Animated Solution for Physics - Current Electricity: In the circuit shown $A$ and $B$ are two cells of same emf $E$ but different internal resistances $r_1$ and $r_2$ ($r_1>r_2$) respectively. Find the value of $R$ such that the potential difference across the terminals of cell $A$ is zero, a long time after the key $K$ is closed.

Imagine you are tracing the path of current in a complex network of resistors, capacitors, and inductors. The problem states that a long time has passed since the key $K$ was closed. This phrase is a classic trigger in physics, signaling that the circuit has reached a steady state.

In a DC circuit, once the steady state is achieved, the components behave in very predictable ways. An inductor, which initially opposes changes in current, eventually acts as a perfect conducting wire. This means its resistance drops to zero ($R=0$). Conversely, a capacitor, which initially allows current to flow as it charges, becomes fully charged and acts as an open circuit. It blocks any further direct current from flowing through its branch, effectively having infinite resistance ($R=\infty$).

By replacing the inductor with a plain wire and removing the capacitor branch entirely, our complex circuit simplifies significantly. Resolving the remaining parallel and series combinations of the resistors yields the net external resistance connected across our two batteries. For this specific topology, the equivalent external resistance is:

Now, let's focus on the main branch containing the two identical cells, $A$ and $B$. Since they are connected in series with the same polarity, their electromotive forces add up. The total emf is $E+E=2E$. Similarly, their internal resistances add up to $r_1+r_2$.

Using Ohm's law, the total current $i$ flowing through the batteries is the total emf divided by the total resistance of the circuit (which is the sum of the external resistance and the internal resistances):

The problem introduces a fascinating constraint: the potential difference across the terminals of cell $A$ is exactly zero. When a cell is discharging, the terminal potential difference is its emf minus the voltage drop across its internal resistance. Therefore, we can write:

This immediately tells us that the emf $E$ must be perfectly balanced by the internal voltage drop:

We now have two powerful equations. Let's substitute the expression for the total current $i$ into our voltage balance equation:

Notice the elegance here: the emf $E$ appears on both sides of the equation and can be cleanly canceled out. This leaves us with a purely algebraic relationship between the resistances:

Cross-multiplying to clear the fraction gives:

Subtracting $r_1$ and $r_2$ from both sides allows us to isolate the term containing $R$:

Finally, multiplying both sides by $\frac{4}{3}$ yields the exact value of $R$ required to satisfy the condition:

And there we have it! A beautiful interplay of steady-state circuit theory and algebraic manipulation.