A) 4W
B) 2W
C) 1 W
D) 5 W
Correct Answer: D
Solution :
Voltage across \[\,2\,\Omega \]is the same as voltage across arm containing\[\,1\,\Omega \]and\[\,5\,\Omega \]resistances. Voltage across \[\,2\,\Omega \] resistance, \[V=2\times 3=6V\] So, voltage across lowest arm, \[{{V}_{1}}=6V\] Current across\[5\,\Omega ,I=\frac{6}{1+5}=1\,A\] Thus, power across \[5\,\Omega ,\] \[P={{I}^{2}}R={{(1)}^{2}}\times 5=5\,W\]You need to login to perform this action.
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