Question:
A series $L C R$ circuit of $R=5 \Omega, L=20 \mathrm{mH}$ and $\mathrm{C}=0.5 \mu \mathrm{F}$ is connected across an AC supply of $250 \mathrm{~V}$, having variable frequency. The power dissipated at resonance condition is___________ $\times 10^{2} \mathrm{~W}$.
Solution:
$\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ (due to resonance)
$\mathrm{Z}=\mathrm{R}$ so $\mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{\mathrm{V}}{\mathrm{R}}$
$\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{250 \times 250}{5}=125 \times 10^{2} \mathrm{~W}$
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