Question

Two cells with the same emf $ E $ and 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. \( \sqrt{r_1 r_2} \)
• B. \( r_1 + r_2 \)
• C. \( r_1 - r_2 \)
• D. \( \frac{r_1 r_2}{2} \)

Hint:

When two cells are connected in series, the condition that the potential difference across one cell is zero helps determine the external resistance required to balance the internal resistances.

Answer 1

The Correct Option is C

Let's analyze the problem given two cells with the same electromotive force (emf), \( E \), and different internal resistances, \( r_1 \) and \( r_2 \), connected in series with an external resistance, \( R \). The potential difference across the first cell is zero, and we are required to find the value of \( R \).

When the cells are in series, the total emf of the connection is \( E + E = 2E \), and the total internal resistance is \( r_1 + r_2 \).

The condition given is that the potential difference across the first cell is zero, which implies that the current through the first cell does not result in any potential drop across its resistance. Hence:

\[ V_1 = E - Ir_1 = 0 \]

Here, \( I \) is the current flowing through the circuit. Solving for \( I \):

\[ E = Ir_1 \]

\[ I = \frac{E}{r_1} \]

The same current \( I \) flows through the entire series circuit, and by Ohm’s law, the potential difference across the entire circuit is:

\[ 2E = I(R + r_1 + r_2) \]

Substituting the expression for \( I \), we have:

\[ 2E = \frac{E}{r_1}(R + r_1 + r_2) \]

Rearranging terms gives:

\[ 2r_1 = R + r_1 + r_2 \]

\[ R = 2r_1 - r_1 - r_2 \]

\[ R = r_1 - r_2 \]

Thus, the value of \( R \) is \( r_1 - r_2 \).

Answer 2

To solve the problem, begin by considering the setup where two cells with emf \( E \) and internal resistances \( r_1 \) and \( r_2 \) are connected in series. The total emf of the combined cells in series is \( 2E \) and the total internal resistance is \( r_1 + r_2 \).

The external resistance \( R \) is connected in series, forming the complete circuit. Given that the potential difference across the first cell is zero, the terminal potential difference across it is zero. Thus, the total voltage across the first cell must be entirely dropped across its internal resistance \( r_1 \), due to the current \( I \) flowing through it.

Let's express the current \( I \) as follows:

\[ I = \frac{2E}{R + r_1 + r_2} \]

The potential difference across the first cell (terminal voltage zero) is given by \( E - I \cdot r_1 = 0 \). Rearranging gives:

\[ E = I \cdot r_1 \]

Substituting the expression for \( I \) gives:

\[ E = \frac{2E \cdot r_1}{R + r_1 + r_2} \]

Solving for \( R \), multiply both sides by \( R + r_1 + r_2 \):

\[ E(R + r_1 + r_2) = 2E \cdot r_1 \]

This simplifies to:

\[ R + r_1 + r_2 = 2r_1 \]

Therefore:

\[ R = 2r_1 - r_1 - r_2 \]

\[ R = r_1 - r_2 \]

Thus, the value of \( R \) when the potential difference across the first cell is zero is \( r_1 - r_2 \).