An RC circuit has R = 10 kΩ and C = 100 μF. What is the time constant, and how long until the capacitor reaches 95% of its final charge?
Think about your answer, then reveal below.
Model answer: τ = RC = (10×10³ Ω)(100×10⁻⁶ F) = 1 s. To reach 95%: 1 − e^(−t/τ) = 0.95, so e^(−t/τ) = 0.05, t = −τ ln(0.05) ≈ 3τ = 3 s.
A quick rule of thumb: 1τ ≈ 63%, 2τ ≈ 86%, 3τ ≈ 95%, 5τ ≈ 99%. For most practical purposes, the circuit has 'settled' after 3–5 time constants. The logarithm gives the exact answer; the rule of thumb builds intuition for rapid estimation.
Question 2 Short Answer
At t = 0 a fully charged capacitor (V₀ = 12 V) begins discharging through R = 3 kΩ. What is the initial current, and why does current decrease over time?
Think about your answer, then reveal below.
Model answer: Initial current I₀ = V₀/R = 12/3000 = 4 mA. Current decreases because as charge flows off the capacitor, V_C falls, reducing the driving voltage and therefore the current. The decay follows I(t) = I₀e^(−t/RC).
The discharging capacitor is the driving source; as it empties, it drives less current. This feedback — less charge → less voltage → less current → charge decreases more slowly — is exactly what produces exponential decay rather than linear decay.