Two cells of emf 2E and E with internal resistance r1 and r2 respectively are connected in series to

Q: Why do we add the EMFs of the two cells?

The cells are connected in series with their polarities aligned (negative terminal of the first connected to the positive terminal of the second), so their electromotive forces add up to drive current in the same direction.

Q: Can the potential difference across a cell really be zero?

Yes! If the external circuit allows a current $I = E/r$ to flow, the voltage drop across the internal resistance equals the cell's EMF, resulting in a net terminal voltage of zero. It's equivalent to short-circuiting that specific cell.

Q: What if the cells were connected in opposition?

If they were opposing each other, the net EMF would be the difference between their EMFs (e.g., $2E - E = E$), and the current direction would be determined by the stronger cell.

Q: Why is the internal resistance added to the external resistance?

Internal resistance acts just like a regular resistor in series with the ideal voltage source of the cell. Therefore, it adds to the total resistance of the circuit.

Q: Is it possible for $R$ to be negative?

Mathematically, yes, if $r_2 > r_1/2$. However, physically, a standard resistor cannot have a negative resistance. If the formula yields a negative $R$, it means it's impossible to achieve zero potential difference across the first cell with a passive external resistor.

Question

Two cells of emf $2E$ and $E$ with internal resistance $r_1$ and $r_2$ respectively are connected in series to an external resistor $R$ (see figure). The value of $R$, at which the potential difference across the terminals of the first cell becomes zero is

Options:

(A) $r_1 + r_2$

(B) $\frac{r_1}{2} - r_2$

(C) $\frac{r_1}{2} + r_2$

(D) $r_1 - r_2$

Sigma Percentile · Accepted Answer

The problem presents a classic scenario involving cells in series and the concept of terminal potential difference. Let's break down the physics and the math behind it.

### Analyzing the Setup
We are given a single-loop circuit containing two cells and an external resistor $R$. The first cell has an electromotive force (EMF) of $2E$ and an internal resistance $r_1$. The second cell has an EMF of $E$ and an internal resistance $r_2$.

Because the cells are connected in series and their polarities are aligned (they are aiding each other), their EMFs simply add up. 
$$E_{eq} = 2E + E = 3E$$

Similarly, the total resistance of the circuit is the sum of the external resistance and the internal resistances of both cells.
$$R_{eq} = R + r_1 + r_2$$

### The Master Equation
Using Ohm's law, the current $I$ flowing through the entire circuit is the total EMF divided by the total resistance:
$$I = \frac{3E}{R + r_1 + r_2}$$

Now, let's focus on the first cell. When a cell is discharging (supplying current), the potential difference $V_1$ across its terminals is less than its EMF due to the voltage drop across its own internal resistance. The formula is:
$$V_1 = E_1 - I r_1$$

The problem states a very specific condition: the potential difference across the first cell becomes zero. 
$$0 = 2E - I r_1$$
$$I r_1 = 2E$$

### Final Calculation
Now, we substitute our expression for the current $I$ into this condition:
$$\left( \frac{3E}{R + r_1 + r_2} \right) r_1 = 2E$$

Notice how the $E$ beautifully cancels out from both sides, leaving us with a purely algebraic equation:
$$\frac{3 r_1}{R + r_1 + r_2} = 2$$

Cross-multiplying to eliminate the fraction, we get:
$$3 r_1 = 2(R + r_1 + r_2)$$
$$3 r_1 = 2R + 2r_1 + 2r_2$$

Our goal is to isolate the external resistance $R$. Let's move the internal resistance terms to one side:
$$2R = 3r_1 - 2r_1 - 2r_2$$
$$2R = r_1 - 2r_2$$

Dividing by 2, we arrive at our final expression:
**$$R = \frac{r_1}{2} - r_2$$**

This elegant result tells us exactly how to tune the external resistor to effectively "short-circuit" the terminal voltage of the first cell!

Select Answer:

Visualized Solution

Circuit Analysis

Current in the Circuit

Potential Difference Across First Cell

Substituting the Current

Solving for R

Final Expression for R

Physical Constraint

The Sigma Insight: Cells, EMF, and Internal Resistance

Analyzing the Setup We are given a single-loop circuit containing two cells and an external resistor $R$

The Master Equation

Final Calculation

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