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Classical: i1, i2 with two R-L branches

Short answer

The two identical branches share current equally. KVL gives di/dt + 30i = 100 with i(0)=0. Solution: i1(t) = i2(t) = 3.33(1 − e^(−30t)) A.

In the network shown, the switch is closed at t = 0, with the network previously unenergised. For the element values shown on the diagram, find i1(t) and i2(t) by classical method for t > 0. [Figure: 100 V source with a switch (closes at t = 0) and a 10 Ω resistor in series, feeding two parallel branches: each is a 10 Ω resistor in series with a 1 H inductor; the branch currents are i1 and i2]

Assumptions: 100 V source, switch, and a 10 Ω resistor in series feed two identical parallel branches, each 10 Ω + 1 H. The network is initially unenergised (i = 0).

Because the two branches are identical, i₁ = i₂ = i, and the current through the series 10 Ω is 2i.

Differential equation (KVL)

100 = 10(2i) + 10 i + 1·di/dt
di/dt + 30 i = 100

Complete solution

  • Particular (steady state): i_p = 100/30 = 3.333 A
  • Complementary: characteristic equation s + 30 = 0, so i_c = K e^(−30t)
i = 3.333 + K e^(−30t)
i(0⁺) = 0  →  K = −3.333

Result

i₁(t) = i₂(t) = 3.333 (1 − e^(−30t)) A,   t > 0

Time constant τ = L/R_eq = 1/30 s = 33.3 ms. Check: substituting back, di/dt + 30i = 100 e^(−30t) + 100(1 − e^(−30t)) = 100 ✓.

Answer: i₁(t) = i₂(t) = 3.333(1 − e^(−30t)) A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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