A 10 muF capacitor is in series with a parallel combination of a 15 muF an 5muF

Question

A 10 muF capacitor is in series with a parallel combination of a 15 muF an 5muF capacitor as shown at the right. What is the effective capacitance of this combination in units of muF? Suppose that a 3.0 volt battery is now connected to this arrangement of capacitors by connecting the points A and B to the + and - terminals of the battery. How much charge will be on plates of the 10 muF capacitor when it is fully charged? What will be the voltage drop across the 5.0 muF capacitor when fully charged?

Explanation / Answer

In parallel combination of capacitors , total effective capacitance is given by

C = C1 + C2

For combination of 5 & 15 uF

C = 5 + 15 = 20 uF

For series combination

1/ C = 1/ C1 + 1/ C 2

1/ C = 1/ 10 + 1/ 20

C = 20/3 = 6.7 uF

Ans (b) 6.7 uF

4 ) The setup in the question is equivalent to two capacitors of 10 and 20 uF in series.

In series combination the charge on each plate of every capacitors remains same.

Charge Q = C × V

Effective capacitance of combination

C = 20/3 *10^-6

V= 3 V

Q = 20/ 3 *10^-6 × 3 = 20*10^-6

Ans (b) 20 uC

5 ) ln series combination the voltage V = V 1 + V2

Where V 1= voltage drop across capacitor C1

V2 = voltage drop across capacitor C2

Voltage drop V = Q / C

V across the combination of 5 & 15 uF is

Q= 20*10^-6 uC

C = 20*10^-6 uF

V = 20*10^-6 / 20*10^-6 =1 V

In parallel combination voltage drop is same across both capacitors . So V across 5 uF capacitor = 1 V

Ans ( c) = 1.0 V

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