QJEEM21Phc695: JEE Main 2021 Physics - Class 12 JEEM21Phc695 Solution: A series LCR circuit of R = 5 , L = 20 mH and C = 0.5 F is…

Question

Verified Solution • 2025–26 Session moderate

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.

Accepted Answer

X_L = X_C (due to resonance)

Z = R so

{i_{rms}} = {V \over Z} = {V \over R}

{{{V^2}} \over R} = {{250 \times 250} \over 5} = 125 \times {10^2}W

Step-by-Step Verified Solution