A cell E1 of emf 6V and internal resistance 2 ohm is connected with another cell E2 of emf 4V and in

Q: Why do we subtract the EMFs instead of adding them?

Look at how the cells are connected. The negative terminal of the 6V cell faces the negative terminal of the 4V cell. Because they are pushing current in opposite directions, they oppose each other, so we subtract their EMFs to find the net driving force.

Q: How do we know the direction of the current?

The cell with the larger EMF dominates the circuit. Here, the 6V cell overpowers the 4V cell. Since the 6V cell pushes current out of its positive terminal (to the left), the current flows clockwise around the loop.

Q: Why is the formula for potential difference $V = E + Ir$ and not $V = E - Ir$?

Because the current is flowing *into* the positive terminal of the 4V cell, it is being charged by the stronger 6V cell. When a cell is charging, its terminal voltage is greater than its EMF, hence the plus sign.

Q: What would happen if the cells were connected positive-to-negative?

If they were connected positive-to-negative, they would aid each other. The net EMF would be $6V + 4V = 10V$, and the current would be much larger. The 4V cell would then be discharging, and its terminal voltage would be $V = E - Ir$.

Q: Can we use Kirchhoff's Voltage Law (KVL) instead?

Absolutely! If you traverse the loop and apply KVL, you will arrive at the exact same current. KVL is the fundamental principle behind the net EMF formula we used.

Question

A cell $E_1$ of emf 6V and internal resistance $2\Omega$ is connected with another cell $E_2$ of emf 4V and internal resistance $8\Omega$ (as shown in the figure). The potential difference across points X and Y is

Options:

(A) 2.0 V

(B) 3.6 V

(C) 5.6 V

(D) 10.0 V

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are tracing the path of a single closed loop circuit. We have two cells connected in this loop: $E_1$ with an EMF of $6	ext{ V}$ and an internal resistance of $2\,\Omega$, and $E_2$ with an EMF of $4	ext{ V}$ and an internal resistance of $8\,\Omega$.

The most critical step is to look closely at their polarities. The negative terminal of the $6	ext{ V}$ cell faces the negative terminal of the $4	ext{ V}$ cell, meeting at point X. Because they are pushing current in opposite directions, they are in direct opposition to each other.

### The Master Equation

Since the cells are opposing, the net electromotive force (EMF) driving the circuit will be the difference between their individual EMFs. The total resistance of the circuit is simply the sum of their internal resistances, as they are connected in series within the loop.

We can use Ohm's law for the entire loop to find the current:

$$I = \frac{\varepsilon_{net}}{R_{eq}} = \frac{E_1 - E_2}{r_1 + r_2}$$

### Calculating the Current

Let's substitute the given values into our master equation. The $6	ext{ V}$ cell is stronger than the $4	ext{ V}$ cell, so the net EMF is $6 - 4 = 2	ext{ V}$. The total resistance is $2 + 8 = 10\,\Omega$.

$$I = \frac{6 - 4}{2 + 8} = \frac{2}{10} = 0.2	ext{ A}$$

Now, which way does this current flow? The $6	ext{ V}$ cell dominates the circuit, pushing current out of its positive terminal. Therefore, the current flows clockwise, meaning it travels from Y to X through the top branch.

### The Charging Cell

We need to find the potential difference across points X and Y, which is exactly the terminal voltage across cell $E_2$. Notice a fascinating detail: the current of $0.2	ext{ A}$ is entering the positive terminal of the $4	ext{ V}$ cell.

This means the cell is being **charged**, not discharged! When a cell is being charged by an external source, its terminal voltage is greater than its EMF. The formula we must use is:

$$V_{XY} = E_2 + I r_2$$

### Final Calculation

Let's plug in the numbers for our charging cell. The EMF is $4	ext{ V}$, the current is $0.2	ext{ A}$, and its internal resistance is $8\,\Omega$.

$$V_{XY} = 4 + (0.2)(8)$$

Multiplying $0.2$ by $8$ gives us $1.6	ext{ V}$. Adding this to the $4	ext{ V}$ of EMF, we get:

$$V_{XY} = 4 + 1.6 = 5.6	ext{ V}$$

The total potential difference across points X and Y is exactly $5.6	ext{ V}$.

Animated Solution for Physics - Current Electricity: A cell $E_{1}$ of emf 6V and internal resistance $2Ω$ is connected with another cell $E_{2}$ of emf 4V and internal resistance $8Ω$ (as shown in the figure). The potential difference across points X and Y is

Select Answer:

Visualized Solution

Circuit Analysis

Net EMF and Current

Substituting Values

Calculating Current

Potential Difference Across X and Y

Terminal Voltage Calculation

Final Calculation

Conclusion

The Sigma Insight: Cells, EMF, and Internal Resistance

Analyzing the Setup

The Master Equation

Calculating the Current

The Charging Cell

Final Calculation

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Select Answer:

Visualized Solution

Circuit Analysis

Net EMF and Current

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Potential Difference Across X and Y

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Conclusion

The Sigma Insight: Cells, EMF, and Internal Resistance

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