Animated Solution for Physics - Current Electricity: Find the emf ($V$) and internal resistance ($r$) of a single battery which is equivalent to a parallel combination of two batteries of emfs $V_1$ and $V_2$ and internal resistances $r_1$ and $r_2$ respectively, with polarities as shown in figure

Imagine you are standing at terminal A, looking into a complex electrical network. You see two parallel branches stretching out towards terminal B. Our goal is to replace this entire dual-battery setup with a single, elegant equivalent battery.

To do this, we must first pay very close attention to the polarities of the batteries. In the top branch, we have a resistor $r_2$ and a battery $V_2$. Notice that the positive terminal of $V_2$ is facing towards terminal B. In the bottom branch, we have a resistor $r_1$ and a battery $V_1$, but here, the positive terminal is facing towards terminal A.

What does this mean physically? If you trace a closed loop through these two branches, you will find that both batteries are actually pushing current in the exact same counter-clockwise direction! They are not fighting each other; they are working together within that specific inner loop.

Now, let's assume that terminals A and B are completely open. No current is flowing out into any external circuit. However, because we have a closed loop formed by the two branches, a circulating current, let's call it $i$, will flow continuously inside.

Since both batteries aid each other in this loop, the net Electromotive Force (EMF) driving this circulating current is simply the sum of their individual EMFs: $V_1+V_2$. The total resistance impeding this flow is the sum of the internal resistances: $r_1+r_2$.

By applying Ohm's Law to this closed loop, we can easily determine the magnitude of this circulating current:

The fundamental definition of a battery's EMF is the potential difference across its terminals when it is in an open circuit (drawing zero external current). Therefore, to find the equivalent EMF ($V_{eq}$) of our entire setup, we just need to calculate the open-circuit potential difference between points A and B, which is $V_A-V_B$.

We can find this by taking a mathematical walk from terminal B to terminal A along the bottom branch. As we move from B to A, we are moving with the direction of our circulating current $i$.

When we cross the resistor $r_1$ in the direction of the current, the potential drops by $i{r}_1$. Then, as we cross the battery $V_1$ from its negative to its positive terminal, the potential increases by $V_1$.

Putting this together, we get:

Now, we substitute the expression for $i$ that we found earlier:

Let's perform some algebraic magic. We take a common denominator and expand the numerator:

Notice how beautifully the $V_1{r}_1$ terms cancel each other out! We are left with a pristine, elegant formula for the equivalent EMF:

Finding the equivalent internal resistance ($r_{eq}$) is much simpler. We rely on a powerful concept from circuit theory (often associated with Thevenin's Theorem): to find the equivalent resistance of a network containing ideal voltage sources, we simply turn those sources off.

Turning off an ideal voltage source means setting its voltage to zero, which effectively replaces it with a short circuit (a plain wire).

If we short-circuit both $V_1$ and $V_2$, and look into the circuit from terminals A and B, what do we see? We see that the current would split between the top branch and the bottom branch. This means the resistors $r_1$ and $r_2$ are connected perfectly in parallel!

The equivalent resistance of two parallel resistors is the product of their resistances divided by their sum:

We have successfully conquered the problem! The entire complex arrangement of two parallel batteries can be seamlessly replaced by a single equivalent battery.

The Equivalent EMF is:

The Equivalent Internal Resistance is:

This is a highly versatile result. Whenever you encounter parallel batteries in complex JEE circuit problems, you can use this derivation to instantly simplify the network and save precious time!