Questions: Transient Response in RC Circuits

Question 1

A capacitor is 90% charged. Roughly how many time constants have elapsed since charging began?

• A. 0.9 time constants — the capacitor charges linearly at 100%/τ
• B. 1 time constant — the capacitor is fully charged after one τ
• C. About 2.3 time constants
• D. Exactly 10 time constants

Answer: (C)

Charging follows V(t) = V₀(1 − e^(−t/τ)). Setting 0.90 = 1 − e^(−t/τ) gives e^(−t/τ) = 0.10, so t/τ = −ln(0.10) ≈ 2.3. The charging curve is exponential, not linear — it rushes quickly at first and slows as it approaches the asymptote. One τ only gets you to about 63%, not 90%, and the capacitor never formally 'fully charges' in finite time.

Question 2

Doubling the resistance in an RC circuit while keeping capacitance fixed will double the time constant and therefore halve the initial charging current.

• T. True
• F. False

Answer: False

Doubling R does double the time constant τ = RC — the circuit now charges more slowly. But the initial charging current I₀ = V/R is halved by doubling R. The statement says this 'halves' the current, which is actually true numerically, but the claim combines two separate effects as if they follow the same logic. More importantly, a common confusion is thinking τ increasing means the circuit reaches steady state faster. The opposite is true: τ doubling means the circuit takes twice as long to reach any given fraction of its final charge.

Question 3

At t = τ (one time constant) during RC discharge, what percentage of the initial charge remains on the capacitor?

• T. True
• F. False

Answer: False

At t = τ, Q(τ) = Q₀e^(−1) ≈ 0.368 Q₀, meaning about 37% of the initial charge remains — not 50%. A common misconception is that τ is the 'half-life.' The half-life of an RC discharge is t₁/₂ = τ ln(2) ≈ 0.693τ. The statement is false; after one time constant, roughly 63% has discharged, not 50%.

Question 4

If you increase both R and C by a factor of 10 each, what happens to the time constant?

• A. It stays the same — the increases cancel each other out
• B. It increases by a factor of 10
• C. It increases by a factor of 100
• D. It decreases by a factor of 10, since more resistance reduces current flow

Answer: (C)

τ = RC, so if R → 10R and C → 10C, then τ → (10R)(10C) = 100RC = 100τ. The time constant is a product of resistance and capacitance, not a ratio. Both increases multiply together. The misconception in option A (that they cancel) would apply if τ were R/C or C/R, but it is their product.

Question 5

Why does the voltage across a capacitor change exponentially rather than linearly during charging, and what physical constraint causes this shape?

Model answer: The charging current equals (V_battery − V_C)/R. As the capacitor charges, V_C rises, which reduces the voltage difference driving the current, which slows the rate of charging. The rate of change of charge is proportional to the remaining 'gap' — exactly the condition that produces exponential growth toward an asymptote. The same feedback logic appears in any first-order system: the closer you get to the target, the slower you approach it.

The exponential shape is not arbitrary; it is the mathematical solution to dQ/dt = −Q/(RC), where the rate of change is proportional to the current state. In charging: the driving force (battery EMF minus capacitor voltage) shrinks as charge accumulates, continuously slowing the rate. In discharging: the driving force (the capacitor voltage itself) shrinks as charge leaves, again slowing the rate. The time constant τ = RC quantifies this self-limiting timescale.