Animated Solution for Physics - Current Electricity: In a circuit shown in the figure, the capacitor C is initially uncharged and the key K is open. In this condition, a current of 1 A flows through the $1\Omega$ resistor. The key is closed at time $t=t_0$. Which of the following statement(s) is(are) correct? [Given: $e^{-1}=0.36$]

Before the key $K$ is closed, the circuit consists of three active parallel branches. The bottom branch containing the capacitor is completely disconnected. To systematically analyze this multi-loop circuit, nodal analysis is our best friend. Let's assign a reference potential of $0\text{V}$ to the entire left common wire. Consequently, let the potential of the right common wire be $V$.

We are given a crucial piece of information: a current of $1\text{A}$ flows through the $1\Omega$ resistor towards the left. This allows us to immediately determine the potential $V$. Starting from the left node at $0\text{V}$ and traversing the middle branch to the right, we cross the $5\text{V}$ battery from its negative to positive terminal, gaining $5\text{V}$. Next, we cross the $1\Omega$ resistor. Since we are moving against the $1\text{A}$ current, the potential increases by $I\times R=1\text{A}\times 1\Omega =1\text{V}$.

Therefore, the potential at the right node is:

Now that we know the potential difference across all parallel branches is $6\text{V}$, we can easily find the unknowns in the other branches. Let's look at the third branch containing the $3\Omega$ resistor. The potential drops from $6\text{V}$ on the right to $0\text{V}$ on the left. By Ohm's law, the current $I_1$ flowing to the left is: This confirms that Option (B) is correct.

Next, we apply Kirchhoff's Current Law (KCL) at the right node. The total current leaving the node towards the left through the middle and bottom branches is $1\text{A}+2\text{A}=3\text{A}$. This entire current must be supplied by the top branch. So, the current $I_R$ flowing to the right in the top branch is $3\text{A}$.

For the top branch, the potential drops from the $15\text{V}$ battery across the resistor $R$ to reach the $6\text{V}$ node: This confirms that Option (A) is correct.

Enter Thevenin

When the key $K$ is closed at $t=t_0$, the capacitor begins to charge. Analyzing a four-branch circuit with a charging capacitor can be mathematically tedious. However, we can elegantly simplify the entire active network (the top three branches) into a single Thevenin equivalent battery and resistance connected across the capacitor branch.

Using the parallel battery formula, the equivalent resistance $R_{eq}$ is:

The equivalent EMF ${\epsilon }_{eq}$ is:

Notice how beautifully the Thevenin voltage matches the open-circuit nodal voltage we calculated earlier! Our complex circuit is now just a $6\text{V}$ battery in series with a $0.6\Omega$ internal resistance, connected to the $2\mu \text{F}$ capacitor and its $3\Omega$ series resistor.

After a very long time ($t\to \infty$), the capacitor becomes fully charged and acts as an open circuit. No current flows, so there is no voltage drop across the resistors. The voltage across the capacitor equals the Thevenin EMF. The maximum charge is: This confirms that Option (D) is correct.

Finally, let's evaluate the transient current. The total resistance in our simplified charging loop is $R_{total}=R_{eq}+R_c=0.6\Omega +3\Omega =3.6\Omega$. The time constant $\tau$ of the circuit is:

The charging current as a function of time is given by $i(t)=\frac{{\epsilon }_{eq}}{R_{total}}{e}^{-t/\tau }$. We need to find the current at exactly $t=7.2\mu \text{s}$, which is exactly one time constant $\tau$. This confirms that Option (C) is correct.

Every single statement holds true, making this a masterclass problem in circuit reduction and transient analysis.