JEE Mains · Physics · STD 12 -7. Alternating current
A series \(LCR\) circuit is designed to resonate at an angular frequency \(\omega_{0}=10^{5} \,rad / s\). The circuit draws \(16\, W\) power from \(120\, V\) source at resonance. The value of resistance \('R'\) in the circuit is ...... \(\Omega\).
- A \(1200\)
- B \(600\)
- C \(800\)
- D \(900\)
Answer & Solution
Correct Answer
(D) \(900\)
Step-by-step Solution
Detailed explanation
At resonance \(P =\frac{ V ^{2}}{ R }\) \(R =\frac{ V ^{2}}{ P } =\frac{(120)^{2}}{16}\) \(=900 \,\Omega\)
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