Question

The circuit shown in the figure consists of a charged capacitor of capacity $3\mu F$ and a charge of $30\mu C.$ At the time $t=0$when the key is closed, the value of current flowing through the$5M\Omega$ resistor is$‘x’\mu A$. The value of $‘x’$ to the nearest integer is _______.

Answer 1

Step 1: Given data:

The capacity of a charged capacitor $C=3\mu F=3\times {10}^{-6}F$

Resistance $R=5\mathrm{M\Omega }=5\times {10}^6\Omega$

Value of charge $q=30\mu C=30\times {10}^{-6}C$

The closing time $\mathrm{t}=0$

Current flowing through the circuit$‘x’\mu A$

Step 2: Formula used:

The relation between the charge on the capacitor and voltage across the capacitor-

$q=CV$

Ohm's law-

$V=IR$

Step 3: Calculating the value of voltage $V$

The relation between the charge on the capacitor and voltage-

$$\begin{gathered} q=CV \\ \left(30\times {10}^{-6}\right)=\left(3\times {10}^{-6}\right)\times V \\ V=10V \end{gathered}$$

Step 4: Calculate the value of current

From ohm's law-

When the key is closed,

$$\begin{gathered} V=IR \\ 10=\left(x\times {10}^{-6}\right)\times \left(5\times {10}^6\right) \\ 10=5x \\ x=2 \end{gathered}$$

Therefore the value of $x$ is $2$.