Question

Initially switch $S$ is in open state. How much charge flows through switch $S$ when it is closed? Assume steady state condition to be achieved after $S$ closed.

• A. $64\mu \text{C}$
• B. $-34\mu \text{C}$
• C. $-68\mu \text{C}$
• D. $-54\mu \text{C}$

Answer 1

The correct option is D $-54\mu \text{C}$

Initially switch ${}^{\prime }{S}^{\prime }$ is open, then at the steady state the current will not flow through capacitor branch.


Let the charge on both capacitors will be equal to $\left(q\right)$

$q=C_{net}\times V$

$\Rightarrow q=\left(\frac{6\times 3}{6+3}\right)\times (18-0)$

$\therefore q=\frac{18}{9}\times 18=36\mu \text{C}$

When switch ${}^{\prime }{S}^{\prime }$ is closed for long time again at the steady state condition current will not flows through capacitors.



Current through the circuit, $i=\frac{V}{R_{eq}}$

$\Rightarrow i=\frac{18}{6+3}=2\text{A}$

Potential difference across capacitance $C_1$

$V_1=i{R}_{6\Omega }$

$\Rightarrow V_1=2\times 6=12\text{V}$

$\Rightarrow q_1=C_1{V}_1=6\times 12=72\mu \text{C}$

Similarly potential diffreence across capacitor $C_2$

$V_2=i{R}_{3\Omega }$

$\Rightarrow V_2=2\times 3=6\text{V}$

Charge on capacitor $C_2$;

$q_2=C_2{V}_2$

$\Rightarrow q_2=3\times 6=18\mu \text{C}$

$\text{Charge flown through switch}S$
$=\text{(Final charge on plates of 1 and 2)-(Initial charge on plates of 1 and 2)}$

$\Rightarrow q_{flown}=(-72+18)\mu \text{C}-(-36+36)\mu \text{C}$

$\therefore q_{flown}=-54\mu \text{C}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(d\right)$ is correct.