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, \[{{V}_{A}}-{{V}_{C}}={{E}_{1}}+i{{r}_{1}}\] \[=2+0.25\times 1=2.25\,V\]You need to login to perform this action.
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