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 \[{{V}_{B}}={{\varepsilon }_{B}}+i{{r}_{B}}=1.5+0.025\times 5\] \[=1.5+0.0125=1.625\,V\]You need to login to perform this action.
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