1. A solid copper sphere, 10 cm in diameter is deprived of 10^{20} electrons by a charging scheme. The charge on the sphere is _________

a) 160.2 C

b) -160.2 C

c) 16.02 C

d) -16.02 C

Explanation: n = 10

^{20}, Q = ne = e 10

^{20}= 16.02 C.

Charge on sphere will be positive.

2. A lightning bolt carrying 15,000 A lasts for 100 s. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is _________

a) 13.33 C

b) 75 C

c) 1500 C

d) 1.5 C

Explanation: dQ = i dt = 15000 x 100µ = 1.5 C.

3. If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is _________

a) 0.5 A

b) 2 A

c) 3.33 mA

d) 0.3 mA

Explanation: i = dQ/dt = 120/60 = 2A.

4. The energy required to move 120 coulomb through 3 V is _________

a) 25 mJ

b) 360 J

c) 40 J

d) 2.78 mJ

Explanation: W = Qv = 360 J.

5. Consider the circuit graph shown in figure below. Each branch of circuit graph represent a circuit element. The value of voltage V1 is

a) 30 V

b) 25 V

c) 20 V

d) 15 V

Explanation: 100 = 65 + V2 => V2 = 35 V

V3 – 30 = V2 => V3 = 65 V

105 – V3 + V4 – 65 = 0 => V4 = 25 V

V4 + 15 – 55 + V1 = 0 => V1 = 15 V.

6. What will be the value of Req in the following Circuit?

a) 11.86 ohm

b) 10 ohm

c) 25 ohm

d) 11.18 ohm

Explanation:

We infer from the given diagram that the same pattern is followed after every 10ohm resistor. The infinite pattern in parallel as a whole is considered as to be Rx.

Thus, Rx = R + (R||Rx)

Solving for Rx, we get Rx = 1.62R, where R=10ohm.

So we have Rx = (1.62ohm)*(10ohm), which gives Rx=16.2ohm

Therefore, Req = 5 + (10||16.2)

=> 5 + [(10*16.2)/(10+16.2)] => Req = 5 + (162/26.2) which gives Req=11.18 ohm.

7. In the circuit the dependent source __________

a) supplies 16 W

b) absorbs 16 W

c) supplies 32 W

d) absorbs 32 W

Explanation: P = VIx = 2Ix Ix = 2 x 16 or 32 watt (absorb).

8. Twelve 6 ohm resistors are used to form an edge of a cube. The resistance between two diagonally opposite corner of the cube (in ohm) is ________

a) 5/6

b) 6/5

c) 5

d) 6

Explanation:

The current I will be distributed to the cube branches symmetrically.

Connecting a voltage source Vab across the terminals and applying kvl in the outer loop, we have

=> -Vab + (i/3+i/6+i/3)*R = 0 (where R is given as 6ohm)

=> Vab = 5i

=> Req=Vab/i, which gives Req = 5ohm.

9. The energy required to charge a 10 µF capacitor to 100 V is ________

a) 0.1 J

b) 0.05 J

c) 5 x 10^{(-9)} J

d) 10 x 10^{(-9)} J

Explanation: Energy provided is equal to 0.5 CVxV.

10. A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec interval. The value of capacitance is ________

a) 0.75 mF

b) 1.33 mF

c) 0.6 mF

d) 1.67 mF

Explanation: The capacitor current is given as i=C*(dv/dt), where dv/dt is the derivative of voltage, dt=t2-t1 given as 10 sec and dv is the change in voltage which is given as 12V.

So, we have C=i/(dv/dt)

=> C = 2mA/(12/10) = 2mA/(1.2).

Hence C = 1.67mF.