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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2023
  • Medium
  • Numerical

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 \(................A.\)

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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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