Time for battery current to reach 500 mA

Assumptions: 10 V battery and switch in series with 70 Ω; the 50 Ω resistor and the 100 μF capacitor are in parallel after the 70 Ω; the capacitor is initially uncharged; the switch closes at t = 0.

Battery current as a function of time

Let v be the voltage across the parallel combination: i = (10 − v)/70 and

C dv/dt + v/50 = (10 − v)/70

Equivalent: v rises from 0 to the final value with time constant τ = (70 ∥ 50) C.

R_eq = 70 × 50 /(120) = 29.17 Ω
τ = R_eq C = 29.17 × 100×10⁻⁶ = 2.917 ms
v(∞) = 10 × 50/120 = 4.167 V
v(t) = 4.167 (1 − e^(−t/τ))

Battery current:

i(t) = (10 − v)/70 = 83.3 mA + (142.9 − 83.3) mA · e^(−t/τ)
     = 0.0833 + 0.0595 e^(−343 t) A

Check of the 500 mA value

The battery current starts at 10/70 = 142.9 mA (capacitor acts as a short) and decays to 10/120 = 83.3 mA. It lies between these two limits at all times, so with the data as given (10 V) the current can never reach 500 mA. The value is probably misprinted in the question (for example, 100 mA with 10 V, or 500 mA with a 50 V battery), so the general formula is shown:

t = τ ln[(I₀ − I∞)/(I − I∞)]

For I = 100 mA with the 10 V data: t = 2.917 ms × ln[(142.86 − 83.33)/(100 − 83.33)] = 2.917 ms × ln(3.571) = 3.71 ms.

(For 500 mA with a 50 V battery the ratio is identical, so t = 3.71 ms again.)

Answer: i(t) = 83.3 mA + 59.5 mA e^(−343t); the current is always below 142.9 mA, so 500 mA is never reached with 10 V. The corresponding time for the likely intended 100 mA is t ≈ 3.71 ms.