Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Current Electricity

Question:

A cell of emf 3 V and internal resistance 0.1 Ω is connected to a 3.9 Ω external resistance. What will be the potential difference across the terminals of the cell ? 

Options:

1.925 V

2.925 V

2.529 V

2.295 V

Correct Answer:

2.925 V

Explanation:

Electric Current $I = \frac{E}{r+R} = \frac{3}{0.1+3.9} = 0.75A$

Potential difference across terminal of cell is $ V = E - Ir = 3 - 0.75\times 0.1 = 2.925V$