A series LCR circuit is connected to a 45 sin(ωt) Volt source. The resonant angular frequency of the circuit is 10⁵ rad s⁻¹ and current amplitude at resonance is I₀. When the angular frequency of the source is ω = 8 × 10⁴ rad s⁻¹, the current amplitude in the circuit is 0.05 I₀. If L = 50 mH, match each entry in List-I with an appropriate value from List-II and choose the correct option.

List-I	List-II
(P) $I_0\text{in mA}$	(1) $44.4$
(Q) $\text{The quality factor of the circuit}$	(2) $18$
(R) $\text{The bandwidth of the circuit in rad}{s}^{-1}$	(3) $400$
(S) $\text{The peak power dissipated at resonance in Watt}$	(4) $2250$
	(5) $500$

The correct option is P → 3, Q → 1, R → 4, S → 2.

Given Data:
Peak Voltage of the AC source, _V0 = 45 V
Resonant angular frequency, ω₀ = 10⁵ rad s⁻¹
Inductance, L = 50 mH = 50 × 10⁻³ H = 0.05 H
At angular frequency ω = 8 × 10⁴ rad s⁻¹, current amplitude I = 0.05 I₀

Step-by-step Derivation:

1. Finding the Net Reactance at ω = 8 × 10⁴ rad s⁻¹:
The inductive reactance is X_L = ωL and capacitive reactance is X_C = 1/(ωC).
Since ω₀² = 1/(LC), we have 1/C = ω₀²L.

Substituting 1/C in the net reactance X:

$X=\left|\omega L-\frac{1}{\omega C}\right|=L\left|\omega -\frac{{\omega }_0^2}{\omega }\right|$

Plugging in the given values:

$X=0.05\times \left|8\times {10}^4-\frac{{10}^{10}}{8\times {10}^4}\right|$

$X=0.05\times \left|80000-125000\right|=0.05\times 45000=2250 \Omega$

2. Determining Resistance R and Current Amplitude at Resonance I₀ (P):
The current amplitude at frequency ω is given by:

$I=\frac{V_0}{\sqrt{R^2+X^2}}$

At resonance, I₀ = V₀ / R. Given I = 0.05 I₀:

$\frac{V_0}{\sqrt{R^2+X^2}}=0.05\times \frac{V_0}{R}\Rightarrow \sqrt{R^2+X^2}=20R$

$R^2+X^2=400R^2\Rightarrow 399R^2=X^2$

$R=\frac{X}{\sqrt{399}}\approx \frac{2250}{20}=112.5 \Omega$

Now, calculate peak current I₀:

$I_0=\frac{V_0}{R}=\frac{45}{112.5}=0.4\text{A}=400\text{mA}$

Thus, (P) → (3).

3. Calculating Quality Factor Q (Q):
The Quality factor of the circuit is given by:

$Q=\frac{{\omega }_{0}L}{R}=\frac{{10}^5\times 0.05}{112.5}=\frac{5000}{112.5}=44.4$

Thus, (Q) → (1).

4. Calculating Bandwidth (R):
Bandwidth (Δω) is given by:

$\Delta \omega =\frac{R}{L}=\frac{112.5}{0.05}=2250\text{rad s}{}^{-1}$

Thus, (R) → (4).

5. Calculating Peak Power Dissipated at Resonance (S):
At resonance, maximum/peak power dissipated is:

$P_0=V_0{I}_0=45\times 0.4=18\text{Watt}$

Thus, (S) → (2).

Conclusion:
Matching the entries:
P → 3
Q → 1
R → 4
S → 2