Question

The circuit shown in the figure consists of a battery of EMF $\epsilon =10V$; a capacitor of capacitance $C=1.0\mu F$ and three resistors of values $R_1=2\Omega$, $R_2=2\Omega$ and $R_3=1\Omega$. Initially, the capacitor is completely uncharged and the switch $S$ is open. The switch $S$ is closed at $t=0$. Which of the given options is correct?

• A. The current through resistor $R_3$ at the moment the switch closed is zero.
• B. The current through resistor $R_3$ a long time after the switch closed is $5A$.
• C. The ratio of current through $R_1$ and $R_2$ is always constant.
• D. The maximum charge on the capacitor during the operation is $5\mu C$.

Answer 1

The correct option is D

The maximum charge on the capacitor during the operation is $5\mu C$.

Step 1: Given Data,

$\epsilon =10V$,$\epsilon$is the EMF induced

$R_1=2\Omega$

$R_2=2\Omega$

$R_3=1\Omega$

Step 2: Formula used,

If Two resistors $R_1$ and $R_2$ are connected in series,

Equivalent resistance, $R={R^{}}_1+{R^{}}_2$

If Two resistors $R_1$ and $R_2$ are connected in parallel,

Equivalent resistance,$R=\frac{R_1{R}_2}{R_1+R_2}$

Charge in a capacitor, $Q=CV$

$C$is the capacitance

According to Ohm's law, $V=IR$

$R$ is resistance,$I$ is current and $V$ is the voltage

Step 3. Explanation,

In case of option A,

1. When the switch is just closed, $t=0$, $t$ is the time.
2. So, the capacitor will start charging through the resistors $R_1$ and $R_2$, $R_1$ is the resistor of $2\Omega$, $R_2$ is the resistor of $2\Omega$
3. And there will be no current passing through resistor $R_3$, where $R_3$ is the resistor of $1\Omega$ in series
4. Thus, this option is correct.

In case of option B,

So, the equivalent resistance of the three resistors $R_1=2\Omega$, $R_2=2\Omega$ and $R_3=1\Omega$ will be $R=\frac{R_1{R}_2}{R_1+R_2}+R_3$, since $R_1$ and $R_2$ are in parallel and $R_3$ is connected in series with them.

Current is given by $I=\frac{\epsilon }{R}$, where $\epsilon =10V$ is EMF and $R$ is resistance.

So, find the current by substituting the values.

$\begin{array}{rcl}I& =& \frac{\epsilon }{R}\\ & =& \frac{\epsilon }{{\displaystyle \frac{R_1{R}_2}{R_1+R_2}+R_3}}\\ & =& \frac{10V}{\frac{2\Omega \times 2\Omega }{2\Omega +2\Omega }+1\Omega }\\ & =& \frac{10V}{{\displaystyle 2\Omega }}\\ & =& 5A\end{array}$

Thus, option B is correct.

In case of option C,

1. According to Kirchhoff's second law, “The algebraic sum of potential drops in a circuit is zero.”
2. That is $\sum _{}^{}V=V_1+V_2=0$, which can be further written as $V_1=-V_2$.
3. Since $V=IR$, $I_1{R}_1=-I_2{R}_2$. ${\mathrm{I}}_{1,}{\mathrm{I}}_2\mathrm{are}\mathrm{the}\mathrm{current}\mathrm{in}\mathrm{resistor}\mathrm{arm}{\mathrm{R}}_1,{\mathrm{R}}_2$
4. So, it can be further written as $\frac{I_1}{I_2}=\frac{R_2}{R_1}$. Since the ratio of the current will be constant, $\frac{R_2}{R_1}=cons\tan t$.
5. Thus, this option is correct.

In case of option D,

When two resistors $R_1$ and $R_2$ are connected in parallel their equivalent resistance is $R=\frac{R_1{R}_2}{R_1+R_2}$.

So, the equivalent resistance of the two resistors $R_1=2\Omega$ and $R_2=2\Omega$ will be $R=\frac{R_1{R}_2}{R_1+R_2}$, since $R_1$ and $R_2$ are in parallel.

The voltage is given by $V=IR$, where $I=5A$ is current and $R=\frac{R_1{R}_2}{R_1+R_2}$ is resistance.

The charge in a capacitor is given by $Q=CV$, where $C=1\mu F$ is capacitance.

So, find the charge by substituting the values.

$\begin{array}{rcl}Q& =& CV\\ & =& C\left(IR\right)\\ & =& C\times I\times \frac{R_1{R}_2}{R_1+R_2}\\ & =& 1\mu F\times 5A\times \frac{2\Omega \times 2\Omega }{2\Omega +2\Omega }\\ & =& 1\mu F\times 5A\times 1\Omega \\ & =& 5\mu C\end{array}$

Thus, this option is correct.

Hence, options A, B, C and D are correct.