Skip to main content

Source current expression and time to reach 25 mA

Short answer

Source current i_s(t) = 20 + 14.286e^(−14.286t) mA. Setting 25 = 20 + 14.286e^(−14.286t) gives e^(−14.286t)=0.35, so t = ln(1/0.35)/14.286 = 0.0735 s.

Using classical method, find the expression for the current supplied by the source in the network shown in figure. Also find the time taken by the source current to reach 25 mA. [Figure: 10 V source with a switch (t = 0) in the top wire; a 500 Ω resistor as a shunt; then a 700 Ω resistor in series with a 100 μF capacitor in the right branch]

Assumption: ideal 10 V source; switch closes at t = 0; capacitor initially uncharged. The 500 Ω resistor and the 700 Ω + 100 μF branch are both connected directly across the source.

Branch currents

Because the source is ideal, the two branches are independent.

  • Resistor branch: i₁ = 10/500 = 20 mA (constant, from t = 0)
  • RC branch: loop equation 700 i₂ + (1/C)∫i₂ dt = 10, i.e. di₂/dt + i₂/(RC) = 0
τ = RC = 700 × 100×10⁻⁶ = 0.07 s
i2(0⁺) = 10/700 = 14.286 mA    (capacitor initially a short)
i2(t) = 14.286 e^(−t/0.07) mA = 14.286 e^(−14.286 t) mA

Source current

i_s(t) = i₁ + i₂ = 20 + 14.286 e^(−14.286t) mA, t ≥ 0

At t = 0, i_s = 34.29 mA; it falls to 20 mA as t → ∞.

Time to reach 25 mA

25 = 20 + 14.286 e^(−14.286 t)
e^(−14.286 t) = 5/14.286 = 0.35
t = ln(1/0.35)/14.286 = 1.0498/14.286

Answer: t = 0.0735 s ≈ 73.5 ms

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Circuit Theory chapter 3 questions →