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Laplace: i1, i2 in unenergised two-mesh network

Short answer

Solving s-domain mesh equations yields i1(t)=6.667−5e^(−10t)−1.667e^(−30t) A and i2(t)=3.333−5e^(−10t)+1.667e^(−30t) A, found by partial fractions and inverse Laplace transform.

Using Laplace transform method, find the currents i1(t) and i2(t) for t > 0 in the circuit shown in the figure below. Assume the network is unenergised before the switch is closed. [Figure: 100 V source with a switch (closes at t = 0), a 10 Ω resistor and a 1 H inductor in the top wire of the left mesh (i1); a 10 Ω resistor in the middle shared vertical branch; a 1 H inductor and a 10 Ω resistor in the right mesh (i2)]

Initial currents and energy are zero, so only the s-domain impedances are needed: R = 10 Ω, sL = s Ω, source 100/s. Both mesh currents are taken clockwise.

s-domain mesh equations

Mesh 1: (10 + s + 10) I1 − 10 I2 = 100/s
Mesh 2: −10 I1 + (10 + s + 10) I2 = 0

Mesh 2 gives I2 = 10 I1/(s + 20). Substituting in mesh 1:

[(s + 20) − 100/(s + 20)] I1 = 100/s

Using the determinant Δ = (s + 20)² − 100 = (s + 10)(s + 30):

I1(s) = 100 (s + 20) / [ s (s + 10)(s + 30) ]
I2(s) = 1000 / [ s (s + 10)(s + 30) ]

Partial fractions

I1 = 6.667/s − 5/(s + 10) − 1.667/(s + 30)
     (K1 = 100×20/(10×30);  K2 = 100×10/((−10)(20));  K3 = 100×(−10)/((−30)(−20)))
I2 = 3.333/s − 5/(s + 10) + 1.667/(s + 30)

Inverse Laplace transform

i1(t) = 6.667 − 5e^(−10t) − 1.667e^(−30t) A, t > 0

i2(t) = 3.333 − 5e^(−10t) + 1.667e^(−30t) A, t > 0

Check: i1(0) = 0 and i2(0) = 0 (inductors open the circuit initially); final values i1 = 100/(10 + 5) = 6.667 A and i2 = i1/2 = 3.333 A ✓. Values at t = 0.1 s: i1 = 4.744 A, i2 = 1.577 A, which match a numerical inverse Laplace transform.

✓ Answer checked by Claude Opus 5.5 (high effort)

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