JEE Mains · Physics · STD 12 -7. Alternating current
A series \(LCR\) circuit consists of \(R =80\,\Omega X _{ L }=100\, \Omega\) and \(X _{ C }=40\, \Omega\) The input voltage is \(2500 \cos (100 \pi t )\,V\). The amplitude of current, in the circuit, is \(................A.\)
- A \(24\)
- B \(23\)
- C \(25\)
- D \(22\)
Answer & Solution
Correct Answer
(C) \(25\)
Step-by-step Solution
Detailed explanation
\(X = X_L - X_C = 100 - 40 = 60\,\Omega\) \(Z = \sqrt{R^2 + X^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega\) \(I_0 = \frac{V_0}{Z} = \frac{2500}{100} = 25\,A\)
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