Question

Two cells with same emf E but different internal resistances, $r_1$ and $r_2$ are connected in series to an external resistance R. If the potential difference across the first cell is zero then the value of R is

• A. $\frac{r_1 - r_2}{2}$
• B. $\frac{r_1 + r_2}{2}$
• C. $r_1 - r_2$
• D. $(r_1 + r_2)$

Hint:

The potential difference across the terminals of a cell is given by $V = E - Ir$, where $E$ is the EMF, $I$ is the current flowing through the cell, and $r$ is the internal resistance. In a series combination, the current is the same through all components.

Answer 1

The Correct Option is C

Step 1: Determine the total EMF and total resistance of the circuit.
The total EMF of the series combination of the two cells is $E_{total} = E + E = 2E$. The total resistance of the circuit, including the internal resistances and the external resistance, is $R_{total} = r_1 + r_2 + R$.
Step 2: Calculate the current in the circuit.
Using Ohm's law, the current $I$ flowing through the circuit is: $$I = \frac{E_{total}}{R_{total}} = \frac{2E}{r_1 + r_2 + R}$$
Step 3: Calculate the potential difference across the first cell.
The potential difference $V_1$ across the terminals of the first cell (with EMF $E$ and internal resistance $r_1$) is given by: $$V_1 = E - Ir_1$$ Substitute the expression for the current $I$: $$V_1 = E - \left(\frac{2E}{r_1 + r_2 + R}\right) r_1 = E \left(1 - \frac{2r_1}{r_1 + r_2 + R}\right)$$ $$V_1 = E \left(\frac{r_1 + r_2 + R - 2r_1}{r_1 + r_2 + R}\right) = E \left(\frac{R + r_2 - r_1}{R + r_1 + r_2}\right)$$
Step 4: Use the given condition $V_1 = 0$ to find R.
We are given that the potential difference across the first cell is zero, $V_1 = 0$. $$E \left(\frac{R + r_2 - r_1}{R + r_1 + r_2}\right) = 0$$ Since $E \neq 0$, we must have: $$\frac{R + r_2 - r_1}{R + r_1 + r_2} = 0$$ This implies that the numerator is zero: $$R + r_2 - r_1 = 0$$ Solving for $R$: $$R = r_1 - r_2$$ Final Answer: The final answer is $\boxed{r_1 - r_2}$