Question

Three condensers are connected as shown in series. If the insulated plate of $C_1$ is at $45\text{V}$ and one plate of $C_3$ is earthed, the P.D. between the plates of $C_2$ is

• A. $30\text{V}$
• B. $45\text{V}$
• C. $20\text{V}$
• D. $10\text{V}$

Answer 1

The correct option is D $10\text{V}$


Using the definition of potential difference between A and B,

$V_{\text{A}}-\frac{q}{C_1}-\frac{q}{C_2}-\frac{q}{C_3}=V_{\text{B}}$

Since, $V_{\text{A}}-V_{\text{B}}=45\text{V}$

$45-q[\frac{1}{c_1}+\frac{1}{c_2}+\frac{1}{c_3}]=0$

$45=q[\frac{1}{4}+\frac{1}{7}+\frac{1}{4}]$

$45=q\left[\frac{7+2}{14}\right]$

$q=\frac{45\times 14}{9}=70\mu \text{C}$

So, P.D. between the plates of $C_2$ is, $V_2=\frac{q}{C_2}=\frac{70\times {10}^{-6}}{7\times {10}^{-6}}=10\text{V}$