = + . = = . = . , .
3. The current i4 in the circuit of Figure is equal to
i1 = 5A i2 = 3A i4 = ? i3 = 4A i0 = 7A I (a) 12 A (b) -12 A (c)
4 A (d) None of these
[Gate 1997: 1 Mark] Ans. (b) = + = =
