A) 1.75V
B) 2.25V
C) \[\frac{\text{5}}{\text{4}}\text{V}\]
D) \[\frac{4}{5}\text{V}\]
Correct Answer: B
Solution :
Emfs \[{{E}_{1}}\] and \[{{E}_{2}}\] are opposing each other. Since,\[{{E}_{2}}>{{E}_{1}}\] current will move from right to left. Current in circuit \[i=\frac{{{E}_{2}}-{{E}_{1}}}{R+{{r}_{1}}+{{r}_{2}}}=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25\,A\] The potential drop between points A and C is \[{{V}_{A}}-{{V}_{C}}={{E}_{1}}+i\,{{r}_{1}}\] \[=2+(0.25\times 1)\] \[=2.25\,V\]You need to login to perform this action.
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