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

Transient RC, LR, and LC Circuits: Solving first-order differential equations representing the growth/decay of current in LR circuits and charging/discharging in RC circuits

1. What is a Transient Circuit?

A transient circuit is a circuit in which current or voltage changes with time immediately after switching ON or OFF.

Transient response: The temporary response of a circuit before it reaches its steady state.

The three important circuits are:

• RC circuit → resistor + capacitor
• LR circuit → inductor + resistor
• LC circuit → inductor + capacitor

2. RC Circuit — Charging of Capacitor

Consider a battery of emf E, resistor R and capacitor C connected in series.

Using Kirchhoff's loop law:

E - iR - q/C = 0

Since:

i = dq/dt

we get:

E - R dq/dt - q/C = 0

Therefore:

R dq/dt = E - q/C

3. Differential Equation of Charging RC Circuit

Rearranging:

dq/dt = (E/R) - q/(RC)

The solution of this first-order differential equation is:

q = CE(1 - e^-t/RC)

Therefore capacitor voltage is:

V_C = E(1 - e^-t/RC)

4. Current During Charging

Current is:

i = dq/dt

Differentiating:

i = (E/R)e^-t/RC

Therefore:

i = I₀e^-t/RC

where:

I₀ = E/R

At t=0, current is maximum. As time increases, current decreases exponentially.

5. Time Constant of RC Circuit

The quantity RC is called the time constant.

τ = RC

At:

t = τ = RC

the charge becomes:

q = CE(1-e^-1)

Approximately:

q ≈ 0.632 CE

Thus, after one time constant, the capacitor becomes about 63.2% charged.

6. RC Circuit — Discharging

Suppose a charged capacitor is connected across a resistor without a battery.

Using Kirchhoff's law:

q/C + iR = 0

Since:

i = -dq/dt

we get:

q/C + R(-dq/dt)=0

Therefore:

dq/dt = -q/(RC)

7. Solution of RC Discharging

The solution of:

dq/dt = -q/(RC)

is:

q = Q₀e^-t/RC

Current magnitude:

i = (Q₀/RC)e^-t/RC

Since:

Q₀/C = V₀

we can write:

i = V₀/R × e^-t/RC

8. RC Charging and Discharging

Quantity	Charging	Discharging
Charge	q=CE(1-e-t/RC)	q=Q₀e-t/RC
Current	i=(E/R)e-t/RC	i=(V₀/R)e-t/RC
Voltage	V=E(1-e-t/RC)	V=V₀e-t/RC
Time constant	τ=RC

9. LR Circuit — Growth of Current

Consider a battery of emf E, resistance R and inductor L connected in series.

Kirchhoff's loop law gives:

E - iR - L di/dt = 0

Therefore:

L di/dt + Ri = E

or:

di/dt + (R/L)i = E/L

This is a first-order differential equation.

10. Solution of LR Growth

For an initially unenergized inductor:

i = (E/R)(1-e^-Rt/L)

At long time:

i_∞=E/R

Therefore current gradually rises from zero to E/R.

11. Time Constant of LR Circuit

τ = L/R

At:

t=τ=L/R

current becomes:

i = (E/R)(1-e^-1)

Therefore:

i ≈ 0.632(E/R)

After one time constant, current reaches approximately 63.2% of its final value.

12. LR Circuit — Decay of Current

Suppose the battery is removed and the resistor and inductor form a closed circuit.

Kirchhoff's law:

L di/dt + Ri = 0

Therefore:

di/dt = -(R/L)i

Solution:

i = I₀e^-Rt/L

or:

i=I₀e^-t/τ

where:

τ=L/R

13. RC vs LR Circuits

Feature	RC	LR
Storage element	Capacitor	Inductor
Stored quantity	Charge / Electric energy	Magnetic energy
Time constant	RC	L/R
Growth	q=Qmax(1-e-t/RC)	i=Imax(1-e-Rt/L)
Decay	q=Q₀e-t/RC	i=I₀e-Rt/L

14. LC Circuit

An ideal LC circuit contains an inductor L and capacitor C with no resistance.

Energy continuously transfers between the capacitor and inductor.

U_C=q²/(2C)

Magnetic energy:

U_L=½Li²

Total energy:

U = q²/(2C) + ½Li² = constant

15. Differential Equation of LC Circuit

Using Kirchhoff's law:

q/C + L di/dt = 0

Since:

i=dq/dt

we obtain:

L d²q/dt² + q/C = 0

Therefore:

d²q/dt² + q/(LC)=0

This is the equation of simple harmonic motion.

16. LC Oscillation

The angular frequency of LC oscillations is:

ω = 1/√(LC)

Frequency:

f = 1/(2π√LC)

Time period:

T = 2π√LC

17. Charge and Current in LC Circuit

If initially the capacitor has maximum charge Q₀:

q = Q₀ cos(ωt)

Current:

i=dq/dt

Therefore:

i=-ωQ₀ sin(ωt)

The current amplitude is:

I₀=ωQ₀

18. Energy Exchange in LC Circuit

At maximum capacitor charge:

U_C,max=Q₀²/(2C)

Current is zero.

When capacitor charge becomes zero:

U_L,max=½LI₀²

Current is maximum.

Energy continuously changes between electric energy in the capacitor and magnetic energy in the inductor.

19. Solved Example — RC Charging

A 10 V battery is connected to a 2 MΩ resistor and a 5 μF capacitor. Find the time constant.

τ=RC

τ=(2×10⁶)(5×10⁻⁶)

τ=10 s

20. Solved Example — LR Growth

An LR circuit has L=4 H and R=2 Ω. Find its time constant.

τ=L/R

τ=4/2

τ=2 s

21. Solved Example — LC Frequency

An LC circuit has L=1 H and C=1 μF. Find its angular frequency.

ω=1/√LC

ω= 1/√(1×10⁻⁶)

ω=1000 rad/s

22. Important Formula Sheet

Circuit	Important Formula
RC charging	q=CE(1-e-t/RC)
RC charging current	i=(E/R)e-t/RC
RC discharging	q=Q₀e-t/RC
RC time constant	τ=RC
LR growth	i=(E/R)(1-e-Rt/L)
LR decay	i=I₀e-Rt/L
LR time constant	τ=L/R
LC differential equation	d²q/dt²+q/(LC)=0
LC angular frequency	ω=1/√LC
LC frequency	f=1/(2π√LC)
LC time period	T=2π√LC

Section A — 1 Mark MCQs
10 Questions

Q1. The time constant of an RC circuit is: 1 Mark

A) R/C

B) RC

C) 1/RC

D) C/R

Answer: B

Q2. The time constant of an LR circuit is: 1 Mark

A) LR

B) R/L

C) L/R

D) 1/LR

Answer: C

Q3. During RC charging, the final charge is: 1 Mark

A) CE

B) E/C

C) CR

D) ER

Answer: A

Q4. During LR growth, final current is: 1 Mark

A) ER

B) E/R

C) R/E

D) EL

Answer: B

Q5. At t=0, the current in an initially unenergized LR circuit is: 1 Mark

A) E/R

B) Zero

C) Infinite

D) R/E

Answer: B

Q6. After one time constant during RC charging, capacitor charge is approximately: 1 Mark

A) 37%

B) 50%

C) 63.2%

D) 100%

Answer: C

Q7. The angular frequency of an ideal LC circuit is: 1 Mark

A) √LC

B) 1/√LC

C) L/C

D) C/L

Answer: B

Q8. The energy stored in an inductor is: 1 Mark

A) LI²

B) ½LI²

C) ½CV²

D) L/I²

Answer: B

Q9. The charge on a discharging capacitor varies as: 1 Mark

A) e^t/RC

B) e^-t/RC

C) t²

D) t

Answer: B

Q10. An ideal LC circuit exhibits: 1 Mark

A) Damped oscillations

B) SHM-like oscillations

C) No energy transfer

D) DC only

Answer: B

Section B — 2 Marks
10 Questions

Q11. Define time constant of an RC circuit. 2 Marks

The time constant is the time required for a charging capacitor to reach 63.2% of its final charge.

τ=RC

Q12. Write the equation for charge during RC charging. 2 Marks

q=CE(1-e^-t/RC)

Q13. Write the equation for current during LR growth. 2 Marks

i=(E/R)(1-e^-Rt/L)

Q14. Why does current in an LR circuit not increase suddenly? 2 Marks

The inductor opposes any sudden change in current by producing a back EMF. Hence current rises gradually.

Q15. What happens to the capacitor current after a long time of charging? 2 Marks

The current approaches zero because the capacitor becomes fully charged and behaves like an open circuit for DC.

Q16. Write the equation for current decay in an LR circuit. 2 Marks

i=I₀e^-Rt/L

Q17. What is the time constant of an LR circuit? 2 Marks

τ=L/R

Q18. Write the differential equation of an LC circuit. 2 Marks

d²q/dt²+q/(LC)=0

Q19. Write the frequency of an LC oscillator. 2 Marks

f=1/(2π√LC)

Q20. What happens to the energy in an ideal LC circuit? 2 Marks

Energy continuously transfers between the capacitor's electric field and the inductor's magnetic field. Total energy remains constant.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the expression for charge during charging of an RC circuit. 3 Marks

Kirchhoff's law:

E-iR-q/C=0

Since:

i=dq/dt

Therefore:

R dq/dt=E-q/C

Solving this first-order differential equation:

q=CE(1-e^-t/RC)

Q22. Derive the current during charging of an RC circuit. 3 Marks

Charge:

q=CE(1-e^-t/RC)

Current:

i=dq/dt

Therefore:

i=(E/R)e^-t/RC

Q23. Derive the equation for discharge of a capacitor through a resistor. 3 Marks

Kirchhoff's law:

q/C+Ri=0

Since:

i=-dq/dt

Therefore:

dq/dt=-q/(RC)

Solving:

q=Q₀e^-t/RC

Q24. Derive the current growth equation for an LR circuit. 3 Marks

Kirchhoff's law:

E-Ri-Ldi/dt=0

Therefore:

Ldi/dt+Ri=E

For i=0 at t=0, solution is:

i=(E/R)(1-e^-Rt/L)

Q25. Derive the decay equation for current in an LR circuit. 3 Marks

After removing the battery:

Ldi/dt+Ri=0

Thus:

di/dt=-(R/L)i

Solving:

i=I₀e^-Rt/L

Q26. A 20 V battery is connected to a 4 kΩ resistor and a 5 μF capacitor. Find the time constant and maximum charge. 3 Marks

Time constant:

τ=RC

τ=(4000)(5×10⁻⁶)=0.02 s

Maximum charge:

Q=CE

Q=(5×10⁻⁶)(20)

τ=0.02 s,   Q=100 μC

Q27. An LR circuit has L=2 H and R=4 Ω. Find its time constant and the fraction of final current after one time constant. 3 Marks

τ=L/R=2/4=0.5 s

After one time constant:

i/I_∞=1-e^-1

τ=0.5 s and i≈0.632 i_∞

Q28. An LC circuit has L=0.5 H and C=2 μF. Find angular frequency and time period. 3 Marks

Angular frequency:

ω=1/√LC

ω= 1/√[(0.5)(2×10⁻⁶)]

ω=1000 rad/s

Time period:

T=2π/ω

T≈6.28×10⁻³ s

Q29. A capacitor initially has charge 100 μC. It discharges through a resistor. If RC=2 s, find the charge after 2 s. 3 Marks

q=Q₀e^-t/RC

Here:

t=RC

Therefore:

q=100e^-1 μC

q≈36.8 μC

Q30. Explain the analogy between RC charging and LR current growth. 3 Marks

In both circuits, the response changes exponentially with time. For RC charging:

q=Q_max(1-e^-t/RC)

For LR growth:

i=I_max(1-e^-Rt/L)

Both have a characteristic time constant.

RC: τ=RC    |    LR: τ=L/R

23. Quick Revision

RC Circuit

q=CE(1-e^-t/RC)

i=(E/R)e^-t/RC

τ=RC

---

LR Circuit

i=(E/R)(1-e^-Rt/L)

i=I₀e^-Rt/L

τ=L/R

---

LC Circuit

d²q/dt²+q/(LC)=0

ω=1/√LC

T=2π√LC

f=1/(2π√LC)

Transient RC, LR and LC Circuits
CBSE + JEE Competitive Physics | Theory + Differential Equations + Numericals + 30 Questions

Mutual Inductance of Concentric and Non-Coplanar Systems: Deriving the coefficient of mutual inductance between a very large coplanar loop (e.g., a square) and a tiny loop

1. Introduction

When two coils or loops are placed near each other, a changing current in one loop can produce magnetic flux through the other loop. The corresponding induced EMF is described using mutual inductance.

M = Φ₂₁ / I₁

where:

• M = mutual inductance
• Φ₂₁ = magnetic flux through coil 2 due to current I₁ in coil 1
• I₁ = current in coil 1

2. Mutual Inductance

If current in coil 1 changes, the induced EMF in coil 2 is:

ε₂ = -M dI₁/dt

The negative sign represents Lenz's law.

Important: Mutual inductance depends on the geometry, relative orientation, distance and number of turns of the two circuits.

3. Large Loop and Tiny Loop

Consider a very large square loop carrying current I. A tiny loop of area A is placed near its centre.

If the magnetic field produced by the large loop is approximately uniform over the small loop, then:

Φ = BA cosθ

For the small loop whose plane is perpendicular to the magnetic field:

Φ = BA

Therefore:

M = Φ/I

Hence:

M = BA/I

4. Magnetic Field at the Centre of a Square Loop

Let the side of the large square loop be a and current I flows through it. The magnetic field at the centre is obtained by adding the fields due to all four sides.

For a finite straight conductor:

B = μ₀I/(4πr)(sinθ₁ + sinθ₂)

At the centre of a square:

• Distance from centre to each side = a/2
• Angle made by each side at the centre = 45° on each end

Thus for one side:

B₁ = μ₀I/(4π(a/2)) (sin45° + sin45°)

Since:

sin45° = 1/√2

We get:

B₁ = μ₀I√2/(2πa)

There are four identical sides:

B = 2√2 μ₀I/(πa)

5. Mutual Inductance with a Tiny Loop

Let the small loop have area A and be sufficiently small that the magnetic field can be considered uniform over it.

The magnetic flux through the small loop is:

Φ = BA

Substituting the field at the centre of the square:

Φ = [2√2 μ₀I/(πa)]A

Therefore:

M = Φ/I

Hence:

M = 2√2 μ₀A/(πa)

This is the important result for a tiny loop placed at the centre of a large square loop, with the small loop oriented so that its normal is along the magnetic field.

6. If the Small Loop is Tilted

If the normal to the small loop makes an angle θ with the magnetic field, the magnetic flux is:

Φ = BA cosθ

Therefore:

M = BA cosθ / I

For the square loop:

M = 2√2 μ₀A cosθ/(πa)

7. Non-Coplanar Arrangement

When the small loop is not in the same plane as the large loop, the mutual inductance depends on the angle between the magnetic field at the small loop and the normal to the small loop.

M = Φ/I

If the field is approximately uniform across the small loop:

M = BA cosθ/I

Therefore:

M = BAI⁻¹ cosθ

8. Effect of Distance

The approximation:

Φ ≈ BA

is valid when the small loop is sufficiently small and lies in a region where the magnetic field does not change appreciably.

If the small loop is moved away from the centre, the magnetic field at its position changes and the simple centre-field expression may no longer be used directly.

For a tiny loop: M ≈ B(position) A cosθ / I

9. Reciprocity of Mutual Inductance

For two passive circuits in a linear medium:

M₁₂ = M₂₁ = M

This means mutual inductance of coil 1 with coil 2 is equal to that of coil 2 with coil 1.

10. SI Unit

The SI unit of mutual inductance is:

Henry (H)

One henry is the mutual inductance when a current changing at 1 A/s in one circuit produces an induced EMF of 1 V in the other circuit.

11. Solved Example 1

A square loop of side 2 m carries a current of 5 A. A tiny loop of area 0.01 m² is placed at its centre with its normal along the magnetic field. Find the mutual inductance.

For a square:

B = 2√2 μ₀I/(πa)

Mutual inductance:

M = 2√2 μ₀A/(πa)

Putting:

a=2 m,   A=0.01 m²

M = 2√2(4π×10⁻⁷)(0.01)/(π×2)

M ≈ 5.66 × 10⁻⁹ H

12. Solved Example 2 — Tilted Loop

For the same arrangement, if the normal of the small loop makes an angle 60° with the magnetic field, find the new mutual inductance.

M = M₀ cos60°

Since:

cos60° = 1/2

Therefore:

M = M₀/2

13. Solved Example 3 — Induced EMF

A tiny loop has mutual inductance M = 2 mH with a large loop. If current in the large loop changes at 500 A/s, find the induced EMF.

ε = -M dI/dt

Magnitude:

|ε|=(2×10⁻³)(500)

|ε| = 1 V

14. Solved Example 4 — Zero Mutual Inductance

If the normal of the small loop is perpendicular to the magnetic field, then θ=90°.

M = BA cos90°/I

Since:

cos90°=0

Therefore:

M = 0

15. Important Formula Sheet

Quantity	Formula
Mutual inductance	M=Φ/I
Induced EMF	ε=-M dI/dt
Flux	Φ=BA cosθ
Small loop	M=BA cosθ/I
Centre field of square	B=2√2 μ₀I/(πa)
Square + tiny loop	M=2√2 μ₀A cosθ/(πa)
SI unit	Henry (H)
Reciprocity	M₁₂=M₂₁

Section A — 1 Mark MCQs
10 Questions

Q1. The SI unit of mutual inductance is: 1 Mark

A) Tesla

B) Weber

C) Henry

D) Ohm

Answer: C

Q2. Mutual inductance is defined as: 1 Mark

A) I/Φ

B) Φ/I

C) ΦI

D) I²/Φ

Answer: B

Q3. The induced EMF due to mutual induction is: 1 Mark

A) M/I

B) MI

C) -M dI/dt

D) M dI/dt²

Answer: C

Q4. For a small loop in a uniform magnetic field, flux is: 1 Mark

A) BA

B) BA cosθ

C) B/A

D) A/B

Answer: B

Q5. If the normal of the small loop is parallel to B, θ is: 1 Mark

A) 0°

B) 30°

C) 60°

D) 90°

Answer: A

Q6. If θ=90°, mutual inductance is: 1 Mark

A) Maximum

B) Zero

C) Infinite

D) Negative always

Answer: B

Q7. Mutual inductance of two passive circuits satisfies: 1 Mark

A) M₁₂ = -M₂₁

B) M₁₂ = M₂₁

C) M₁₂ = 0 always

D) M₁₂ = 2M₂₁

Answer: B

Q8. The magnetic field at the centre of a square loop is proportional to: 1 Mark

A) I/a

B) Ia

C) a/I

D) I/a²

Answer: A

Q9. Mutual inductance increases if the area of the small loop: 1 Mark

A) Increases

B) Decreases

C) Becomes zero

D) Has no effect

Answer: A

Q10. Lenz's law determines the: 1 Mark

A) Magnitude only

B) Direction of induced current

C) Resistance

D) Area

Answer: B

Section B — 2 Marks
10 Questions

Q11. Define mutual inductance. 2 Marks

Mutual inductance is the flux linked with one circuit per unit current in the other circuit:

M=Φ/I

Q12. Write the relation between induced EMF and mutual inductance. 2 Marks

ε=-M dI/dt

Q13. What factors affect mutual inductance? 2 Marks

It depends on the geometry, size, relative position, orientation, number of turns and magnetic properties of the medium.

Q14. What happens to M when the small loop is tilted through 60°? 2 Marks

M=M₀cos60°=M₀/2

Q15. Why can the magnetic field be considered uniform over a tiny loop? 2 Marks

Because the dimensions of the tiny loop are much smaller than the scale over which the magnetic field changes appreciably.

Q16. State the reciprocity relation for mutual inductance. 2 Marks

M₁₂=M₂₁

Q17. What is the mutual inductance if the magnetic flux through the second loop is zero? 2 Marks

M=Φ/I=0

Therefore mutual inductance is zero.

Q18. Write the magnetic field at the centre of a square loop of side a. 2 Marks

B=2√2 μ₀I/(πa)

Q19. What happens to mutual inductance if the area of the small loop is doubled? 2 Marks

Since M is directly proportional to A:

M∝A

Doubling A doubles M, under the small-loop approximation.

Q20. Why does a changing current produce induced EMF in the second loop? 2 Marks

A changing current produces a changing magnetic field. This changes the magnetic flux through the second loop and hence induces EMF according to Faraday's law.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the mutual inductance between a large square loop and a tiny loop at its centre. 3 Marks

For a square loop of side a:

B=2√2 μ₀I/(πa)

Flux through tiny loop of area A:

Φ=BA

Therefore:

M=Φ/I

Hence:

M=2√2 μ₀A/(πa)

Q22. Derive the mutual inductance when the tiny loop is tilted at angle θ. 3 Marks

Flux:

Φ=BAcosθ

Therefore:

M=BAcosθ/I

For a square loop:

M=2√2 μ₀A cosθ/(πa)

Q23. A square loop of side 1 m carries current I. Find its magnetic field at the centre. 3 Marks

B=2√2 μ₀I/(πa)

For a=1 m:

B=2√2 μ₀I/π

Q24. A tiny loop of area 2×10⁻³ m² is placed at the centre of a square loop of side 2 m. Find M. 3 Marks

M=2√2 μ₀A/(πa)

Putting values:

M= 2√2(4π×10⁻⁷)(2×10⁻³)/(π×2)

M≈1.13×10⁻⁹ H

Q25. If mutual inductance is 5 mH and current changes at 200 A/s, find induced EMF. 3 Marks

|ε|=M dI/dt

|ε|=(5×10⁻³)(200)

|ε|=1 V

Q26. Explain why mutual inductance becomes zero when the normal of the small loop is perpendicular to B. 3 Marks

Flux:

Φ=BAcosθ

For θ=90°:

Φ=BAcos90°=0

Therefore:

M=Φ/I=0

Q27. A small loop is rotated from θ=0° to θ=60°. By what fraction does its mutual inductance change? 3 Marks

Initially:

M₀=BA/I

Finally:

M=M₀cos60°=M₀/2

Therefore the mutual inductance becomes half its initial value.

Decrease = 50%

Q28. A large square loop has side 4 m and carries 10 A current. A tiny loop of area 0.02 m² is at its centre. Find the flux through the tiny loop. Take μ₀=4π×10⁻⁷ H/m. 3 Marks

Field at centre:

B=2√2 μ₀I/(πa)

B= 2√2(4π×10⁻⁷)(10)/(π×4)

B≈2.83×10⁻⁶ T

Flux:

Φ=BA

Φ=(2.83×10⁻⁶)(0.02)

Φ≈5.66×10⁻⁸ Wb

Q29. Show that the mutual inductance between a square loop and a tiny loop is independent of the current in the large loop. 3 Marks

Magnetic field at centre:

B=2√2 μ₀I/(πa)

Flux:

Φ=BA

Therefore:

Φ= 2√2 μ₀IA/(πa)

Mutual inductance:

M=Φ/I

Thus:

M=2√2 μ₀A/(πa)

The current I cancels out.

Q30. Explain the physical meaning of mutual inductance and state two factors affecting it. 3 Marks

Mutual inductance measures how effectively a changing current in one circuit produces magnetic flux and induced EMF in another circuit.

M=Φ/I

Two important factors are:

• Relative position and orientation of the loops.
• Size/geometry and number of turns of the loops.

The magnetic properties of the medium can also affect mutual inductance.

16. Quick Revision

Remember these results:

M=Φ/I

ε=-M dI/dt

Φ=BAcosθ

B_square=2√2 μ₀I/(πa)

M_square-small = 2√2 μ₀A cosθ/(πa)

M₁₂=M₂₁

Special cases:

θ=0° → M is maximum.

θ=90° → M=0.

If A doubles → M doubles.

If side a of the large square increases → M decreases as 1/a, within the centre-field/tiny-loop approximation.

Mutual Inductance of Concentric and Non-Coplanar Systems
CBSE + JEE Competitive Physics | Concept + Derivation + Numericals + 30 Questions