Question

1. Two capacitors C; (12uf) and C2 (24uf) are in series connected across a 360 volts d.c.
supply. Calculate the charges on C, and C, respectively.
A. 2880 x 10-6 C, 2880 x 10-C
B. 7170 x 10-6 C, 8140 x 10-6C
C. 4770 x 10-6C, 4770 x 10-6C
D. 9090 x 10-6 C, 8880 x 10-6 C
E. 5810 x 10-6 C, 6610 x 10-6 C
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Answer 1

12th/Physics

Capacitor and capacitance

Answer :

Two capacitora of 12uF(C₁) and 24uF(C₂) are connected im series with a battery of 360V.

We have to find charge on each capacitor$.$

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◈ First of all, we need to find equivalent capacitance of the series connection.

⇒ 1/C = 1/C₁ + 1/C₂

⇒ 1/C = 1/12 + 1/24

⇒ 1/C = 2/24 + 1/24

⇒ C = 24/3

⇒ C = 8 uF

◈ Net charge flow in circuit :

➝ C = Q/V

➝ 8 = Q/360

➝ Q = 8 × 360

➝ Q = 2880 uC

➝ $\bf{Q=2880\times 10^{-6}\:C}$

We know that, charge remains same on each capacitor in series connection. Hence, charge on each caoacitor will be equal to the net charge flow in circuit!

Option - (1) is the correct answer.

Cheers!