Question

The capacitors A and B are connected in series with a battery as shown in the figure. When the switch S is closed and the two capacitors get charged fully, then-

• A. The potential difference across the plates of A is $4V$ and across the plates of B is $6V$
• B. The potential difference across the plates of A is $6V$ and across the plates of B is $4V$
• C. The ratio of electric energies stored in A and B is 2:3
• D. The ratio of charges on A and B is 3:2

Answer 1

The correct option is B The potential difference across the plates of A is $6V$ and across the plates of B is $4V$

Given,

Both capacitors are in series

Capacitance of A & B, $C_A=2\mu F\&C_B=3\mu F$

Total Potential is $V_T=(V_A+V_B)=10V$

Let,

Potential difference on capacitor A & B, $V_A,V_B$

In series, Capacitor have equal charge

$Q_A=Q_B$

$C_A{V}_A=C_B{V}_B$

$\frac{V_A}{V_B}=\frac{C_B}{C_A}$

$\frac{V_A+V_B}{V_B}=\frac{C_B+C_A}{C_A}$

$V_B=\frac{C_A(V_A+V_B)}{C_B+C_A}=\frac{2\times 10}{5}=4V$

$V_B=4V$

Similarly, $\frac{V_A}{V_B}=\frac{C_B}{C_A}\Rightarrow V_A=\frac{V_B\times C_B}{C_A}$

$\Rightarrow V_A=\frac{4\times 3}{2}=6V$

Potential difference on capacitor A & B, $V_A=6V,V_B=4V$