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 .............
- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
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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