Question

Two cells of emf \( E_1 \) and \( E_2 \), and internal resistances \( r_1 \) and \( r_2 \), respectively, are connected in parallel as shown in the figure. The equivalent emf of the combination is \( E_{\text{eq}} \). Then 

• A. \( E_{\text{eq}} < E_1 \)
• B. \( E_{\text{eq}} > E_1 \) and \( E_2 \) is nearer \( E_1 \)
• C. \( E_{\text{eq}} < E_1 \) and \( E_2 \) is nearer \( E_1 \)
• D. \( E_{\text{eq}} = E_1 \)

Hint:

When cells are connected in parallel, the equivalent emf is weighted based on the internal resistances of the cells. It lies between the two emfs, closer to the cell with the lower internal resistance.

Answer 1

The Correct Option is C

For two cells connected in parallel, the equivalent emf \( E_{\text{eq}} \) is given by the weighted average of the individual emfs: \[ E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \] This formula accounts for the internal resistances of the cells. Since the cells are connected in parallel, the equivalent emf lies between the emfs of the two cells. Thus, \( E_{\text{eq}} \) is generally less than both \( E_1 \) and \( E_2 \), and it is closer to the emf of the cell with the lower internal resistance. Thus, the correct answer is: \[ E_{\text{eq}} < E_1 \text{ and } E_2 \text{ is nearer } E_1 \]