Sharpness of Resonance and Bandwidth Derivations: Mathematical derivation of half-power frequencies, bandwidth, and Quality Factor (Q-factor) in a series LCR resonance circuit

1. What is Resonance?

A series LCR circuit contains a resistor R, inductor L and capacitor C connected in series with an AC source.

The impedance of the circuit is:

Z = √[R² + (X_L - X_C)²]

where:

• X_L = ωL = inductive reactance
• X_C = 1/(ωC) = capacitive reactance

At resonance:

X_L = X_C

Therefore:

ω₀ = 1/√LC

The resonant frequency is:

f₀ = 1/(2π√LC)

2. Current in Series LCR Circuit

For an AC source of RMS voltage V:

I = V/Z

Therefore:

I = V / √[R² + (ωL - 1/ωC)²]

At resonance:

ωL = 1/ωC

Therefore:

Z = R

and maximum current is:

I₀ = V/R

3. Sharpness of Resonance

Sharpness of resonance tells us how sharply the current reaches its maximum value near the resonant frequency.

A circuit with a large Q-factor has a sharp resonance and a narrow bandwidth.

A circuit with a small Q-factor has a less sharp resonance and a wider bandwidth.

4. Half-Power Frequencies

At resonance, current is maximum:

I₀ = V/R

The average power in the resistor is:

P = I²R

At half-power frequency:

P = P₀/2

Therefore:

I = I₀/√2

Since:

I = V/Z

we get:

Z = √2 R

5. Derivation of Half-Power Condition

For a series LCR circuit:

Z² = R² + (ωL - 1/ωC)²

At half-power:

Z = √2 R

Squaring:

Z² = 2R²

Therefore:

R² + (ωL - 1/ωC)² = 2R²

Hence:

(ωL - 1/ωC)² = R²

Therefore:

ωL - 1/ωC = ±R

These two frequencies are called the half-power frequencies or cut-off frequencies.

6. Derivation of Lower and Upper Half-Power Frequencies

Lower Frequency ω₁

For the lower frequency:

ω₁L - 1/(ω₁C) = -R

Multiplying by ω₁:

ω₁²L + Rω₁ - 1/C = 0

Solving the quadratic equation:

ω₁ = [-R + √(R² + 4L/C)]/(2L)

Upper Frequency ω₂

For the upper frequency:

ω₂L - 1/(ω₂C) = R

Multiplying by ω₂:

ω₂²L - Rω₂ - 1/C = 0

Therefore:

ω₂ = [R + √(R² + 4L/C)]/(2L)

7. Bandwidth

Bandwidth is the difference between upper and lower half-power frequencies.

BW = ω₂ - ω₁

Using the above expressions:

ω₂ - ω₁ = R/L

Therefore:

Bandwidth in angular frequency: Δω = R/L

In ordinary frequency:

Δf = R/(2πL)

8. Important Relation Between Half-Power Frequencies

The product of the lower and upper half-power angular frequencies is:

ω₁ω₂ = 1/(LC)

But:

ω₀² = 1/(LC)

Therefore:

ω₀ = √(ω₁ω₂)

Thus resonant angular frequency is the geometric mean of the half-power frequencies.

9. Quality Factor (Q-Factor)

The quality factor of a series LCR circuit is defined as:

Q = ω₀L/R

Since:

ω₀ = 1/√LC

we obtain:

Q = 1/R √(L/C)

Another important relation is:

Q = f₀/Δf

or:

Q = ω₀/Δω

10. Derivation of Q = f₀/Δf

We know:

Δω = R/L

Also:

ω₀ = 1/√LC

Therefore:

Q = ω₀L/R

But:

Δω=R/L

Hence:

Q = ω₀/Δω

Since:

ω=2πf

we get:

Q = f₀/Δf

11. Relation Between Sharpness and Q-Factor

Quality factor is a measure of the sharpness of resonance.

Q = f₀/Δf

Therefore:

• Large Q → narrow bandwidth → sharp resonance
• Small Q → wide bandwidth → less sharp resonance

For a highly selective tuning circuit, a high Q-factor is desirable.

12. Resonance Curve

The current-frequency curve of a series LCR circuit has a maximum at the resonant frequency.

At f₁ and f₂, current becomes:

I = I₀/√2

These points correspond to half of the maximum power.

Frequency	Current	Power
f₀	I₀	P₀
f₁	I₀/√2	P₀/2
f₂	I₀/√2	P₀/2

13. Physical Meaning of Bandwidth

Bandwidth represents the range of frequencies over which the circuit operates effectively around resonance.

Bandwidth = f₂ - f₁

For a series LCR circuit:

Δf = R/(2πL)

Thus, increasing resistance increases bandwidth.

14. Effect of Resistance on Resonance

If resistance R is increased:

• Bandwidth increases.
• Q-factor decreases.
• Resonance becomes less sharp.
• Maximum current decreases.

If R is decreased:

• Bandwidth decreases.
• Q-factor increases.
• Resonance becomes sharper.

15. Effect of Inductance

For a series LCR circuit:

Δf = R/(2πL)

Therefore, increasing L decreases bandwidth if R remains constant.

Also:

Q = ω₀L/R

Hence L affects the sharpness of resonance.

16. Effect of Capacitance

The resonant frequency is:

f₀=1/(2π√LC)

Therefore:

• Increasing C decreases resonant frequency.
• Decreasing C increases resonant frequency.

17. Solved Example 1 — Resonant Frequency

A series LCR circuit has L=0.5 H and C=20 μF. Find the resonant frequency.

f₀=1/(2π√LC)

f₀= 1/[2π√(0.5×20×10⁻⁶)]

f₀ ≈ 50.3 Hz

18. Solved Example 2 — Bandwidth

A series LCR circuit has R=10 Ω and L=0.5 H. Find its bandwidth.

Δf=R/(2πL)

Δf=10/(2π×0.5)

Δf≈3.18 Hz

19. Solved Example 3 — Q-Factor

A series LCR circuit has L=2 H, R=10 Ω and resonant angular frequency ω₀=100 rad/s. Find Q.

Q=ω₀L/R

Q=(100)(2)/10

Q=20

20. Solved Example 4 — Half-Power Current

At resonance, the current in a series LCR circuit is 10 A. Find the current at half-power frequency.

I=I₀/√2

I=10/√2

I≈7.07 A

21. Solved Example 5 — Half-Power Frequencies

The half-power angular frequencies of a circuit are 80 rad/s and 120 rad/s. Find the resonant angular frequency and bandwidth.

ω₀=√(ω₁ω₂)

ω₀=√(80×120)

ω₀≈97.98 rad/s

Bandwidth:

Δω=ω₂-ω₁

Δω=40 rad/s

22. Important Formula Sheet

Quantity	Formula
Impedance	Z=√[R²+(ωL-1/ωC)²]
Resonant angular frequency	ω₀=1/√LC
Resonant frequency	f₀=1/(2π√LC)
Maximum current	I₀=V/R
Half-power current	I=I₀/√2
Half-power condition	|ωL-1/ωC|=R
Lower frequency	ω₁=[√(R²+4L/C)-R]/2L
Upper frequency	ω₂=[√(R²+4L/C)+R]/2L
Bandwidth	Δω=R/L
Bandwidth in Hz	Δf=R/(2πL)
Q-factor	Q=ω₀L/R
Alternative Q	Q=(1/R)√(L/C)
Q and bandwidth	Q=f₀/Δf
Resonant frequency relation	ω₀=√(ω₁ω₂)

Section A — 1 Mark MCQs
10 Questions

Q1. At resonance in a series LCR circuit: 1 Mark

A) X_L > X_C

B) X_L < X_C

C) X_L = X_C

D) X_L = R

Answer: C

Q2. The impedance of a series LCR circuit at resonance is: 1 Mark

A) Zero

B) R

C) XL

D) XC

Answer: B

Q3. At half-power frequency, current is: 1 Mark

A) I₀

B) I₀/2

C) I₀/√2

D) √2I₀

Answer: C

Q4. Bandwidth of a series LCR circuit in angular frequency is: 1 Mark

A) R/L

B) L/R

C) RC

D) 1/RC

Answer: A

Q5. Q-factor of a series LCR circuit is: 1 Mark

A) R/ω₀L

B) ω₀L/R

C) RC

D) RLC

Answer: B

Q6. A high Q-factor indicates: 1 Mark

A) Broad resonance

B) Sharp resonance

C) No resonance

D) Zero current

Answer: B

Q7. Resonant angular frequency is: 1 Mark

A) √LC

B) 1/√LC

C) L/C

D) C/L

Answer: B

Q8. Increasing R in a series LCR circuit: 1 Mark

A) Decreases bandwidth

B) Increases bandwidth

C) Makes bandwidth zero

D) Has no effect

Answer: B

Q9. The two half-power frequencies are also called: 1 Mark

A) Resonant frequencies

B) Cut-off frequencies

C) Natural frequencies only

D) Zero frequencies

Answer: B

Q10. The relation between resonant and half-power angular frequencies is: 1 Mark

A) ω₀=ω₁+ω₂

B) ω₀=ω₁-ω₂

C) ω₀=√(ω₁ω₂)

D) ω₀=ω₁/ω₂

Answer: C

Section B — 2 Marks
10 Questions

Q11. Define bandwidth of a series LCR circuit. 2 Marks

Bandwidth is the difference between the upper and lower half-power frequencies.

Δf=f₂-f₁

Q12. What is meant by sharpness of resonance? 2 Marks

Sharpness indicates how rapidly the current falls when the frequency moves away from resonance. Higher Q means sharper resonance.

Q13. What is the half-power condition? 2 Marks

At half-power frequency:

I=I₀/√2

and:

|X_L-X_C|=R

Q14. Write the expression for bandwidth in Hz. 2 Marks

Δf=R/(2πL)

Q15. Write two expressions for Q-factor. 2 Marks

Q=ω₀L/R

Q=f₀/Δf

Q16. How does increasing resistance affect resonance? 2 Marks

Increasing R increases bandwidth and decreases Q-factor. Hence resonance becomes less sharp.

Q17. What is the current at resonance? 2 Marks

At resonance impedance is R:

I₀=V/R

This is the maximum current.

Q18. Why is power maximum at resonance? 2 Marks

At resonance impedance is minimum and equal to R. Hence current is maximum, and power P=I²R becomes maximum.

Q19. What is the significance of Q=f₀/Δf? 2 Marks

It shows that a high Q-factor corresponds to a small bandwidth relative to the resonant frequency, giving sharp resonance.

Q20. State the relation between half-power frequencies and resonant frequency. 2 Marks

ω₀=√(ω₁ω₂)

The resonant angular frequency is the geometric mean of the two half-power angular frequencies.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the half-power condition for a series LCR circuit. 3 Marks

At resonance:

I₀=V/R

At half power:

I=I₀/√2

Since I=V/Z:

Z=√2R

But:

Z²=R²+(ωL-1/ωC)²

Therefore:

|ωL-1/ωC|=R

Q22. Derive the bandwidth of a series LCR circuit. 3 Marks

The half-power frequencies satisfy:

ω₁L-1/(ω₁C)=-R

and:

ω₂L-1/(ω₂C)=R

Solving:

ω₁=[√(R²+4L/C)-R]/2L

ω₂=[√(R²+4L/C)+R]/2L

Subtracting:

Δω=ω₂-ω₁=R/L

Q23. Derive Q=ω₀L/R. 3 Marks

For a series LCR circuit:

Q=ω₀/Δω

Bandwidth:

Δω=R/L

Therefore:

Q=ω₀/(R/L)

Hence:

Q=ω₀L/R

Q24. Show that Q=f₀/Δf. 3 Marks

We know:

Q=ω₀/Δω

Since:

ω₀=2πf₀

and:

Δω=2πΔf

Therefore:

Q=f₀/Δf

Q25. A series LCR circuit has R=20 Ω and L=2 H. Find its bandwidth. 3 Marks

Δf=R/(2πL)

Δf=20/(2π×2)

Δf≈1.59 Hz

Q26. A series LCR circuit has resonant frequency 1000 Hz and bandwidth 20 Hz. Find Q-factor. 3 Marks

Q=f₀/Δf

Q=1000/20

Q=50

Q27. The half-power frequencies of a series LCR circuit are 900 Hz and 1100 Hz. Find resonant frequency and bandwidth. 3 Marks

Bandwidth:

Δf=f₂-f₁

Δf=1100-900=200 Hz

Resonant frequency:

f₀=√(f₁f₂)

f₀=√(900×1100)

f₀≈995 Hz,   Δf=200 Hz

Q28. At resonance, current in a series LCR circuit is 8 A. Find current and power ratio at half-power frequency. 3 Marks

At half power:

I=I₀/√2

I=8/√2

I≈5.66 A

Power ratio:

P/P₀=1/2

Therefore power is 50% of maximum power.

Q29. A series LCR circuit has L=1 H and R=10 Ω. Its resonant angular frequency is 100 rad/s. Find Q-factor and bandwidth. 3 Marks

Q-factor:

Q=ω₀L/R

Q=(100)(1)/10=10

Bandwidth:

Δω=R/L

Q=10,   Δω=10 rad/s

Q30. Explain why a high-Q series LCR circuit is preferred for frequency selection. 3 Marks

For a series LCR circuit:

Q=f₀/Δf

A high Q means Δf is small. Therefore the circuit responds strongly to a narrow range of frequencies around resonance and rejects frequencies away from resonance.

High Q → Narrow Bandwidth → Sharp Resonance → Better Frequency Selection

23. Quick Revision

Resonance

ω₀=1/√LC

f₀=1/(2π√LC)

Half Power

I=I₀/√2

|ωL-1/ωC|=R

Bandwidth

Δω=R/L

Δf=R/(2πL)

Q-Factor

Q=ω₀L/R

Q=f₀/Δf

Sharpness

High Q → Sharp Resonance → Small Bandwidth

Low Q → Broad Resonance → Large Bandwidth

Sharpness of Resonance and Bandwidth
Series LCR Circuit | Derivations + Formulae + Numericals + 30 CBSE/JEE Questions