Q: In the shown circuit, the switch S has been open for a long time. It is then suddenly closed. Take ε = 20 V, R₁ = 500 Ω, R₂ = 250 Ω, and C = 0.002 F.
Determine the following:
- (a) Time constant before the switch is closed
- (b) Time constant after the switch is closed
- (c) Charge on the capacitor at t = 3.5τ after the switch is closed
- (d) Current in the circuit at t = 3.5τ after the switch is closed
🔹 (a) Time Constant Before Switch is Closed
Before closing the switch, the capacitor is charging through resistor R₁ only.
τbefore = R₁ × C = 500 × 0.002 = 1 s
🔹 (b) Time Constant After Switch is Closed
After the switch is closed, the capacitor discharges through both R₁ and R₂ in series. Total resistance becomes:
Rtotal = R₁ + R₂ = 500 + 250 = 750 Ω
Therefore, the new time constant is:
τafter = Rtotal × C = 750 × 0.002 = 1.5 s
🔹 (c) Charge on Capacitor at t = 3.5τ
Initially, the charge on the capacitor is:
Q₀ = C × ε = 0.002 × 20 = 0.04 C
At time t = 3.5τ, the charge is given by:
Q(t) = Q₀ × e−t/τ = 0.04 × e−3.5 ≈ 0.04 × 0.0302 = 1.208 × 10⁻³ C
🔹 (d) Current in the Circuit at t = 3.5τ
Using the formula for current in an RC discharging circuit:
I(t) = (Q₀ / RC) × e−t/τ
Substituting values:
I(3.5τ) = (0.04 / (750 × 0.002)) × 0.0302 = (0.04 / 1.5) × 0.0302 ≈ 0.0267 × 0.0302 = 8.06 × 10⁻⁴ A = 0.806 mA
✅ Final Answers
- (a) Time constant before switch is closed: 1 s
- (b) Time constant after switch is closed: 1.5 s
- (c) Charge at t = 3.5τ: 1.208 mC
- (d) Current at t = 3.5τ: 0.806 mA