Question

Three condensers are connected as shown in series. If the insulated plate of $C_1$ is at $45V$ one plate of $C_3$ is earthed, find the p.d between the plate of $C_2$.

• A. $10V$
• B. $20V$
• C. $30V$
• D. $45V$

Answer 1

The correct option is A $10V$

Net capacitance = C

then $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{20}+\frac{1}{30}+\frac{1}{15}$

$\Rightarrow \frac{1}{C}=\frac{3+2+4}{60}=\frac{9}{60}=\frac{3}{20}$

$\Rightarrow C=\frac{20}{3}\mu f$

total charge drawn $=Cv=\frac{20}{3}\times 45\mu C=300\mu C$

as all are in series so every plate shed have same magnitude of charge i.e

$300\mu C$ or every capacitor shed have charge = $300\mu C$

So voltage across $C_2$ will be $\Delta V=\frac{Q}{C_2}=\frac{300\mu C}{30\mu f}$

$\Rightarrow \Delta V=10volt$