Question

Exercise 2 In the figure the battery supplies a voltage ofV-9 Volts, C Ca 30 uF and C Ca 15 uF. Find: HH CA CB 1) The charge stored at each of the four capacitors. 2) The voltage difference between the plates of capacitors C, Ca and C 3) The total charge supplied by the battery 4) V C2 The net capacitance, and total energy stored in the circuit.

Answer 1

d)

net capacitance of circuit

C= ( 1/ 30 + 1/ 30 + 1/30)^-1

C = 10 uF

energy stored in circuit

U = 0.5 CV^2 = 0.5* 10* 10^-6* 9^2

U = 4.05* 10^-4 J

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a)

charge through C1

q1 = CV = 10* 9 = 90 uC

charge through C2

q2 = q1 = 90 uC

charge through C3

q3 = q1 / 2 = 45 uC

charge through C4

q3 = q4 = 45 uC

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b)

V1 = q/C1 = 90/30 = 3 V

V2 = V1 = 3 V

V4 = 3 V

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c)

U = 4.05* 10^-4 J

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comment in case any doubt, will reply for sure.. goodluck

Answer 2

SOLUTION :

C3 and C4 are in parallel, so equivalent capacitance = C3 + C4 = 15 + 15 = 30 µF

Now, C1 , (C3 + C4) and C2 are in series.

So, equivalent capacitance of the circuit , C

= 1/ (1/C1  + 1/(C3 + C4) + 1/C2)

= 1/(1/30 + 1/30 + 1/30) 

= 1/(3/30)

= 10 µF

1.

Charge stored in each of the capacitors in the series circuit = C V = 10 * 9 = 90 µC.

As C3 = C4 , so charge stored at C3 and C4 = 90/2 = 45 µC

Hence,

Charge q1 stored at C1 = 90  µC(ANSWER)

Charge q2 stored at C2 = 90 µC (ANSWER)

Charge q4 stored at C4 = 45 µC (ANSWER)

2.

Voltage difference V1  between plates of C1 = q1/C1 = 90/30 = 3 V (ANSWER).

Voltage difference V2 between plates of C2 = q2/C2 = 90/30 = 3 V (ANSWER).

Voltage difference V4 between plates of C4 = q4/C4 = 45/15 = 3 V (ANSWER).

3.

Total charge supplied by the battery = 90 µC 

4,

Net capacitance of the circuit = 10 µF (as calculated above) (ANSWER).

Total energy stored in the circuit

= 0.5 C V^2

= 0.5 * 10 * 10^(-6) * 9^2

= 4.05 * 10^4) J (ANSWER).