A series LCR circuit of R = 5$\Omega$, L = 20 mH and C = 0.5 $\mu$F is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______________ $\times$ 102 W.
Answer (integer)
125
Solution
X<sub>L</sub> = X<sub>C</sub> (due to resonance)<br><br>Z = R so ${i_{rms}} = {V \over Z} = {V \over R}$<br><br>${{{V^2}} \over R} = {{250 \times 250} \over 5} = 125 \times {10^2}W$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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