Question

The circuit shown in the figure consists of a battery of emf $\mathrm{\epsilon }=10\mathrm{V}$; a capacitor of capacitance $\mathrm{C}=1.0\mu \mathrm{F}$ and three resistors of values ${\mathrm{R}}_1=2\Omega$, ${\mathrm{R}}_2=2\Omega$ and ${\mathrm{R}}_3=1\Omega$. Initially, the capacitor is completely uncharged and the switch $\mathrm{S}$ is open. The switch $\mathrm{S}$ is closed at $\mathrm{t}=0$

• A. The current through resistor R₃ at the moment the switch closed is zero.
• B. The current through resistor R₃ a long time after the switch closed is 5A
• C. The ratio of current through R₁ and R₂ is always constant.
• D. The maximum charge on the capacitor during the operation is 5 µC

Answer 1

The correct option is D

The maximum charge on the capacitor during the operation is 5 µC

Step 1. Given data

${\mathrm{R}}_1=2\Omega$

${\mathrm{R}}_2=2\Omega$

${\mathrm{R}}_3=1\Omega$

$\mathrm{\epsilon }=10\mathrm{V}$

$\mathrm{C}=1.0\mu \mathrm{F}$

$\mathrm{t}=0$

We have to find the maximum charge on capacitor.

Step 2. Concept to be used

The capacitance of a capacitor is defined as the charge stored per unit potential difference change.

$\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$

Here, $\mathrm{C}$ capacitance of capacitor which is measured in farads $\mathrm{F}$, $\mathrm{Q}$ is the charge stored which are measured in coulombs $\left(\mathrm{C}\right)$ and voltage across capacitor which are measured in volts $\left(\mathrm{V}\right)$.

The Kirchhoff's voltage law states that the algebraic sum of all voltages is equal to zero around a close path.

So,

$\mathrm{\epsilon }-\mathrm{IR}-\frac{\mathrm{Q}}{\mathrm{C}}=0$

Here, $\mathrm{I}$ is current and $\mathrm{R}$ is resistance.

Step 3. Find the current through third resistor.

At $\mathrm{t}=\infty$, the equivalent circuit is,

So,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{\epsilon }}{{\mathrm{R}}_3+{\displaystyle \frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}}} \\ \Rightarrow \mathrm{I}=\frac{10}{1+{\displaystyle \frac{2\times 2}{2+2}}} \\ \Rightarrow \mathrm{I}=\frac{10}{1+{\displaystyle 1}} \\ \Rightarrow \mathrm{I}=5A \end{gathered}$$

Also,

${\mathrm{I}}_{{\mathrm{R}}_1}{\mathrm{R}}_1={\mathrm{I}}_{{\mathrm{R}}_2}{\mathrm{R}}_2$

Or

$\frac{{\mathrm{I}}_{{\mathrm{R}}_1}}{{\mathrm{I}}_{{\mathrm{R}}_2}}=\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}$

So,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{AB}}={\mathrm{IR}}_3 \\ \Rightarrow {\mathrm{V}}_{\mathrm{AB}}=5\times 1 \\ \Rightarrow {\mathrm{V}}_{\mathrm{AB}}=5\mathrm{V} \end{gathered}$$

Step 4. Find the maximum charge on capacitor

The maximum charge on capacitor is,

$$\begin{gathered} {\mathrm{Q}}_{\max }={\mathrm{CV}}_{\mathrm{AB}} \\ \Rightarrow {\mathrm{Q}}_{\max }=1\mu F\times 5 \\ \Rightarrow {\mathrm{Q}}_{\max }=5\mu F \end{gathered}$$

Hence, option $\mathrm{D}$ is correct answer.