Question

Two cells connected in series have electromotive force of 1.5 V each. Their internal resistance are $0.5\Omega$ . and $0.25\Omega$ respectively. This combination is connected to a resistance of $2.25\Omega$ . Potential difference across the terminals of each cell

• A. IV, 0.25V
• B. IV, 1.25V
• C. 1.5V, 2.25V
• D. 1.5V, 2.56V

Answer 1

Answer: (B)

The arrangement is shown in the figure. The effective emf in the circuit is $E=1.5+1.5=3.0V$ and the total resistance is $R=0.5+0.25+2.25=3.0\Omega .$ Hence, the current in the circuit is $I=\frac{E}{R}=\frac{3.0}{3.0}=1.0A$ Potential difference across the terminals of the first cell is $V_1=E-I{r}_1=1.5-(1.0)\times (0.5)=1.0V$ Potential difference across the terminals of the second cell is $V_2=E-I{r}_2=1.5-(1.0)\times (0.25)=1.25V$