Explanation:
A Given,
Alternating voltage, $\varepsilon=30 \sin 200 \mathrm{t} \mathrm{V}$
.....(i)
Resistance, $(\mathrm{R})=10 \Omega$,
Inductance, $(\mathrm{L})=0.05 \mathrm{H}$,
Capacitance $(\mathrm{C})=500 \mu \mathrm{F}=500 \times 10^{-6} \mathrm{~F}$

Standard equation of AC voltage, $\varepsilon=\varepsilon_{\mathrm{m}} \sin \omega \mathrm{t}$
From equation (i) and (ii), we get-
$\varepsilon_{\mathrm{m}}=30 \mathrm{~V}, \omega=200 \mathrm{rad} / \mathrm{sec}$
We know that,
Inductive reactance,
$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=200 \times 0.05=10 \Omega$
And, Capacitive reactance,
$\mathrm{X}_{\mathrm{C}}=\frac{\mathrm{I}}{\omega \mathrm{C}}=\frac{1}{200 \times 500 \times 10^{-6}}=10 \Omega$
Now, Impedance of the LCR circuit,
$Z =\sqrt{(R)^{2}+\left(X_{L}-X_{C}\right)^{2}}$
$=\sqrt{(10)^{2}+(10-10)^{2}}=\sqrt{100+0}$
$Z =10 \Omega$
So, Amplitude of the current is,
$\mathrm{I}_{\max }=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{Z}}=\frac{\varepsilon_{\mathrm{m}}}{\mathrm{Z}}=\frac{30}{10}=3 \mathrm{~A}$
$\mathrm{I}_{\max }=3 \mathrm{~A}$