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.
Answer (integer)
25
Solution
$\omega=100 \pi$
<br/><br/>$$
\begin{aligned}
& \text { So } Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\\\
& =\sqrt{80^{2}+(100-40)^{2}} \\\\
& =100 \Omega \\\\
& i_{0}=\frac{V_{0}}{Z}=\frac{2500}{100} \mathrm{~A}=25 \mathrm{~A}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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