A) IV, 0.25V
B) IV, 1.25V
C) 1.5V, 2.25V
D) 1.5V, 2.56V
Correct Answer: B
Solution :
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 \[R=0.5+0.25+2.25=3.0\text{ }\Omega .\] 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 \[{{V}_{1}}=E-I{{r}_{1}}=1.5-(1.0)\times (0.5)=1.0\,V\] Potential difference across the terminals of the second cell is \[{{V}_{2}}=E-I{{r}_{2}}=1.5-(1.0)\times (0.25)=1.25\,V\]You need to login to perform this action.
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