An inductor 20mH , a capacitor 50 and a resistor 40Ω are connected...

Given :& $\text{R=40Ω , L= 20 mH , C=50 \mu F}$

Comparing the given equation for voltage with general equation~$V = {V}_{0}sin\omega t$

$\omega = 340rads^{- 1}$~ & & & and& & & & & & & & & & & & & & peak voltage& & ${V}_{0} = 10V$

Inductive reactance~${X}_{L}\omegaL = 340 \times 20 \times {10}^{- 3} = 68 \times {10}^{- 1} = 6.8Ω$
Capacitive reactance~${X}_{C}$
$\dfrac{1}{\omegaC} = \dfrac{1}{340 \times 50 \times {10}^{- 6}} = \dfrac{{10}^{4}}{34 \times 5} = \dfrac{2}{34} \times {10}^{3}$

$= 0.0588 \times {10}^{3} = 58.82Ω$

$\text{Z=}\sqrt{\left({X}_{L} - {X}_{C}\right)2{+R}^{2}}\text{⇒Z} = \sqrt{\left(\omegaL - \dfrac{1}{ωC}\right)2 + R^{2}}$
$\text{Z=}\sqrt{\left(58 . 8 - 6 . 8\right)2\text{+}\left(40\right)2}\text{⇒ Z} = \sqrt{2704 + 1600} \approx 65.6Ω$

${I}_{rms} = \dfrac{Vrms}{Z} = \dfrac{10}{65 .6 \times \sqrt{2}}( {V}_{rms} = \dfrac{V0}{2} )$

Power $= {E}_{rms}{I}_{rms}cosϕ = {I}_{rms}^{2}R$~

\left(∵ cosϕ = \dfrac{R}{Z}, {V}_{rms} = {I}_{rms}Z\right)$</p><p>$= \dfrac{100 \times 40}{{\left(65.6\right)}^{2} \times 2} = \dfrac{2000}{{\left(65.6\right)}^{2}}$<br /><br />$= 0.51W$