Question

The emf of a Daniel cell is 1.08 V. When the terminals of the cells are connected to a resistance of 3 $ \Omega $ , the potential difference across the terminals is found to be 0.6 V. Then the internal resistance of the cell is:

• A. 1.8 $ \Omega $
• B. 2.4 $ \Omega $
• C. 3.24 $ \Omega $
• D. 0.2 $ \Omega $

Answer 1

Answer: (B)

The emf of cell is $ E=V+ir $ Where $ V=0.6V,\, $ P.D. between terminal of the cell connected to a resistance $ R=3\Omega $ $ E=V+\left( \frac{V}{R} \right)r $ $ 1.08=0.6+\left( \frac{0.6}{3} \right)r $ Where V = 0.6 V PD between terminal of the cell conected a resistance $ R=3\Omega $ $ 0.2r=1.08-0.6=0.48 $ $ r=\frac{0.48}{0.2}=2.4\Omega $