91Ó°ÊÓ

Consider the given values of the resistances and emf is:

$$\begin{gathered} \text{ε}=30\text{V} \\ {\text{R}}_1=20k\mathrm{Ω} \\ {R}_2=10k\mathrm{Ω} \end{gathered}$$

According to Ohm’s law, the direct current flowing in a conductor is directly proportional to the potential difference between its ends.

Formulae:

$$\begin{gathered} I=\frac{V}{R} \\ {R}_{eq}=R_1+R_2 \end{gathered}$$

Initially,thecapacitor is uncharged, sothevoltage across resistancewould be zero, and the voltage across$R_1$is 30 V.

By Ohm’s law:

Current in resistance$R_1$

$\begin{array}{c}{i}_1=\frac{30}{20×{10}^3}\\ =1.5×{10}^{−3}A\end{array}$

The current in resistance$R_1$ is$i_1=1.5×{10}^{-3}A$

Current in resistance $R_2$is $i_2=0$this is because the voltage drop across this resistance is 0.

A long time later, capacitor reduces to zero, so resistance $R_1$and$R_2$will be in series.

Hence,

$\begin{array}{c}{R}_{eq}=R_1+R_2\\ =(20×{10}^3)+(10×{10}^3)\\ =30×{10}^3\mathrm{Î\copyright }\end{array}$

Solve for the value of the current as:

$\begin{array}{c}i=\frac{30}{30×{10}^3}\\ =1.0×{10}^{−3}A\end{array}$

Current in resistance $R_2$after long time later is$i=1.0×{10}^{−3}A$