Question

40. Two capacitors of capacitance 6 uF and 4 uF are connected in parallel with a battery. The charge on 6 uF
capacitor is 24 uC, then find the charge on another capacitor and voltage of battery.
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Answer 1

Answer:

The charge on second capacitor = 16 µC

The voltage of the battery is 4 V.

Explanation:

Two capacitors of capacitance 6 µF and 4 µF are connected in parallel with a battery.

We know that the charge of the capacitor is given by,

q = CV

where

C is the capacitance

V is the voltage

Given that the charge on 6 uF capacitor is 24 µC.

24 µC = 6 µF × V

V = 24/6

V = 4 V

The voltage is same for both the capacitors as that of the battery in parallel connection.

Hence, the voltage of the battery is 4 V.

For the second capacitor :

q = ?

C = 4 µF

V = 4 V

→ q = CV

q = 4 × 4

q = 16 µC

Therefore the charge on the second capacitor of capacitance 4 µF is 16 µC.