A series LCR circuit is connected to a 45 ( ) Volt source. The...

Resonant angular frequency is given by, $\dfrac{1}{\sqrt{LC}} = {10}^{5}$

$\dfrac{1}{\sqrt{\left(50 \times {10}^{- 3}\right)C}} = {10}^{5} ⇒ C = 2 \times {10}^{- 9}F$

Given: $V = 45\sin \left(\omegat\right)$. Therefore, ${V}_{0} = 45$.

Now, ${I}_{0} = \dfrac{{V}_{0}}{R} = \dfrac{45}{R} . . . \left(ii\right)$

Inductive reactance, ${X}_{L} = \omegaL = \left(8 \times {10}^{4}\right) \times \left(50 \times {10}^{- 3}\right) = 4000Ω$.

and capacitive reactance, ${X}_{C} = \dfrac{1}{\omegaC} = \dfrac{1}{\left(8 \times {10}^{4}\right) \times \left(2 \times {10}^{- 9}\right)} = 6250Ω$.

For new current amplitude, we can write

$0 . 05{I}_{0} = \dfrac{45}{\sqrt{{R}^{2} + {\left({X}_{L} - {X}_{C}\right)}^{2}}} ⇒ 0 . 05{I}_{0} = \dfrac{45}{\sqrt{{R}^{2} + {\left(6250 - 4000\right)}^{2}}} ⇒ 0 . 05 \times \dfrac{45}{R} = \dfrac{45}{\sqrt{{R}^{2} + {\left(6250 - 4000\right)}^{2}}} ⇒ {R}^{2} + {\left(6250 - 4000\right)}^{2} = \dfrac{{R}^{2}}{{\left(0 . 05\right)}^{2}} ⇒ {R}^{2} + {\left(2250\right)}^{2} = 400{R}^{2} ⇒ R = \dfrac{2250}{\sqrt{399}} = 112 . 67Ω$

Where ${X}_{{L}_{0}} = {X}_{{C}_{0}}$ are at resonant frequencies

On solving, $⇒ {I}_{0} = \dfrac{45}{R} ≃ 400mA$

Quality factor $Q = \dfrac{{X}_{L}}{R} ≃ 44 . 44$

Now, $Q = \dfrac{{\omega}_{0}}{∆ \omega} ⇒ ∆ \omega ≃ 2250rad{s}^{- 1}$.

Peak power $= 45 \times \dfrac{400}{1000}W = 18W$

Therefore, $P \to 3, Q \to 1, R \to 4, S \to 2$.