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JEE Mains · Physics · STD 12 -7. Alternating current

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 .............

  1. A \(3\)
  2. B \(4\)
  3. C \(5\)
  4. D \(6\)
Verified Solution

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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