A) 5.1 A
B) 0.51 A
C) 1.5 A
D) 0.15 A
Correct Answer: C
Solution :
Given that emf \[{{E}_{N}}={{1.5}_{rN}}\] where \[{{r}_{N}}\] is the internal resistance of nth cell. Total emf \[E={{E}_{1}}+{{E}_{2}}+{{E}_{3}}+......{{E}_{n}}\] \[=1.5[{{r}_{1}}+{{r}_{2}}+{{r}_{3}}+.....+{{r}_{n}}]\] Total internal resistance \[r={{r}_{1}}+{{r}_{2}}+{{r}_{3}}+.....+{{r}_{n}}\] \[\therefore \] current \[i=\frac{{{E}_{total}}}{{{r}_{total}}}\] \[i=\frac{1.5[{{r}_{1}}+{{r}_{2}}+{{r}_{3}}+.....+{{r}_{n}}]}{[{{r}_{1}}+{{r}_{2}}+{{r}_{3}}+......+{{r}_{n}}]}\] Hence, \[i=1.5A\]
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