JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2021
- Medium
- Numerical
A sinusoidal voltage of peak value \(250\, V\) is applied to a series \(LCR\) circuit, in which \(R =8 \Omega, L =24\, mH\) and \(C =60 \mu F\). The value of power dissipated at resonant condition is \('x'\, kW\). The value of \(x\) to the nearest integer is .............
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
At resonance power \((P)\)
\(P=\frac{\left(V_{ rms }\right)^{2}}{ R }\)
\(P=\frac{(250 / \sqrt{2})^{2}}{8}=3906.25 W\)
\(\approx 4 kW\)
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