Question

In the circuit shown in figure, if $C_1=C_2=2\mu \text{F}$ ,then charges present on $C_1\text{and}{C}_2$ will be

• A. $0,40\text{mC}$
• B. $40\text{mC},0$
• C. $20\text{mC},10\text{mC}$
• D. $30\text{mC},60\text{mC}$

Answer 1

The correct option is B $40\text{mC},0$

In steady state


$I=\frac{V}{R_{\text{eff}}}$

From the given figure,

$I=\frac{120}{3}=40\text{A}$

$V_{\text{A}}-V_{\text{B}}=20\times 1=20\text{V};V_{\text{A}}-V_{\text{E}}=20\times 2=40\text{V}$

$(V_{\text{A}}-V_{\text{E}})-(V_{\text{A}}-V_{\text{B}})=40-20$

$V_{\text{B}}-V_{\text{E}}=20\text{V}$

$\therefore$ Charge on

$Q_1=C_1{V}_{\text{BE}}=2\times 20=40\text{mC}$

$V_{\text{C}}-V_{\text{D}}=3\times 20=60\text{V}$

$V_{\text{F}}-V_{\text{D}}=20\times 3=60\text{V}$

$\therefore V_{\text{C}}-V_{\text{F}}=0\Rightarrow$ charge on $C_2$ is zero.

Hence, option (b) is the correct answer.