Understand RC charging - discharging time constants: Question bank

0. The facts every trap rests on (build them here first)

Before hunting traps we need every symbol earned. Look at the circuit below: a source pushes charge through a resistor into a capacitor.

Figure — Understand RC charging - discharging time constants

The two responses look like this:

And the direction current flows reverses between the two cases — keep this sign picture in mind for the edge-case traps:

True or false — justify

TRUE/FALSE: After one time constant

τ

, the capacitor is fully charged.

False. After

1 τ

it is only at

1 - e^{- 1} = 63.2%

(see figure s02); the curve is exponential, so it keeps creeping upward and needs about

5 τ

to reach

> 99%

TRUE/FALSE: A larger resistor makes the capacitor charge faster.

False. Larger

R

limits current, so

τ = R C

grows and charging gets slower — a thinner straw fills the bucket slower.

TRUE/FALSE: The time constant

τ = R C

depends on the source voltage

V_{s}

False.

τ = R C

is set purely by the circuit (

R

and

C

V_{s}

only scales how high the curve ends up, not how fast it gets there.

TRUE/FALSE: During charging, the current is largest right at

t = 0

True. At

t = 0

the capacitor voltage is

0

, so all of

V_{s}

sits across

R

, giving the maximum

I = V_{s} / R

; current then decays as

e^{- t / τ}

TRUE/FALSE: A capacitor's voltage can jump instantly if you switch fast enough.

False.

I = C d V_{C} / d t

, so an instant jump needs infinite current, which any real resistance forbids — the voltage must ramp smoothly.

TRUE/FALSE: In one time constant, charging always adds the same

63.2%

of whatever gap remains.

True. The exponential is memoryless: from any starting point, each further

τ

closes

63.2%

of the remaining distance to the target — because

d V_{C} / d t \propto (V_{s} - V_{C})

TRUE/FALSE: Doubling both

R

and

C

doubles the time constant.

False.

τ = R C

, so doubling both multiplies

τ

2 \times 2 = 4

. Doubling only one doubles

τ

TRUE/FALSE: The charging and discharging curves are mirror images of the same shape.

True in shape — both solve the same "rate ∝ gap" ODE, so both carry

e^{- t / τ}

with the same

τ

; one rises (

1 - e^{- t / τ}

), the other falls (

e^{- t / τ}

), and they cross the

50%

line at the same

t = τ ln 2

TRUE/FALSE: The units of

τ = R C

are seconds.

True.

Ω \cdot F = \frac{V}{A} \cdot \frac{C}{V} = \frac{C}{A} =

seconds, since a coulomb per amp is a second.

Spot the error

FIND THE ERROR: "To discharge, use

V_{C} (t) = V_{0} (1 - e^{- t / τ})

Discharging falls, so it must be

V_{C} = V_{0} e^{- t / τ}

. The

(1 - e^{- t / τ})

form rises toward a target and belongs to charging (compare the two curves in s02).

FIND THE ERROR: "

R = 1 k Ω

C = 10 μ F

, so

τ = 1000 \times 10 = 10000

s."

They forgot the micro factor.

10 μ F = 10 \times 1 0^{- 6} F

, so

τ = R C = 1000 \times 10 \times 1 0^{- 6} = 0.01

s — a million times smaller.

FIND THE ERROR: "The capacitor charges linearly, gaining

20%

V_{s}

every

τ

Charging is exponential, not linear. It gains

1 - e^{- 1} = 63.2%

in the first

τ

but only

(e^{- 1} - e^{- 2}) = 0.233 \approx 23%

in the second

τ

, because each step covers a fraction of the shrinking gap.

FIND THE ERROR: "As the capacitor charges, current stays constant because

V_{s}

is fixed."

Current is

I = (V_{s} - V_{C}) / R

. As

V_{C}

rises, the resistor's voltage

V_{s} - V_{C}

shrinks, so current decays exponentially — it is not constant.

FIND THE ERROR: "At

t = 5 τ

the capacitor is exactly, mathematically full."

It is never exactly full —

e^{- t / τ}

only reaches zero at

t = \infty

. At

5 τ

it is

1 - e^{- 5} = 99.3%

, which engineers treat as full, but the math never quite lands (see how the curve in s02 hugs but never touches the top line).

FIND THE ERROR: "Since

τ

has units of seconds, waiting

τ

seconds charges any cap by the same voltage."

No — waiting

τ

charges to the same fraction (

63.2%

) of that circuit's target, but the actual volts depend on

V_{s}

, and

τ = R C

itself differs per circuit.

Why questions

WHY does the charging current decay even though the source voltage stays fixed?

Because the voltage across the resistor is

V_{s} - V_{C}

, and as

V_{C}

climbs this difference shrinks — less push across

R

means less current, following

I = (V_{s} / R) e^{- t / τ}

WHY does the same

τ

appear in both the voltage curve and the current curve?

They come from solving the same ODE

R C d V_{C} / d t + V_{C} = V_{s}

; the exponential

e^{- t / τ}

is that equation's natural decay rate, so everything in the circuit shares it.

WHY does an exponential (rather than a straight line or a parabola) show up at all?

Because

d V_{C} / d t = (V_{s} - V_{C}) / R C

— the rate of change is proportional to the remaining gap. The only function whose rate is proportional to how far it still has to go is an exponential; that "self-slowing" is exactly what

e

describes.

WHY do we measure RC circuit progress in multiples of

τ

rather than raw seconds?

Because the fraction charged (

63.2%

86.5%

, ...) depends only on

t / τ

, so quoting time in

τ

units makes every RC circuit obey the same universal table regardless of

R

and

C

WHY does capacitor voltage lag behind a sudden source change?

Charge must physically flow through

R

to build

V_{C}

, and finite current means finite filling rate — the capacitor "resists" voltage change, which is the seed of low-pass filtering.

WHY is

e^{- 1} = 0.368

the magic number behind the

63.2%

figure?

t = τ

, the exponent is exactly

- 1

, so

e^{- t / τ} = 0.368

remains and

1 - 0.368 = 0.632

has charged — the

63.2%

is just "one

e

-fold" of decay.

WHY does bigger capacitance slow the circuit even though it stores more energy?

A bigger

C

needs more charge for each volt (

Q = C V_{C}

), and that extra charge still has to trickle through the same

R

, so it takes longer per volt — hence

τ = R C

grows.

Edge cases

EDGE CASE: What is

τ

R = 0

(ideal wire)?

τ = R C = 0

, so charging is instantaneous in the idealized model — with no resistance to limit current, the capacitor jumps to

V_{s}

immediately (real wires always have some

R

, preventing this).

EDGE CASE: What happens to charging time as

C \to 0

τ = R C \to 0

, so a vanishingly small capacitor charges essentially instantly — it needs almost no charge to reach

V_{s}

, so even tiny current fills it at once.

EDGE CASE: A capacitor already sits at

V_{s}

when you connect the source — what current flows?

Zero. The gap

V_{s} - V_{C} = 0

, so there is nothing to drive current; the circuit is already at its steady state and stays there.

EDGE CASE: What is

V_{C}

t \to \infty

during charging vs discharging?

Charging:

V_{C} \to V_{s}

(the

e^{- t / τ}

term dies, leaving the target). Discharging:

V_{C} \to 0

(all stored charge eventually bleeds through

R

). Neither is reached in finite time — the curve in s02 approaches its line asymptotically.

EDGE CASE: If you disconnect the source with an ideal open circuit (infinite

R

), how fast does a charged cap discharge?

It doesn't —

τ = R C \to \infty

, so

V_{C}

holds its charge indefinitely. This is why real capacitors keep a shock long after power is removed (only leakage slowly drains them).

EDGE CASE: During discharge, is the current direction the same as during charge?

No, it reverses (see the arrow flip in s03). Charging pushes current into the capacitor's plate; discharging lets the stored charge flow back out through

R

, so the current sign flips.

Recall One-line self-test

If you can explain why "63 up, 37 down, 5 to settle" is exponential and not linear — because $d V_{C} / d t \propto (V_{s} - V_{C})$ — you have beaten the biggest trap on this page.

Connections

Parent: RC time constants
Capacitors and stored energy
Kirchhoff's Voltage Law
First-order differential equations
RC low-pass and high-pass filters
555 timer circuits
Digital signal rise time and integrity