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In the circuit shown in figure, if the point $R$ is earthed and point $P$ is given a potential of $+1800 \mathrm{~V}$, then charges on $C_2$ and $C_3$ are respectively
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Verified Answer
The correct answer is:
$2.4 \times 10^{-3} \mathrm{C} ; 1.2 \times 10^{-3} \mathrm{C}$
$\begin{gathered}C_{\mathrm{eq}} \text { of system }=\left(C_2 \text { parallel } C_3\right) \text { series } C_1 \\ =1 /\left(\frac{1}{3}+\frac{1}{(4+2)}\right)=2 \mu \mathrm{F}\end{gathered}$
So, charge taken from source
$$
=q_{\mathrm{eq}}=C_{\mathrm{eq}} \Delta V=1800 \times 2 \times 10^{-6} \mathrm{C}=3600 \mu \mathrm{C}
$$
Potential droop across
$$
C_1=\frac{q_{C_1}}{C_{C_1}}=\frac{3600 \times 10^{-6}}{3 \times 10^{-6}}=1200 \mathrm{~V}
$$
So, potential drop across combination of $4 \mu \mathrm{F}$ and 2 $\mu \mathrm{F}$ capacitors
$$
=1800-1200=600 \mathrm{~V}
$$
Hence,
$$
\begin{aligned}
q_2 & =C_2 V_{Q R}=4 \times 10^{-6} \times 600=2.4 \times 10^{-3} \mathrm{C} \\
\text { and } \quad q_3 & =C_3 V_{Q R}=2 \times 10^{-6} \times 600=1.2 \times 10^{-3} \mathrm{C}
\end{aligned}
$$
So, charge taken from source
$$
=q_{\mathrm{eq}}=C_{\mathrm{eq}} \Delta V=1800 \times 2 \times 10^{-6} \mathrm{C}=3600 \mu \mathrm{C}
$$
Potential droop across
$$
C_1=\frac{q_{C_1}}{C_{C_1}}=\frac{3600 \times 10^{-6}}{3 \times 10^{-6}}=1200 \mathrm{~V}
$$
So, potential drop across combination of $4 \mu \mathrm{F}$ and 2 $\mu \mathrm{F}$ capacitors
$$
=1800-1200=600 \mathrm{~V}
$$
Hence,
$$
\begin{aligned}
q_2 & =C_2 V_{Q R}=4 \times 10^{-6} \times 600=2.4 \times 10^{-3} \mathrm{C} \\
\text { and } \quad q_3 & =C_3 V_{Q R}=2 \times 10^{-6} \times 600=1.2 \times 10^{-3} \mathrm{C}
\end{aligned}
$$
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