Two batteries with emf 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The i

Q: Why can't we just add the voltages of the batteries?

When batteries are in parallel, their voltages don't simply add up like they do in series. Instead, they act together to maintain a common potential difference, and the battery with the higher EMF will even try to push current back into the battery with the lower EMF!

Q: What if the batteries were connected with opposite polarities?

If one battery was reversed, you would simply use a negative sign for its EMF in the Millman's theorem formula. For example, $E_1/r_1 - E_2/r_2$.

Q: Why is Millman's theorem better than Kirchhoff's laws here?

Kirchhoff's laws will absolutely give you the right answer (as shown in the alternate solution), but they require setting up and solving multiple simultaneous linear equations. Millman's theorem condenses all that algebra into two simple, direct formulas, saving you precious time in exams like JEE.

Q: Does the $10\,\Omega$ resistor affect the equivalent EMF of the batteries?

No! The equivalent EMF and equivalent internal resistance are properties of the battery combination itself, completely independent of whatever external load you connect them to.

Q: How do we know the current direction in the load?

Since both batteries have their positive terminals connected to the same side (let's say the left node), the equivalent battery will also have its positive terminal on the left. Therefore, the current will flow from left to right through the external $10\,\Omega$ load resistor.

Question

Two batteries with emf $12	ext{ V}$ and $13	ext{ V}$ are connected in parallel across a load resistor of $10\,\Omega$. The internal resistances of the two batteries are $1\,\Omega$ and $2\,\Omega$, respectively. The voltage across the load lies between

Options:

(A) $11.6	ext{ V}$ and $11.7	ext{ V}$

(B) $11.5	ext{ V}$ and $11.6	ext{ V}$

(C) $11.4	ext{ V}$ and $11.5	ext{ V}$

(D) $11.7	ext{ V}$ and $11.8	ext{ V}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an electrical engineer tasked with analyzing a power supply system. You have two batteries connected in parallel, driving current through a single $10\,\Omega$ load resistor.

The first battery is a $12	ext{ V}$ source with an internal resistance of $1\,\Omega$. The second is a slightly stronger $13	ext{ V}$ source with an internal resistance of $2\,\Omega$.

When batteries are connected in parallel, they don't just independently push current to the load. They interact with each other. The battery with the higher EMF will actually try to push current **back** into the weaker battery!

To find the voltage across the load, we could use Kirchhoff's Voltage and Current Laws, setting up multiple loops and solving simultaneous equations. But there is a much more elegant and powerful tool at our disposal.

### The Master Equation: Millman's Theorem

Instead of wrestling with complex algebra, we can use **Millman's Theorem** (also known as the parallel combination of cells formula).

This theorem allows us to replace any number of parallel voltage sources with a **single equivalent battery**. This equivalent battery will have an EMF ($E_{eq}$) and an internal resistance ($r_{eq}$) given by:

$$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$$

$$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$$

This is a massive shortcut that condenses the entire parallel network into one simple series circuit!

### Calculating the Equivalent Battery

Let's plug our values into the master equation. For the equivalent EMF, we substitute $E_1 = 12	ext{ V}$, $r_1 = 1\,\Omega$, $E_2 = 13	ext{ V}$, and $r_2 = 2\,\Omega$:

$$E_{eq} = \frac{\frac{12}{1} + \frac{13}{2}}{\frac{1}{1} + \frac{1}{2}}$$

Calculating the numerator, $12 + 6.5$ gives us $18.5$. The denominator is $1 + 0.5$, which is $1.5$.

$$E_{eq} = \frac{18.5}{1.5} = \frac{37}{3}	ext{ V}$$

Now, let's find the equivalent internal resistance. The reciprocal of $r_{eq}$ is the sum of the reciprocals of the individual internal resistances:

$$\frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2}$$

Taking the reciprocal of both sides, we get:

$$r_{eq} = \frac{2}{3}\,\Omega$$

We have successfully reduced our complex two-battery system into a single battery of $\frac{37}{3}	ext{ V}$ with an internal resistance of $\frac{2}{3}\,\Omega$.

### Final Calculation: Voltage Across the Load

Now, visualize the simplified circuit: our equivalent battery is connected directly in series with the $10\,\Omega$ load resistor.

To find the voltage $V$ across the load, we can use Ohm's law. The total current $I$ in the circuit is the equivalent EMF divided by the total resistance ($R + r_{eq}$). The voltage across the load is simply this current multiplied by the load resistance $R$:

$$V = I R = \left(\frac{E_{eq}}{R + r_{eq}}ight) R$$

Let's substitute our calculated values into this equation:

$$V = \frac{\frac{37}{3}}{10 + \frac{2}{3}} 	imes 10$$

First, let's simplify the denominator. $10 + \frac{2}{3}$ becomes $\frac{32}{3}$.

$$V = \frac{\frac{37}{3}}{\frac{32}{3}} 	imes 10$$

Notice how the $3$ in the denominators beautifully cancel out! We are left with:

$$V = \frac{37}{32} 	imes 10 = \frac{370}{32}$$

Evaluating this fraction gives us our final answer:

$$V = 11.5625	ext{ V}$$

Looking at our options, this value perfectly lies between **$11.5	ext{ V}$ and $11.6	ext{ V}$**. The elegance of Millman's theorem has led us straight to the correct answer without a single simultaneous equation!

Animated Solution for Physics - Current Electricity: Two batteries with emf $12 V$ and $13 V$ are connected in parallel across a load resistor of $10 Ω$ . The internal resistances of the two batteries are $1 Ω$ and $2 Ω$ , respectively. The voltage across the load lies between

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The Sigma Insight: Cells, EMF, and Internal Resistance

Analyzing the Setup

The Master Equation

Calculating the Equivalent Battery

Final Calculation

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