A series LCR circuit is connected to a 45 sin(ωt) Volt source. The resonant angular frequency of the

Q: Why did we use $Z = 20R$?

Because the current at the new frequency is $1/20$th of the resonant current. Since $I = V/Z$ and $I_0 = V/R$, $V/Z = V/(20R)$, which means $Z = 20R$.

Q: Why is the circuit capacitive at the lower frequency?

At a frequency lower than the resonant frequency, the capacitive reactance $X_C = 1/(\omega C)$ increases, while the inductive reactance $X_L = \omega L$ decreases. Thus, $X_C > X_L$.

Q: How did we approximate $\sqrt{399}$?

$\sqrt{399}$ is extremely close to $\sqrt{400}$, which is $20$. This approximation simplifies the calculation for $R$ significantly without losing much precision.

Q: What is the physical meaning of the Quality Factor?

The Quality Factor (Q) represents the sharpness of the resonance. A higher Q means a sharper peak, indicating less energy loss relative to the stored energy in the circuit.

Q: Why is peak power calculated as $I_0^2 R$?

At resonance, the inductive and capacitive reactances cancel each other out ($X_L = X_C$), making the circuit purely resistive. Therefore, all the power is dissipated across the resistor.

Question

A series LCR circuit is connected to a $45 \sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^5 	ext{ rad s}^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega = 8 	imes 10^4 	ext{ rad s}^{-1}$, the current amplitude in the circuit is $0.05 I_0$. If $L = 50 	ext{ mH}$, match each entry in List-I with an appropriate value from List-II and choose the correct option.

List-I:

P. $I_0$ in mA

Q. The quality factor of the circuit

R. The bandwidth of the circuit in $	ext{rad s}^{-1}$

S. The peak power dissipated at resonance in Watt

List-II:

1. 44.4

2. 18

3. 400

4. 2250

5. 500

Sigma Percentile · Accepted Answer

The beauty of an LCR circuit lies in its dynamic response to varying frequencies. In this problem, we are given a series LCR circuit and asked to evaluate its behavior at resonance and at a shifted frequency. Let's embark on this journey step by step.

### Analyzing the Setup

We are given an alternating voltage source $V = 45 \sin(\omega t)$. The circuit has an inductor with $L = 50 	ext{ mH} = 50 	imes 10^{-3} 	ext{ H}$.

We are also given the resonant angular frequency, $\omega_0 = 10^5 	ext{ rad s}^{-1}$. At this frequency, the current amplitude is $I_0$.

However, when the frequency is lowered to $\omega = 8 	imes 10^4 	ext{ rad s}^{-1}$, the current amplitude drops significantly to $0.05 I_0$. This is our master clue to unlock the circuit's resistance.

### Uncovering the Hidden Capacitance

Before we analyze the shifted frequency, we need to find the missing piece of our circuit: the capacitance $C$.

We know that the resonant frequency is determined purely by the inductor and capacitor. The formula is:
$$ \omega_0 = \frac{1}{\sqrt{LC}} $$

By squaring both sides and rearranging for $C$, we get:
$$ C = \frac{1}{L \omega_0^2} $$

Substituting the given values:
$$ C = \frac{1}{50 	imes 10^{-3} 	imes (10^5)^2} = 2 	imes 10^{-9} 	ext{ F} $$

Now our circuit is fully defined in terms of its reactive components!

### The Frequency Shift and Impedance

Let's investigate what happens at the new frequency $\omega = 8 	imes 10^4 	ext{ rad s}^{-1}$.

First, we calculate the inductive reactance $X_L$:
$$ X_L = \omega L = (8 	imes 10^4) 	imes (50 	imes 10^{-3}) = 4000 \, \Omega = 4 	ext{ k}\Omega $$

Next, the capacitive reactance $X_C$:
$$ X_C = \frac{1}{\omega C} = \frac{1}{(8 	imes 10^4) 	imes (2 	imes 10^{-9})} = 6250 \, \Omega = 6.25 	ext{ k}\Omega $$

Notice that $X_C > X_L$. This means the circuit is **capacitive** at this lower frequency.

Now, let's use the current condition. The current is $0.05 I_0$, which is $\frac{I_0}{20}$. Since current is inversely proportional to impedance ($I = \frac{V_0}{Z}$), the new impedance $Z$ must be $20$ times the resistance $R$.
$$ Z = 20R $$

Using the impedance triangle formula:
$$ Z^2 = R^2 + (X_C - X_L)^2 $$

Substitute our knowns:
$$ (20R)^2 = R^2 + (6.25 - 4)^2 	ext{ k}\Omega^2 $$
$$ 400R^2 = R^2 + (2.25)^2 	ext{ k}\Omega^2 $$
$$ 399R^2 = 5.0625 	ext{ k}\Omega^2 $$

To simplify, we can approximate $\sqrt{399} \approx 20$. 
$$ R \approx \frac{2.25 	imes 10^3}{20} = 112.5 \, \Omega $$

### Calculating the Core Parameters

With $R$ in our arsenal, we can now evaluate all the entries in List-I.

**1. Peak Current at Resonance ($I_0$)**
At resonance, the impedance is purely resistive ($Z = R$).
$$ I_0 = \frac{V_0}{R} = \frac{45}{112.5} = 0.4 	ext{ A} = 400 	ext{ mA} $$
This matches entry (P) with (3).

**2. Quality Factor ($Q$)**
The Quality Factor determines the sharpness of the resonance.
$$ Q = \frac{1}{R} \sqrt{\frac{L}{C}} $$
$$ Q = \frac{1}{112.5} \sqrt{\frac{50 	imes 10^{-3}}{2 	imes 10^{-9}}} = \frac{1}{112.5} \sqrt{25 	imes 10^6} $$
$$ Q = \frac{5000}{112.5} = 44.4 $$
This matches entry (Q) with (1).

**3. Bandwidth ($\Delta \omega$)**
The bandwidth is the range of frequencies over which the power is at least half of its peak value.
$$ \Delta \omega = \frac{\omega_0}{Q} = \frac{10^5}{44.4} = 2250 	ext{ rad s}^{-1} $$
This matches entry (R) with (4).

**4. Peak Power Dissipated ($P_{	ext{max}}$)**
At resonance, all power is dissipated across the resistor.
$$ P_{	ext{max}} = I_0^2 R = (0.4)^2 	imes 112.5 = 0.16 	imes 112.5 = 18 	ext{ W} $$
This matches entry (S) with (2).

### Final Matrix Match

We have successfully decoded the entire circuit! The final mapping is:
- (P) $ightarrow$ (3)
- (Q) $ightarrow$ (1)
- (R) $ightarrow$ (4)
- (S) $ightarrow$ (2)

This problem beautifully intertwines the concepts of resonance, impedance, and power in AC circuits. Always remember to visualize how the reactances battle each other as the frequency shifts!

List-I

List-II

Select Matching Pairs:

Visualized Solution

$Series LCR Circuit Setup$

$Finding Capacitance C$

$Reactances at ω = 8 \times 1 0^{4} rad/s$

$Finding Resistance R$

$Calculating I_{0} (List-I: P)$

$Calculating Quality Factor Q (List-I: Q)$

$Calculating Bandwidth Δ ω (List-I: R)$

$Calculating Peak Power P_{max} (List-I: S)$

$Final Matrix Match$

$The Way Forward$

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

Uncovering the Hidden Capacitance

The Frequency Shift and Impedance

Calculating the Core Parameters

Final Matrix Match

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List-I

List-II

Select Matching Pairs:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Series LCR Circuit Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Capacitance C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reactances at ω=8×104 rad/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Resistance R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating I0​ (List-I: P)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Quality Factor Q (List-I: Q)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Bandwidth Δω (List-I: R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Peak Power Pmax​ (List-I: S)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Matrix Match

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward