A) 5.1 A
B) 0.51 A
C) 1.5 A
D) 0.15 A
Correct Answer: C
Solution :
Given that emf \[{{E}_{N}}=1.5\,{{r}_{N}}\] 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, \[i=1.5\,\,A\]
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