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 :

                 Key Idea: The two cells in the given circuit are opposing each other. Emf's\[{{E}_{1}}\]and\[{{E}_{2}}\]are opposing each other. Since,\[{{E}_{2}}>{{E}_{1}},\]so current will flow from right to left. Current in the circuit, \[i=\frac{Net\text{ }emf}{Total\text{ }resis\tan ce}=\frac{{{E}_{2}}-{{E}_{1}}}{R+{{r}_{1}}+{{r}_{2}}}\] Given,   \[R=5\,\Omega ,{{r}_{1}}=1\,\Omega ,{{r}_{2}}=2\,\Omega ,\]                 \[{{E}_{1}}=2V,{{E}_{2}}=4V\] \[\therefore \]  \[i=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25A\] The potential drop between points A and C,                 \[{{V}_{A}}-{{V}_{C}}={{E}_{1}}+i{{r}_{1}}\]                                 \[=2+0.25\times 1=2.25\,V\]