Question

Two capacitors of capacity $C_1$ and $C_2$ are connected as shown in the figure. Now the switch is closed. Calculate the charge on each capacitor.

Answer 1

Total charge Before switch is closed

$c_1{v}_1+c_2{v}_2=(2\times {10}^{-6}\times 200)+(3\times {10}^{-6}\times 400)$

$=(4\times {10}^{-4})+(10+{10}^{-4})=16\times {10}^{-4}C$..........(1)

Let the common potential be ${}^{\prime }{V}^{\prime }$ when the switch is closed

As no change is lost then

$c_{1}v+c_{2}v=c_1{v}_1+c_2{v}_2$

$\Rightarrow (c_1+c_2)v=10\times {16}^{-4}$ From eq (1)

$\Rightarrow (2+3)\times {10}^{}\times v=16\times {10}^{}$

$\Rightarrow v=\frac{1600}{5}=320v$

Now charge on each capacitor

${\theta }_1=c_{1}v=2\times {10}^{-6}\times 320=6.4\times {10}^{-4}c$

${\theta }_2=c_{2}v=2\times {10}^{-6}\times 320=9.6\times {10}^{-4}c$