Question

The circuit shown in the figure consists of a charged capacitor of capacity $3\mu \text{F}$ and a charge of $30\mu \text{C}$. At time $t=0$ when the key is closed, the value of current flowing through the $5\text{M}\Omega$ resistor is $x\mu \text{A}$ . The value of $x$ to the nearest integer is .

Answer 1

After closing the switch, capacitor will start discharging.

We know charge on capacitor is given as, $Q=Q_0{e}^{-t/RC}$

Given, $Q_0=30\times {10}^{-6}C$, $C=3\times {10}^{-6}F$ and $R=5\times {10}^6\Omega$

Current, $I=\left|\frac{dQ}{dt}\right|=\frac{Q_0}{RC}{e}^{-t/RC}$

at $t=0,I=\frac{Q_0}{RC}{e}^{-0/RC}$

$I=\frac{Q_0}{RC}=\frac{30\times {10}^{-6}}{5\times {10}^6\times 3\times {10}^{-6}}=2\times {10}^{-6}\text{A}$

$I=2\mu \text{A}$