Animated Solution for Physics - Current Electricity: In the given circuit, $E_1=3E_2=2E_3=6\text{ V}$ and $R_1=2R_4=6\Omega$ $R_3=2R_2=4\Omega$, $C=5\mu \text{F}$. Find the current in $R_3$ and the energy stored in the capacitor.

When dealing with DC circuits containing capacitors, the first rule of thumb is to identify the steady state. In a DC steady state, a capacitor acts as an open circuit, meaning it completely blocks any continuous flow of current through its branch.

Looking at our circuit, the capacitor $C$ is connected in series with the top branch containing $R_1$. Because no current can flow through the capacitor, the current through $R_1$ must also be zero. With zero current, there is zero potential drop across $R_1$, which elegantly tells us that the nodes on either side of it are at the exact same potential ($V_a=V_b$).

At first glance, the circuit looks like a complex multi-loop network. However, if we carefully trace the vertical connecting wires, a beautiful simplification emerges. The vertical wire in the middle connects node $b$ directly to node $e$. Similarly, the vertical wire on the right connects node $c$ directly to node $f$.

This topological arrangement effectively places the battery $E_1$ directly in parallel with the resistor $R_3$! Because components in parallel share the same potential difference, the voltage across $R_3$ is forced to be exactly equal to $E_1$, which is $6\text{ V}$.

Using Ohm's law, the current through $R_3$ is simply:

Since the positive terminal of $E_1$ is on the right, node $f$ is at a higher potential than node $e$, driving the $1.5\text{ A}$ current from right to left.

To find the energy stored in the capacitor, we need the potential difference across its plates, which are connected to nodes $a$ and $d$. We already know $V_a=V_b$. Let's use nodal analysis to find $V_d$.

We can arbitrarily set the potential of node $e$ to $0\text{ V}$. This immediately gives us $V_b=0\text{ V}$ and $V_a=0\text{ V}$. Since node $f$ is $6\text{ V}$ higher than node $e$, $V_f=6\text{ V}$. The right vertical wire ensures that $V_c=V_f=V_h=6\text{ V}$.

Now, let's define the potential of the left vertical wire (nodes $d$ and $g$) as $V_L$. We apply Kirchhoff's Current Law (KCL) at this super-node $L$. The sum of currents leaving node $L$ through the middle and bottom branches must be zero:

Substituting the known values:

Multiplying the entire equation by 6 to clear the denominators:

We have successfully found the potentials on both sides of the capacitor! Node $a$ is at $0\text{ V}$ and node $d$ is at $2.4\text{ V}$. The potential difference across the capacitor is simply $V_{da}=2.4\text{ V}$.

The energy stored in the capacitor is given by the standard formula:

By leveraging nodal analysis and carefully tracing the circuit's topology, a seemingly daunting network was dismantled into a few elegant algebraic steps.