A) 4 W
B) 2 W
C) 1 W
D) 5 W
Correct Answer: D
Solution :
Key Idea: Voltage across \[2\Omega \]is 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 \[5\Omega ,I=\frac{6}{1+5}=1A\] Thus, power across \[5\Omega ,\] \[P={{I}^{2}}R={{(1)}^{2}}\times 5=5W\]You need to login to perform this action.
You will be redirected in
3 sec