A) 5.0 W
B) 2.5 W
C) 1.25 W
D) zero
Correct Answer: D
Solution :
Given, \[{{V}_{0}}=50\,\sin (50t)V\] Maximum voltage, \[{{V}_{0}}=50\,V\] \[i={{i}_{0}}\sin \left( 50t+\frac{\pi }{3} \right)mA\] Maximum current, \[{{i}_{0}}=50\text{ }mA=50\times {{10}^{-3}}A\] Power dissipated, \[P=\frac{{{i}_{0}}}{\sqrt{2}}\times \frac{{{V}_{0}}}{\sqrt{2}}\] \[=\frac{50\times 50\times {{10}^{-3}}}{2}\] \[=\frac{2500\times {{10}^{-3}}}{2}=1.25\,W\]
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