A series LCR circuit is connected to a 45   sin ( ω t ) Volt source. The resonant angular…

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

$$$$\begin{gathered} \frac{1}{\sqrt{\left(50\times {10}^{-3}\right)C}}={10}^5 \\ \Rightarrow C=2\times {10}^{-9} \mathrm{F} \end{gathered}$$

Given: $V=45\sin \left(\omega t\right)$. Therefore, $V_0=45$.

Now, $I_0=\frac{V_0}{R}=\frac{45}{R} ...\left(\mathrm{ii}\right)$

Inductive reactance, $X_L=\omega L=\left(8\times {10}^4\right)\times \left(50\times {10}^{-3}\right)=4000 \Omega$.

and capacitive reactance, $X_C=\frac{1}{\omega C}=\frac{1}{\left(8\times {10}^4\right)\times \left(2\times {10}^{-9}\right)}=6250 \Omega$.

For new current amplitude, we can write

$$\begin{gathered} 0.05I_0=\frac{45}{\sqrt{R^2+{\left(X_L-X_C\right)}^2}} \\ \Rightarrow 0.05I_0=\frac{45}{\sqrt{R^2+{\left(6250-4000\right)}^2}} \\ \Rightarrow 0.05\times \frac{45}{R}=\frac{45}{\sqrt{R^2+{\left(6250-4000\right)}^2}} \\ \Rightarrow R^2+{\left(6250-4000\right)}^2=\frac{R^2}{{\left(0.05\right)}^2} \\ \Rightarrow R^2+{\left(2250\right)}^2=400R^2 \\ \Rightarrow R=\frac{2250}{\sqrt{399}}=112.67 \Omega \end{gathered}$$

Where $X_{L_0}=X_{C_0}$ are at resonant frequencies

On solving, $\Rightarrow I_0=\frac{45}{R}\simeq 400 \mathrm{mA}$

Quality factor $Q=\frac{X_L}{R}\simeq 44.44$

Now, $Q=\frac{{\omega }_0}{∆\omega }\Rightarrow ∆\omega \simeq 2250 \mathrm{rad} {\mathrm{s}}^{-1}$.

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

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