JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 20)
A series $\mathrm{LCR}$ circuit consists of $\mathrm{R}=80 \Omega, \mathrm{X}_{\mathrm{L}}=100 \Omega$, and $\mathrm{X}_{\mathrm{C}}=40 \Omega$. The input
voltage is 2500 $\cos (100 \pi \mathrm{t}) \mathrm{V}$. The amplitude of current, in the circuit, is _________ A.
voltage is 2500 $\cos (100 \pi \mathrm{t}) \mathrm{V}$. The amplitude of current, in the circuit, is _________ A.
Answer
25
Explanation
$\omega=100 \pi$
$$ \begin{aligned} & \text { So } Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\\\ & =\sqrt{80^{2}+(100-40)^{2}} \\\\ & =100 \Omega \\\\ & i_{0}=\frac{V_{0}}{Z}=\frac{2500}{100} \mathrm{~A}=25 \mathrm{~A} \end{aligned} $$
$$ \begin{aligned} & \text { So } Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\\\ & =\sqrt{80^{2}+(100-40)^{2}} \\\\ & =100 \Omega \\\\ & i_{0}=\frac{V_{0}}{Z}=\frac{2500}{100} \mathrm{~A}=25 \mathrm{~A} \end{aligned} $$
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