Question

A $10\mu F$ capacitor is fully charged to a potential difference of $50V$. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference becomes $20V$. The capacitance of the second capacitor.

• A. $15\mu F$
• B. $20\mu F$
• C. $10\mu F$
• D. $30\mu F$

Answer 1

The correct option is A

$15\mu F$

Step 1. Given data

Capacitance of the charged capacitor, $C_1=10\mu F$

Potential difference before removing the source voltage, $V_1=50V$

Step 2. Finding the capacitance of uncharged capacitor, $C_2$

We know that,

$Q=C_1{V}_1$ [Where, $Q$ is the charge on the capacitor]

$Q=10\times 50$

$Q=500\mu C$

Now, using the formula of equivalent capacitance, $C_{eq}$

$C_{eq}=C_1+C_2$

$C_{eq}=10+C_2$

Now again using the formula of charge, $Q$

$Q=C_{eq}\times V$ [Where, $V=20$ is the final potential difference as given in question.]

$500=\left(10+C_2\right)\times 20$

$C_2=\frac{500}{20}-10$

$C_2=25-10$

$C_2=15\mu F$

Hence, the correct option is $\left(A\right)$.