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 \[ E_{eq} = \frac{(1.2 \times 1) + (1.5 \times 2)}{2 + 1} \] \[ E_{eq} = \frac{1.2 + 3}{3} = \frac{4.2}{3} = 1.4 \text{ V} \] Thus, the correct answer is \( 1.4 \) V. \bigskip

Answer 2

Given:
- Cell 1: emf \( E_1 = 1.2 \, V \), internal resistance \( r_1 = 2 \, \Omega \)
- Cell 2: emf \( E_2 = 1.5 \, V \), internal resistance \( r_2 = 1 \, \Omega \)
- Cells connected in parallel with like poles together.

We need to find the emf of the combination, \( E \).

Step 1: When two cells with different emf and internal resistances are connected in parallel with like poles connected, the combined emf is given by:
\[ E = \frac{E_1 / r_1 + E_2 / r_2}{1 / r_1 + 1 / r_2} \]

Step 2: Substitute the values:
\[ E = \frac{\frac{1.2}{2} + \frac{1.5}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{0.6 + 1.5}{0.5 + 1} = \frac{2.1}{1.5} = 1.4 \, V \]

Therefore, the emf of the combination of the two cells is:
\[ \boxed{1.4 \, V} \]