JEE Mains · Physics · STD 12 -7. Alternating current
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 \(.....\,\times 10^{2}\, {W}\)
- A \(150\)
- B \(125\)
- C \(160\)
- D \(200\)
Answer & Solution
Correct Answer
(B) \(125\)
Step-by-step Solution
Detailed explanation
\(X_{L}=X_{c}\) (due to resonance) \(Z=R \text { So } i_{m s}=\frac{V}{Z}=\frac{V}{R}\) \(\frac{V^{2}}{R}=\frac{250 \times 250}{5}=125 \times 10^{2} \,W\)
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