Animated Solution for Physics - Current Electricity: In the circuit shown in figure $E$, $F$, $G$ and $H$ are cells of emf $2$, $1$, $3$ and $1\text{ V}$ respectively, and their internal resistances are $2$, $1$, $3$ and $1\Omega$ respectively. Calculate (a) the potential difference between $B$ and $D$ and (b) the potential difference across the terminals of each cells $G$ and $H$.

Welcome to a fascinating journey through the world of multi-loop circuits! In this problem, we are tasked with deciphering the hidden currents and potential differences in a complex network of cells and resistors. This is a classic application of Kirchhoff's Laws, and it will test not only your algebraic skills but also your physical intuition regarding charging and discharging cells.

Imagine you are standing at node $A$ of this circuit. You have multiple paths to choose from, each guarded by a cell with its own electromotive force (EMF) and internal resistance. To make sense of this chaos, we must impose order using Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL).

Let's define two primary loop currents to minimize our variables. We'll assign a current $i_1$ flowing clockwise in the left loop ($A-B-D-A$) and a current $i_2$ flowing clockwise in the right loop ($B-C-D-B$).

By applying KCL at node $B$, the current entering from $A$ is $i_1$. A portion of this, $i_2$, continues towards $C$. The remainder must flow down the diagonal branch towards $D$. Therefore, the current in the central $2\Omega$ resistor is simply $(i_1-i_2)$.

Now, let's take a walk around our loops and tally up the voltage changes using KVL. Remember our sign convention: moving across a resistor in the direction of current results in a voltage drop ($-ir$), and moving across a cell from its negative to its positive terminal results in a voltage gain ($+E$).

Loop 1 ($A-B-D-A$): Starting at $A$ and moving to $B$, we cross cell $E$ (gaining $2\text{ V}$) and its internal resistance (losing $2i_1$). Moving from $B$ to $D$, we cross the $2\Omega$ resistor (losing $2(i_1-i_2)$). Finally, from $D$ back to $A$, we cross cell $F$ (losing $1\text{ V}$ because we go from $+$ to $-$) and its internal resistance (losing $1i_1$).

Setting the sum to zero:

Simplifying this beautifully yields our first master equation:

Loop 2 ($B-C-D-B$): Similarly, tracing the right loop gives us:

Which simplifies to:

We now have a straightforward system of linear equations. Substituting $i_1$ from equation (ii) into equation (i) allows us to solve for the currents:

With the currents unmasked, the rest of the problem unravels easily.

(a) Potential Difference between $B$ and $D$: The current flowing from $B$ to $D$ is $(i_1-i_2)=\frac{5}{13}-\frac{6}{13}=-\frac{1}{13}\text{ A}$. The negative sign simply means the actual current flows from $D$ to $B$. The potential difference is the current multiplied by the resistance:

(b) Terminal Voltages of Cells $G$ and $H$: This is where physics intuition shines. We must determine if a cell is acting as a source (discharging) or a load (charging).

For Cell $G$, the current $i_2$ flows from $C$ to $D$, leaving the positive terminal. The cell is discharging.

For Cell $H$, the current $i_2$ flows from $B$ to $C$, entering the positive terminal. The cell is being charged by the rest of the circuit!

And there you have it! By methodically applying fundamental laws and interpreting the physical meaning of our mathematical results, we've completely decoded this intricate circuit.