A) 51.5
B) 50.7
C) 51.1
D) 50.3
Correct Answer: D
Solution :
A series resonance circuit admits maximum current, as \[P={{i}^{2}}R\] So, power dissipated is maximum at resonance. So, frequency of the source at which maximum power is dissipated in the circuit is \[v=\frac{1}{2\pi \sqrt{LC}}\] \[=\frac{1}{2\times 3.14\sqrt{25\times {{10}^{-3}}\times 400\times {{10}^{-6}}}}\] \[=\frac{1}{2\times 3.14\sqrt{{{10}^{-5}}}}\] \[=50.3Hz\]You need to login to perform this action.
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