Question

Figure shows two capacitors of capacitance $2\mu F$ and $4\mu F$ and a cell of $90V$. The switch $k$ is such that when it is in position $1$ , the circuit $ABCD$ is closed and when it is in position $2$, the circuit $BCEF$ is closed. The resistance of both the circuits is negligible so that the capacitor gets fully charged instantly. Initially, the switch is in position $1$. Then it is turned in position $2$. This makes one cycle. It is then again turned in position $1$ and then in position $2$. Now two cycles are completed. Find the charge (in $\times 10^2\mu C$ ) on the capacitor of capacitance $4\mu F$ after two cycles.

Answer 1

Answer: 2

Initially when key $k$ is in position $1$ , charge on $C_1$ is
$Q_1=180\mu C$

As the key is pushed to position $2,C_1$ and $C_2$ are in parallel
$\therefore$ Charge on $C_2$ is $q_2=C_2\left(\frac{\text{ net charge on }{C}_1\&C_2}{C_1+C_2}\right)$
$=4\times \frac{180}{6}=120\mu C$
Again when key $k$ is pushed position $1$ , charge on $C_1$ is
$Q_1=180\mu C$
As the key is again pushed to position $2$, charge on
$\therefore$ Charge on $C_2$ is $q_2=C_2\left(\frac{\text{ net charge on }{C}_1\&C_2}{C_1+C_2}\right)$
$=4\times \frac{180+120}{6}=200\mu C$