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 $R=3\Omega$ $0.2r=1.08-0.6=0.48$ $r=\frac{0.48}{0.2}=2.4\Omega$