GATE Electrical Engineering - Complete Master Library

 

📘 Laplace Transform in Network Transient Analysis (RL, RC, RLC)

Laplace Transform is extremely powerful in solving transient problems. Instead of solving differential equations directly, we convert the circuit into s-domain and solve algebraically.

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🔹 1. RL Circuit Transient

Consider series RL circuit with step input V.

Time-domain equation:

L di/dt + Ri = V

Step 1: Take Laplace Transform

L[sI(s) − i(0)] + RI(s) = V/s

Step 2: Solve for I(s)

I(s) = (V/s + Li(0)) / (Ls + R)

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Worked Example 1 – RL Step Response

Given: R = 5Ω L = 1H V = 10V step i(0) = 0

I(s) = (10/s) / (s + 5)

Using partial fractions:

10 / [s(s+5)] = 2/s − 2/(s+5)

Inverse Laplace:

i(t) = 2 − 2e^{-5t}

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🔹 2. RC Circuit Transient

Time-domain equation:

RC dv/dt + v = V

Worked Example 2 – RC Charging

R = 2Ω C = 0.5F V = 10V step v(0) = 0

V(s) = (10/s) / (s + 1)

Partial fraction:

10 / [s(s+1)] = 10/s − 10/(s+1)

Inverse:

v(t) = 10 − 10e^{-t}

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🔹 3. RLC Circuit Transient

Time-domain equation:

L d²i/dt² + R di/dt + (1/C)i = V(t)

Worked Example 3 – Underdamped Case

L = 1H R = 4Ω C = 0.25F Step input 5V

Characteristic equation:

s² + 4s + 4 = 0

Roots:

s = -2 ± j0

Critically damped case.

Inverse form:

i(t) = (A + Bt)e^{-2t}

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🔹 4. Initial & Final Value in Transients

Initial Current in RL:

i(0⁺) = lim s→∞ [sI(s)]

Final Current:

i(∞) = lim s→0 [sI(s)]

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🔹 5. Damping Conditions

• Overdamped → Two real roots
• Critically damped → Equal roots
• Underdamped → Complex roots

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🎯 GATE Important Points

• RL and RC step responses frequently asked
• RLC damping classification important
• Partial fraction mistakes common
• Initial condition handling critical

Laplace Makes Transient Analysis Simple & Powerful