Question

A battery of emf $V$ volts, resistances $R_1$ and $R_2$, a condenser $C$ and switches $S_1$ and $S_2$ are connected in a circuit as shown. The capacitor $C$ gets fully charged to $V$ volts when

• A. $S_1$ and $S_2$ are both closed
• B. $S_1$ and $S_2$ are both open
• C. $S_1$ closed and $S_2$ is open
• D. $S_2$ is closed and $S_1$ open

Answer 1

The correct option is C $S_1$ closed and $S_2$ is open

$\text{Case 1}$

Both switches are open,

Then one end of capacitor will be isolated from the battery and

hence it won't be charged.

$\text{Case 2}$

Both switches are closed,

There will be current flowing in the circuit as $S_2$ makes it

a closed circuit.

And hence the potential difference between the capacitor is less than V.

So it won't be fully charged to CV.

$\text{Case 3}$

$S_2$ closed $S_1$ open,

Then battery won't be connected to the capacitor and hence

capacitor can't be charged.

$\text{Case 3}$

$S_1$ closed $S_2$ open,

Then there is no current in the circuit abnd hence no potential

drop across the resistance $R_1$ and hence voltage across

capacitor is V. And capacitor is charged fully to CV.

Option $\text{C}$ is correct.