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

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, \[\text{i =}\frac{\text{net emf}}{\text{total resistance}}=\frac{{{E}_{2}}-{{E}_{1}}}{R+{{r}_{1}}+{{r}_{2}}}\] Given, \[R=5\Omega ,{{r}_{1}}=1\Omega ,{{r}_{2}}=2\Omega \] \[{{E}_{1}}=2V\] and \[{{E}_{2}}=4V\] \[\therefore \]  \[i=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25A\] The potential drop between points A and C, \[{{V}_{A}}-{{V}_{C}}={{E}_{1}}+i{{r}_{1}}\] \[=2+0.25\times 1=2.25V\]