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) 𝐼0 in mA (1) 44.4 (Q) The quality factor of the circuit (2) 18 (R) The bandwidth of the circuit in rad s −1 (3) 400 (S) The peak power dissipated at resonance in Watt (4) 2250 (5) 500 (A) 𝑃 → 2,𝑄 → 3, 𝑅 → 5, 𝑆 → 1 (B) 𝑃 → 3,𝑄 → 1, 𝑅 → 4, 𝑆 → 2 (C) 𝑃 → 4,𝑄 → 5, 𝑅 → 3, 𝑆 → 1 (D) 𝑃 → 4,𝑄 → 2, 𝑅 → 1, 𝑆 → 5

 To solve this problem, we need to analyze the given LCR circuit and use the provided information to match the entries in List-I with the appropriate values in List-II.

Given:

• Source voltage: $V(t)=45\sin (\omega t)$ V

• Resonant angular frequency: ${\omega }_0=10^5$ rad/s

• Current amplitude at resonance: $I_0$

• Angular frequency when current amplitude is $0.05I_0$: $\omega =8\times 10^4$ rad/s

• Inductance: $L=50$ mH

Step-by-Step Solution:

1. Find $I_0$ (P):

At resonance, the impedance of the circuit is purely resistive, so:

$$I_0=\frac{V}{R}$$

where $V=45$ V (amplitude of the source voltage).

We need to find $R$. At resonance:

$${\omega }_0=\frac{1}{\sqrt{LC}}$$

Given ${\omega }_0=10^5$ rad/s and $L=50$ mH:

$$C=\frac{1}{{\omega }_0^{2}L}=\frac{1}{(10^5{)}^2\times 50\times 10^{-3}}=2\times 10^{-6}\text{F}$$

Now, at $\omega =8\times 10^4$ rad/s, the current amplitude is $0.05I_0$. The impedance at this frequency is:

$$Z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$$

The current amplitude is:

$$I=\frac{V}{Z}=0.05I_0$$

Substituting $I_0=\frac{V}{R}$:

$$\frac{V}{Z}=0.05\frac{V}{R}⟹Z=20R$$

So:

$$\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}=20R$$

Squaring both sides:

$$R^2+{(\omega L-\frac{1}{\omega C})}^2=400R^2$$$${(\omega L-\frac{1}{\omega C})}^2=399R^2$$

Substituting $\omega =8\times 10^4$ rad/s, $L=50$ mH, and $C=2\times 10^{-6}$ F:

$${(8\times 10^4\times 50\times 10^{-3}-\frac{1}{8\times 10^4\times 2\times 10^{-6}})}^2=399R^2$$$${(4000-6250)}^2=399R^2$$$$(-2250{)}^2=399R^2$$$$5062500=399R^2$$$$R^2=\frac{5062500}{399}\approx 12688.47$$$$R\approx 112.6\mathrm{\Omega }$$

Now, $I_0=\frac{V}{R}=\frac{45}{112.6}\approx 0.4$ A or 400 mA.

So, $I_0=400$ mA.

2. Find the Quality Factor $Q$ (Q):

The quality factor $Q$ is given by:

$$Q=\frac{{\omega }_{0}L}{R}$$

Substituting ${\omega }_0=10^5$ rad/s, $L=50$ mH, and $R=112.6$ Ω:

$$Q=\frac{10^5\times 50\times 10^{-3}}{112.6}\approx 44.4$$

3. Find the Bandwidth $\mathrm{\Delta }\omega$ (R):

The bandwidth is given by:

$$\mathrm{\Delta }\omega =\frac{{\omega }_0}{Q}$$

Substituting ${\omega }_0=10^5$ rad/s and $Q=44.4$:

$$\mathrm{\Delta }\omega =\frac{10^5}{44.4}\approx 2250\text{rad/s}$$

4. Find the Peak Power Dissipated at Resonance (S):

At resonance, the power dissipated is:

$$P=I_0^{2}R$$

Substituting $I_0=0.4$ A and $R=112.6$ Ω:

$$P=(0.4{)}^2\times 112.6\approx 18\text{W}$$

Matching List-I with List-II:

• (P) $I_0$ in mA: 400 → (4)

• (Q) Quality factor: 44.4 → (1)

• (R) Bandwidth: 2250 → (4)

• (S) Peak power: 18 → (2)

Correct Option:

(B) $P\to 3,Q\to 1,R\to 4,S\to 2$

However, there seems to be a discrepancy in the matching. Let's re-evaluate:

• $I_0=400$ mA → (4)

• $Q=44.4$ → (1)

• $\mathrm{\Delta }\omega =2250$ → (4)

• $P=18$ W → (2)

Thus, the correct matching is:

• $P\to 4$

• $Q\to 1$

• $R\to 4$

• $S\to 2$

But since $R$ cannot be matched with the same value as $P$, there might be an error in the problem setup or options. Based on the calculations, the closest option is:

(B) $P\to 3,Q\to 1,R\to 4,S\to 2$

However, this does not perfectly match the calculated values. The correct matching should be:

• $P\to 4$

• $Q\to 1$

• $R\to 4$

• $S\to 2$

But since $R$ cannot be matched with the same value as $P$, the problem might have an inconsistency.