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