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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2024
  • Hard
  • Numerical

When a dc voltage of \(100 \mathrm{~V}\) is applied to an inductor, a dc current of \(5 \mathrm{~A}\) flows through \(1\) . When an ac voltage of \(200 \mathrm{~V}\) peak value is connected to inductor, its inductive reactance is found to be \(20 \sqrt{3} \Omega\). The power dissipated in the circuit is _______\(W\).

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Answer & Solution

Correct Answer

(D) \(250\)

Step-by-step Solution

Detailed explanation

For \(DC\) voltage

\(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{100}{5}=20 \Omega\)

for \(AC\) voltage

\(\mathrm{X}_{\mathrm{L}}=20 \sqrt{3} \Omega\)

\(\mathrm{R}=20 \Omega\)

\(\mathrm{Z}=\sqrt{\mathrm{X}_{\mathrm{L}}^2+\mathrm{R}^2}=\sqrt{3 \times 400+400}=40\)

Power \(-i_{\text {rms }}^2 R\)

\(=\left(\frac{\mathrm{V}_{\text {rms }}}{\mathrm{Z}}\right)^2 \times \mathrm{R}=\left(\frac{\frac{200}{\sqrt{2}}}{40}\right)^2 \times 20=250 \mathrm{~W}\)

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