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)  1.75 V          

B)  2.25 V

C)  \[\frac{5}{4}\]v             

D)  \[\frac{4}{5}\]v

Correct Answer: B

Solution :

Emf s \[{{E}_{1}}\] 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                 \[{{V}_{A}}-{{V}_{C}}={{E}_{1}}+i\,\,{{r}_{1}}\] = 2.25V