Question

Three condenser of capacities 3mF , 6mF, 12mF. are connected in series with a battery .If the charge on 12mF condenser is 24mC the P.D across the battery
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Answer 1

$\mathsf{V_{1} = \frac{24}{12} = 2V} \\ \\ \mathsf{V_{2} = \frac{24}{6} = 4V} \\ \\ \mathsf{V_{3} = \frac{24}{3} = 8V} \\ \\ \boxed{\mathsf { V_{1} + V_{2} + V_{3} = 14V}}$

Answer 2

Answer:

The potential difference across the battery is 14V.

Explanation:

Given three condensers connected in series with a battery.

Let the capacitances be $Q_1=3mF , Q_2=6mF~\&~ Q_3=12mF$

Charge on 12mF condenser, $Q= 24mC$

For a condenser, the maximum amount of charge Q that can acquire is proportional to the voltage V.

$Q=CV$  

where C is the proportionality constant, called capacitance.

And from $Q=CV$ we have $V=\dfrac{Q}{C}$

When the condensers are connected in series, the charge on each adjacent condenser must be the same.

Therefore, when the charge on 12mF condenser is 24mC, then the charge remains same for 3mF and 6mF also.

Hence, the voltage across $C_1$ is $V_1=\dfrac{Q}{C_1}=\dfrac{24}{3}=8V$

Voltage across $C_2$ is $V_2=\dfrac{Q}{C_2}=\dfrac{24}{6}=4V$

Voltage across $C_3$ is $V_3=\dfrac{Q}{C_3}=\dfrac{24}{12}=2V$

Thus, the potential difference across the battery is the sum of voltage drops across capacitors, i.e.,

$V=V_1+V_2+V_3=8+4+2=14V$

Therefore, the potential difference across the battery is 14V.