A) \[\frac{{{V}^{2}}}{\omega L-\frac{1}{\omega C}}\]
B) \[{{I}^{2}}\,C\,\omega \]
C) \[{{I}^{2}}R\]
D) \[\frac{{{V}^{2}}}{\omega C}\]
Correct Answer: C
Solution :
Key Idea: In a series L-C-R circuit, resonance occurs when capacitive reactance becomes equal to inductive reactance. |
In series L-C-R circuit at resonance, capacitive reactance \[({{X}_{C}})=\] inductive reactance \[({{X}_{L}})\] |
i.e., \[\frac{1}{\omega C}=\omega L\] |
Total impedance of the circuit |
\[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\] |
\[=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\] |
i.e., \[R=Z\] |
So, power factor |
\[\cos \phi =\frac{R}{Z}=\frac{R}{R}=1\] |
Thus, power loss at resonance is given by |
\[P={{E}_{rms}}\,{{I}_{rms}}\,\cos \,\phi \] |
\[={{E}_{rms}}\,{{I}_{rms}}\times 1\] |
\[=({{I}_{rms}}\,R)\,{{I}_{rms}}\] |
\[={{({{I}_{rms}})}^{2}}\,R\] |
\[={{I}^{2}}R\] |
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