A) 1.75 V
B) 2.25 V
C) \[\frac{5}{4}\]v
D) \[\frac{4}{5}\]v
Correct Answer: B
Solution :
Emf s \[{{E}_{1}}\] 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 \[{{V}_{A}}-{{V}_{C}}={{E}_{1}}+i\,\,{{r}_{1}}\] = 2.25VYou need to login to perform this action.
You will be redirected in
3 sec