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