Current Electricity 1 Question 6
6. Determine the charge on the capacitor in the following circuit
(2019 Main, 9 April I)
(a) $2 \mu C$
(b) $200 \mu C$
(c) $10 \mu C$
(d) $60 \mu C$
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Solution:
- Given circuit is
To find charge on capacitor, we need to determine voltage across it. In steady state, capacitor will acts as open circuit and circuit can be reduced as
In series, $R _{\text {eq }}=2 \Omega+10 \Omega=12 \Omega$
$\Downarrow$
In parallel, $R _{eq}=\frac{4 \times 12}{4+12}=3 \Omega$
In series, $R _{eq}=6 \Omega+3 \Omega=9 \Omega$
So, current in steady state, $I=\frac{V}{R}=\frac{72}{9}=8 A$
Now, by using current division, at point $P$, current in $6 \Omega$ branch is
Current in $4 \Omega$ branch is,
$$ \dot{I} _2=\frac{V _P-0}{4}=\frac{24-0}{4}=6 \Omega $$
So, current in $2 \Omega$ resistance is
$$ \begin{aligned} I _1 & =8-I _2 \quad\left[\because I=I _1+I _2\right] \\ & =8-6=2 A \end{aligned} $$
$\therefore$ Potential difference across $10 \Omega$ resistor is
$$ V _{Q G}=2 A \times 10 \Omega=20 V $$
Same potential difference will be applicable over the capacitor (parallel combination).
So, charge stored in the capacitor will be
$$ \begin{aligned} & Q=C V=10 \times 10^{-6} \times 20 \\ \Rightarrow \quad & Q=2 \times 10^{-4} C=200 \mu C \end{aligned} $$
