A series LCR circuit has L = 0 . 01H, R = 10Ω and C = 1 and it is...

For an LCR circuit. the resonant angular frequency is given by, ${\omega}_{0} = \dfrac{1}{\sqrt{LC}} = {10}^{4}rad{s}^{- 1}$

The given frequency is $60$ lower than resonant frequency. Therefore,

${\omega}^{'} = 0 . 4 \times {10}^{4} = 4000rad{s}^{- 1}$

Reactance of the capacitor at given frequency,

${X}_{C} = {\left({\omega}^{'}C\right)}^{- 1} = 250Ω$.

Reactance of the inductor at given frequency,

${X}_{L} = \left({\omega}^{'}L\right) = 40Ω$

Now the amplitude of the current in the given circuit will be,

${i}_{0} = \dfrac{{V}_{0}}{\sqrt{{R}^{2} + {\left({X}_{C}^{'} - {X}_{L}^{'}\right)}^{2}}} = \dfrac{50}{\sqrt{{10}^{2} + {\left(250 - 40\right)}^{2}}} = 238mA$