Question

A 10uF capacitor is charged to a
potential difference of 50V and is
connected to another uncharged
capacitor in parallel. Now the commo
potential difference becomes 20 volt
The capacitance of second capacitor​

Answer 1

Solution :-

As per the given data ,

Initially ,

• Capacitor ( C1 ) = 10 μ F
• Potential difference ( V ) = 50 V

we know that ,

⇒ C = Q / V

On rearranging ,

⇒ Q = C 1 x V

⇒ Q = 10 x 50

⇒ Q = 500 μC

Now ,

• C 1 is connected to another unknown capacitor ( C 2 ) in parallel combination
• Common potential difference ( V ') = 20 V
• Charge ( Q ) = 500 μC

⇒ C ' = Q / V'

⇒ C 1 + C 2 = 500 / 20

⇒ C 1 + C 2 = 25

⇒ C 2 = 25 - 10

⇒ C 2 = 15 μF

The capacitance of the second capacitor is 15 μF