A) 12 A
B) 14 A
C) 8 A
D) 10 A
Correct Answer: A
Solution :
\[\frac{1}{{{\operatorname{R}}_{eq}}}=\frac{1}{{{\operatorname{R}}_{1}}}+\frac{1}{{{\operatorname{R}}_{2}}}\] \[P=\frac{{{V}^{2}}}{{{\operatorname{R}}_{eq}}}=\frac{{{V}^{2}}}{{{\operatorname{R}}_{1}}}+\frac{{{V}^{2}}}{{{\operatorname{R}}_{2}}}+.......\] \[=15\times 40+5\times 100+80\times 5+1\times 1000=2500w\] \[P=VI\] \[P=2500=220I\] \[I=\frac{250}{22}=\frac{125}{11}\]
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