question_answer

Two cells A and B of emf 2V and 1.5V respectively, are connected as shown in figure through an external resistance \[10\,\Omega .\] The internal resistance of each cell is \[5\,\Omega .\]The  potential difference \[{{E}_{A}}\]and \[{{E}_{B}}\]across the terminals of the cells A and B respectively are

A) \[{{E}_{A}}=2.0V,{{E}_{B}}=1.5\,V\]

B)  \[{{E}_{A}}=2.125\,V,\,{{E}_{B}}=1.375\,V\]

C)  \[{{E}_{A}}=1.875\,V,{{E}_{B}}=1.625\,V\]

D)  \[{{E}_{A}}=1.875\,V,{{E}_{B}}=1.375V\]

Correct Answer: C

Solution :

 The figure can be redrawn as, The current through the circuit \[i=\frac{\text{net}\,\text{emf}}{\text{effective}\,\text{resistance}}\] \[=\frac{2-1.5}{5+5+10}=\frac{0.5}{20}=\frac{1}{40}=0.025\,A\] The terminal potential difference of the  batteries \[{{V}_{A}}={{\varepsilon }_{A}}-i{{r}_{A}}=2-0.025\times 5\] \[=2-0.0125=1.875\,V\] and \[{{V}_{B}}={{\varepsilon }_{B}}+i{{r}_{B}}=1.5+0.025\times 5\] \[=1.5+0.0125=1.625\,V\]