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 𝑽 = 𝑽𝟐𝑨 + 𝟐 × 𝟐 + 𝟓 = 𝑽𝟐𝑨 + 𝟗 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 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]

1. A 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

Ans. The current through the inductor is 𝒊𝑳 = 𝟏𝑳 ∫ 𝒗𝒅𝒕 . 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 𝑳𝒆𝒒 = 𝑳𝟏 + 𝑳𝟐 ± 𝟐𝑴 = 𝟐 + 𝟐 ± 𝟐 × 𝟎. 𝟏√𝟐 × 𝟐 𝑴 = 𝑲√𝑳𝟏𝑳𝟐 = 𝟒 ± 𝟎. 𝟒 = 𝟒. 𝟒, 𝟑. 𝟔 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) 𝑰 = 𝒊𝟎 + 𝒊𝟏 = 𝟏𝟐𝑨 𝒊𝟒 = −𝟏𝟐𝑨