Physics Electromagnetism - How to Solve: Inductance and RL Circuits (Growth/Decay of Current) – IIT JEE Guide

How to Solve: Inductance and RL Circuits (Growth/Decay of Current) – IIT JEE Guide

(For Students & Teachers – Ready-to-Record Script Included)

Introduction

Mastering RL circuits unlocks 8–12 marks in IIT JEE (Main + Advanced) every year—enough to push you into the top 1000. Whether it’s a 3-mark growth/decay question or a 4-mark energy-based problem, this topic is a guaranteed scorer if you follow the exact steps below.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you’re 100% clear on these:
1. Kirchhoff’s Voltage Law (KVL) – Sum of voltage drops = source voltage.
2. Exponential functions – How ( e^{-t/\tau} ) and ( (1 - e^{-t/\tau}) ) behave.
3. Time constant (τ) – Definition and physical meaning in RC/RL circuits.

(If any of these feel shaky, pause here and review them first.)

KEY TERMS & FORMULAS

Key Terms

Formulas

STEP-BY-STEP METHOD

(Follow these exactly—no shortcuts!)

For Growth of Current (Switch Closed at t=0)

1. Identify the circuit:
2. Battery (V), resistor (R), inductor (L) in series.

3. Switch closes at ( t = 0 ).

4. Write KVL: ( V - IR - L \frac{dI}{dt} = 0 ) (Battery - resistor drop - inductor drop = 0)

5. Rearrange for ( \frac{dI}{dt} ): ( \frac{dI}{dt} = \frac{V - IR}{L} )

6. Solve the differential equation:

7. Separate variables: ( \frac{dI}{V - IR} = \frac{dt}{L} )

8. Integrate both sides: ( \int \frac{dI}{V - IR} = \int \frac{dt}{L} ) ( -\frac{1}{R} \ln(V - IR) = \frac{t}{L} + C )

9. Apply initial condition (I=0 at t=0):

10. ( C = -\frac{1}{R} \ln(V) )

11. Substitute back: ( -\frac{1}{R} \ln(V - IR) = \frac{t}{L} - \frac{1}{R} \ln(V) ) ( \ln \left( \frac{V - IR}{V} \right) = -\frac{Rt}{L} ) ( \frac{V - IR}{V} = e^{-Rt/L} ) ( I(t) = \frac{V}{R} (1 - e^{-Rt/L}) )

12. Final formula: ( I(t) = I_0 (1 - e^{-t/\tau}) ) where ( I_0 = \frac{V}{R} ) and ( \tau = \frac{L}{R} ).

For Decay of Current (Switch Opened at t=0)

1. Identify the circuit:
2. Inductor (L) and resistor (R) in series (no battery).

3. Initial current ( I_0 ) flows at ( t = 0 ).

4. Write KVL: ( -IR - L \frac{dI}{dt} = 0 ) (Resistor drop + inductor drop = 0)

5. Rearrange for ( \frac{dI}{dt} ): ( \frac{dI}{dt} = -\frac{IR}{L} )

6. Solve the differential equation:

7. Separate variables: ( \frac{dI}{I} = -\frac{R}{L} dt )

8. Integrate both sides: ( \int \frac{dI}{I} = -\frac{R}{L} \int dt ) ( \ln(I) = -\frac{Rt}{L} + C )

9. Apply initial condition (I=I₀ at t=0):

10. ( C = \ln(I_0) )

11. Substitute back: ( \ln(I) = -\frac{Rt}{L} + \ln(I_0) ) ( \ln \left( \frac{I}{I_0} \right) = -\frac{Rt}{L} ) ( I(t) = I_0 e^{-Rt/L} )

12. Final formula: ( I(t) = I_0 e^{-t/\tau} ) where ( \tau = \frac{L}{R} ).

WORKED EXAMPLES

Example 1 – Basic Growth

Question: An RL circuit has ( R = 10 \, \Omega ), ( L = 5 \, H ), and ( V = 20 \, V ). Find the current at ( t = 1 \, s ) after the switch is closed.

Solution:
1. Identify ( I_0 ) and ( \tau ): ( I_0 = \frac{V}{R} = \frac{20}{10} = 2 \, A ) ( \tau = \frac{L}{R} = \frac{5}{10} = 0.5 \, s )

1. Use growth formula: ( I(t) = I_0 (1 - e^{-t/\tau}) ) ( I(1) = 2 (1 - e^{-1/0.5}) = 2 (1 - e^{-2}) )

2. Calculate: ( e^{-2} \approx 0.135 ) ( I(1) = 2 (1 - 0.135) = 2 \times 0.865 = 1.73 \, A )

What we did and why: - Found ( I_0 ) and ( \tau ) first (always do this!). - Plugged into the growth formula (not decay!). - Used ( e^{-2} ) from memory (common value).

Example 2 – Medium Decay

Question: An inductor ( L = 2 \, H ) with initial current ( I_0 = 3 \, A ) is connected to a resistor ( R = 4 \, \Omega ). Find the time when current drops to ( 1 \, A ).

Solution:
1. Identify ( \tau ): ( \tau = \frac{L}{R} = \frac{2}{4} = 0.5 \, s )

1. Use decay formula: ( I(t) = I_0 e^{-t/\tau} ) ( 1 = 3 e^{-t/0.5} )

2. Solve for t: ( \frac{1}{3} = e^{-2t} ) ( \ln \left( \frac{1}{3} \right) = -2t ) ( t = -\frac{1}{2} \ln \left( \frac{1}{3} \right) = \frac{1}{2} \ln(3) \approx 0.55 \, s )

What we did and why: - Used decay formula (not growth!). - Took natural log to solve for ( t ). - Remembered ( \ln(1/3) = -\ln(3) ).

Example 3 – Exam-Style (Disguised)

Question: A coil of inductance ( L ) and resistance ( R ) is connected to a battery. At ( t = \tau ), the current is ( 2 \, A ). Find the battery voltage ( V ).

Solution:
1. Recognise ( t = \tau ): At ( t = \tau ), ( I = I_0 (1 - e^{-1}) ).

1. Given ( I = 2 \, A ): ( 2 = I_0 (1 - e^{-1}) ) ( I_0 = \frac{2}{1 - e^{-1}} \approx \frac{2}{0.632} \approx 3.16 \, A )

2. Find ( V ): ( I_0 = \frac{V}{R} ) ( V = I_0 R = 3.16 R )

What we did and why: - Spotted ( t = \tau ) → used ( (1 - e^{-1}) ). - Calculated ( I_0 ) first, then ( V ). - Trick: Examiner gave ( t = \tau ) to test if you know ( e^{-1} \approx 0.368 ).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is all you need to remember for RL circuits:
1. Growth: ( I(t) = I_0 (1 - e^{-t/\tau}) ) where ( I_0 = V/R ) and ( \tau = L/R ).
2. Decay: ( I(t) = I_0 e^{-t/\tau} ).
3. At ( t = \tau ): Current is 63.2% of ( I_0 ) (growth) or 36.8% of ( I_0 ) (decay).
4. Energy: ( \frac{1}{2} L I^2 ) (not ( \frac{1}{2} CV^2 )!).
5. Always find ( I_0 ) and ( \tau ) first—no shortcuts!

If the question gives ( t = \tau ), use ( e^{-1} \approx 0.368 ). If it’s a decay problem, start with ( I = I_0 ), not zero. That’s it—go crush it!