question_answer

Determine the potential drop between the points A and C in the following circuit. Resistances \[1\,\,\Omega \] and \[2\,\,\Omega \] are representing the internal resistances of the respective cells.

A)  \[\frac{4}{5}V\]    

B)  1.75 V  

C)  2.25 V  

D)  5 V

Correct Answer: C

Solution :

Emfs \[{{E}_{1}}\] and \[{{E}_{2}}\] are opposing each other, since\[{{E}_{2}}>{{E}_{1}}\], so current will flow from right to left. Current in the circuit,                 \[i=\frac{net\text{ }emf}{total\text{ }resistance}=\frac{{{E}_{2}}-{{E}_{1}}}{R+{{r}_{1}}+{{r}_{2}}}\] Given, \[R=5\,\Omega ,\,{{r}_{1}}=1\,\Omega ,\,{{r}_{2}}=2\,\Omega \]                 \[{{E}_{1}}=2\,V\] and \[{{E}_{2}}=4\,V\] \[\therefore \] \[i=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25\,\,A\] The potential drop between points A and C,                 \[{{V}_{A}}-{{V}_{C}}={{E}_{1}}+i{{r}_{1}}\] \[=2+0.25\times 1=2.25\,V\]