In the circuit shown in figure $R_{1} = R_{2} = 6 R_{3} = 300 M Ω$ $C = 0.01 μ F$ and E = 10V. The switch is closed at t=0, find
(A) Charge on capacitor as a function of time.
(B) Energy of the capacitor at t=20 s.

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Solution

$R_{1} = R_{2} = 6 R_{3} = 300 m Ω$
$\Rightarrow R_{1} = 300 m Ω R_{2} = 300 m Ω$
$R_{3} = \frac{300}{6} = 50 m Ω$
Applying kirchoffs loop law in $1$ circuit
$E - (i_{1} + i_{2}) R_{1} - i_{1} R_{2} = 0$
$\Rightarrow E - i_{1} (R_{1} + R_{2}) - i_{2} R_{1} = 0$ ..... $(i)$
Applying low law in $2$ circuit
$i_{1} R_{2} - i_{2} R_{3} - \frac{q}{C} = 0$ ..... $(i i)$
Now,
from $(i)$
$i_{1} = \frac{E - i_{2} R_{1}}{R_{1} + R_{2}}$
Putting on $(i i)$
$[\frac{E - i_{2} R_{1}}{R_{1} + R_{2}}] R_{2} - i_{2} R_{3} - \frac{q}{c} = 0$
$\frac{E R_{2}}{R_{1} + R_{2}} - i_{2} [\frac{R_{1} R_{2}}{R_{1} + R_{2}} + R_{3}] - \frac{q}{c} = 0$
Let $\frac{R_{1} R_{2}}{R_{1} + R_{2}} = R . \Rightarrow \frac{R_{2}}{R_{1} + R_{2}} = \frac{R}{R_{1}}$
$\frac{E R}{R_{1}} - i_{2} [R + R_{3}] - \frac{q}{c} = 0$
$\Rightarrow i_{2} (R + R_{3}) = \frac{E R C - q R}{R_{1} C}$
$\Rightarrow i_{2} = \frac{E R C - q R_{1}}{R_{1} C (R + R_{3})}$
$∵ i_{2} = \frac{d q}{d t}$
$\Rightarrow \frac{d q}{d t} = \frac{E R C - q R_{1}}{R_{1} C (R + R_{3})}$
$\Rightarrow \int_{0}^{q} \frac{d q}{E R C - q R_{1}} = \int_{0}^{t} \frac{d t}{R_{1} C (R + R_{3})}$
$\Rightarrow q = \frac{E R C}{R_{1}} [1 - e^{- 7 / c} (R + R_{3})]$
Putting value $R$
$q = \frac{E R_{2} C}{R_{1} + R_{2}} [1 - e^{- t / C} (\frac{R_{1} R_{2}}{R_{1} + R_{2}} + R_{3})]$
Putting the value of $R_{1}, R_{2}$ and $R_{9}$
$q = 5 \times 10^{- 8} (1 - e^{- 7 / 2})$ column
$q = 0.05 (1 - e^{- 7 / 2}) μ c$
at $t = 205$
$q = 0.05 (1 - e^{- 20 / 2}) μ C$
$\Rightarrow q ≃ 0.05 μ C$
we know that energy of capacitor $c$
$E = \frac{1}{2} \frac{q^{2}}{C}$
$E = \frac{1}{2} (\frac{0.05 \times 0.05}{0.01}) μ J$
$E = \frac{0.25}{2} = 0.125 μ J$
$\Rightarrow E = 0.125 \times 10^{- 6} J$

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In the circuit shown in figure R1=R2=6R3=300MΩ C=0.01μF and E = 10V. The switch is closed at t=0, find (A) Charge on capacitor as a function of time.(B) Energy of the capacitor at t=20 s.

In the circuit shown in figure $R_{1} = R_{2} = 6 R_{3} = 300 M Ω$ $C = 0.01 μ F$ and E = 10V. The switch is closed at t=0, find
(A) Charge on capacitor as a function of time.
(B) Energy of the capacitor at t=20 s.