A) 125 W
B) 10 W
C) \[\frac{4}{5}\,\,W\]
D) 25 W
Correct Answer: B
Solution :
The given circuit can be redrawn as follows \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{9}{18}=\frac{1}{2}\] and \[i={{i}_{1}}+{{i}_{2}}\] \[\Rightarrow \,\,\,\frac{i}{{{i}_{1}}}\,\,=\,\,1+\frac{{{i}_{2}}}{{{i}_{1}}}\,\,=\,\,1\,+\,2=3\] \[From\,\,P={{i}^{2}}R\,\,\Rightarrow \,\,\frac{{{P}_{10\Omega }}}{{{P}_{18\Omega }}}\,\,=\,\,\left( \frac{i}{{{i}_{1}}} \right){{\,}^{2}}\,\times \,\,\frac{10}{18}\] \[\Rightarrow \,\,\,{{P}_{10\,\Omega }}\,\,=\,\,10\,W\]You need to login to perform this action.
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