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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsEasy2 minQB

PhysicsEasynumerical

A series LCR circuit of $R = 5Ω, L = 20 mH$ and $C = 0.5 μ F$ is connected across an $A C$ supply of $250 V,$ having variable frequency. The power dissipated at resonance condition is ________ $\times 1 0^{2} W .$

Answer:

125.00

Solution:

$X_{L} = X_{C}$ (due to resonance)

$Z = R$ so $i_{r m s} = \frac{V}{Z} = \frac{V}{R}$

$\frac{V ^{2}}{R} = \frac{250 \times 250}{5} = 125 \times 1 0^{2} W$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: QB

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