Capacitors Test

Result

Capacitors Test - 6

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Weekly Quiz Competition

Question 1
4 / -1

The capacitor of capacitance 4µF and 6µF are connected in series. A potential difference of 500 volts applied to the outer plates of the two capacitor system. Then the charge on each capacitor is numerically

A
6000 C

B
1200 C

C
1200 µC

D
6000 µC

Solution

Charge flown = 2.4 × 100 × 10^-6 C = 1200 µC
Question 2
4 / -1

Three capacitors of capacitances 3 µF, 9µF and 18µF are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases will be

A
1 : 15

B
15 : 1

C
1 : 1

D
1 : 3

Solution
Question 3
4 / -1

Two condensers of capacity 0.3 µF and 0.6 µF respectively are connected in series. The combination is connected across a potential of 6 volts. The ratio of energies stored by the condensers will be

A
1/2

B
2

C
1/4

D
4

Solution

In series combination Q is constant, hence according to
Question 4
4 / -1

In the adjoining figure, four capacitors are shown with their respective capacities and the P.D. applied. The charge and the P.D. across the 4 µF capacitor will be

A
600 µC ; 150 volts

B
300 µC ; 75 volts

C
800 µC ; 200 volts

D
580 µC ; 145 volts

Solution
Question 5
4 / -1

A 4µF condenser is connected in parallel to another condenser of 8µF Both the condensers are then connected in series with a 12 µF condenser and charged to 20 volts. The charge on the plate of 4µF condenser is

A
3.3 µC

B
40 µC

C
80 µC

D
240 µC

Solution

Equivalent capacitance of the circuit C_eq = 6 µF

Charge supplied from source Q = 6 × 20 = 120 µC

Hence charge on the plates of 4 µF capacitor
Question 6
4 / -1

In the following circuit, the resultant capacitance between A and B is 1 µF. Then value of C is

A
32/11 µF

B
11/32 µF

C
23/32 µF

D
32/23 µF

Solution

12 µF and 6µF are in series and again are in parallel with 4µF.

Therefore, resultant of these three will be

This equivalent system is in series with 1 µF.

Equivalent of 8µF, 2µF and 2µF

(i) and (ii) are in parallel and are in series with C
Question 7
4 / -1

The mean electric energy density between the plates of a charged capacitor is (here Q = Charge on the capacitor and A = Area of the capacitor plate)

A

B

C

D
None of these

Solution
Question 8
4 / -1

Plate separation of a 15µ F capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between the plates. Then new capacitance is given by

A
15 µ F

B
20µ F

C
30µ F

D
25µ F

Solution

From equation (i) C' = 20 µ F.
Question 9
4 / -1

The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is

A

B

C

D

Solution
Question 10
4 / -1

If a slab of insulating material 4×10^-5 m thick is introduced between the plate of a parallel plate capacitor, the distance between the plates has to be increased by 3.5×10^-5 m to restore the capacity to original value. Then the dielectric constant of the material of slab is

A
10

B
12

C
6

D
8

Solution