question_answer

What is the potential drop between points A and C in the following circuit ? Resistances 1\[\Omega \] and 2\[\Omega \] represent the internal resistances of the respective cells.

A) 1.75V                                   

B) 2.25V

C) \[\frac{\text{5}}{\text{4}}\text{V}\]                                      

D) \[\frac{4}{5}\text{V}\]

Correct Answer: B

Solution :

Emfs \[{{E}_{1}}\] and \[{{E}_{2}}\] are opposing each other. Since,\[{{E}_{2}}>{{E}_{1}}\] current will move from right to left. Current in circuit                 \[i=\frac{{{E}_{2}}-{{E}_{1}}}{R+{{r}_{1}}+{{r}_{2}}}=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25\,A\] The potential drop between points A and C is                                 \[{{V}_{A}}-{{V}_{C}}={{E}_{1}}+i\,{{r}_{1}}\]                                 \[=2+(0.25\times 1)\]                 \[=2.25\,V\]