Alternating Current Ques 1
1. In a series resonant circuit, having $L, C$ and $R$ as its elements, the resonant current is $i$. The power dissipated in the circuit at resonance is
[2002]
(a) $\frac{i^2 \mathrm{R}}{\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)}$
(b) zero
(c) $i^2 \omega \mathrm{L}$
(d) $i^2 \mathrm{R}$
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Answer:
Correct Answer: 1.(d)
Solution: (d)
At resonance $L \omega=\frac{1}{C \omega}, \omega=\frac{1}{\sqrt{L C}}$
Current through circuit $i=\frac{E}{R}$
Power dissipated at Resonance $=i^2 R$
BETA
