A) \[\text{1 }H\]
B) \[\text{1/6 }H\]
C) \[\text{1/3 }H\]
D) \[\text{11 }H\]
Correct Answer: A
Solution :
Minimum value when connected in parallel \[\frac{1}{L}=\frac{1}{{{L}_{1}}}+\frac{1}{{{L}_{2}}}+\frac{1}{{{L}_{3}}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1\]You need to login to perform this action.
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