Question

A cell of emf 1.2 V and internal resistance 2 \( \Omega \) is connected in parallel to another cell of emf 1.5 V and internal resistance 1 \( \Omega \). If the like poles of the cells are connected together, the emf of the combination of the two cells is:

• A. \( 0.8 \) V
• B. \( 3.9 \) V
• C. \( 2.7 \) V
• D. \( 1.4 \) V

Hint:

For cells in parallel, use the formula \( E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \) to find the combined emf.

Answer 1

The Correct Option is D

Step 1: Apply Parallel EMF Formula For two cells connected in parallel, the equivalent emf is given by: \[ E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \] where: \( E_1 = 1.2 V \), \( r_1 = 2 \Omega \), \( E_2 = 1.5 V \), \( r_2 = 1 \Omega \). Step 2: Compute Equivalent EMF \[ E_{eq} = \frac{(1.2 \times 1) + (1.5 \times 2)}{2 + 1} \] \[ E_{eq} = \frac{1.2 + 3}{3} = \frac{4.2}{3} = 1.4 { V} \] Thus, the correct answer is \( 1.4 \) V.