RC circuits — charging, discharging, time constant τ = RC: Question bank

Definition The five symbols on this page (each one earned)

Everything below reuses just these — no new notation sneaks in later.

$R$ — the resistance (in ohms, $Ω$ ): how hard the resistor throttles the flow of charge. Bigger $R$ = a thinner "straw." See Ohm's Law and Resistance.
$C$ — the capacitance (in farads, F): how much charge the capacitor holds per volt. Bigger $C$ = a wider "bucket." See Capacitors and Capacitance.
$Q$ — the charge (in coulombs, C) currently sitting on the capacitor plates. It grows while charging, shrinks while discharging.
$t$ — time (in seconds, s), measured from the moment the switch flips at $t = 0$ .
$τ = R C$ — the time constant built from the two above: a time in seconds that sets how fast everything happens.

Two more we borrow from the parent: $E$ is the battery's emf (its push, in volts), and $V_{C} = Q / C$ is the capacitor's own voltage, which grows as $Q$ piles up and fights back against the battery. Current is the flow of charge, $I = \frac{d Q}{d t}$ .

The two figures below are the mental pictures every question leans on — the circuit and the two curves. Glance back at them whenever a trap feels slippery.

Figure — RC circuits — charging, discharging, time constant τ = RC

True or false — justify

A capacitor charges at a constant rate because the battery voltage is constant.

False. The current is set by the net voltage

E - V_{C}

divided by

R

; as

V_{C}

climbs the net push shrinks, so the rate falls — that is exactly what makes the curve exponential, not a straight line.

t = τ

a charging capacitor is essentially full.

False. At one

τ

it has reached only

1 - e^{- 1} \approx 63.2%

. "Essentially full" (>99%) takes about

5 τ

, since

e^{- 5} \approx 0.7%

remains.

Charging reaches 63% and discharging falls to 63% — same number, same

τ

False. Charging rises to

1 - e^{- 1} \approx 63.2%

; discharging falls to

e^{- 1} \approx 36.8%

. They are complementary (

63.2% + 36.8% = 100%

), not identical.

Doubling both

R

and

C

makes the circuit charge four times slower.

True in timescale.

τ = R C

becomes

4 τ

, so every stage takes four times as long. The shape of the curve is unchanged — only its horizontal stretch.

The charging current and discharging current both decay with the same time constant.

True. Both obey

I \propto e^{- t / τ}

with

τ = R C

; the resistor and capacitor are the same, so the exponential clock is the same in either direction.

A larger capacitor always charges to a higher final voltage.

False. The final voltage is

E

(the battery's push), independent of

C

. A larger

C

only stores more charge (

Q_{m a x} = E C

) and takes longer to get there.

During discharge the current flows in the same direction as during charging.

False. When charging, current flows into the positive plate; when discharging, the capacitor drives current out the same plate, so the current reverses direction through

R

At the very instant charging begins, the resistor carries the maximum possible current.

True. At

t = 0

V_{C} = 0

, so the full emf sits across

R

and

I_{0} = E / R

— the largest current the loop ever sees.

Energy delivered by the battery while charging equals the energy stored in the capacitor.

False. Exactly half the battery's energy is dissipated as heat in

R

; only the other half ends up stored in the capacitor — independent of

R

's value. See Energy Stored in a Capacitor.

Spot the error

"At

t = 0

the capacitor is empty, so its voltage is

E

The error swaps cause and effect. Empty means

Q = 0

, so

V_{C} = Q / C = 0

; the full voltage

E

sits across the resistor, not the capacitor, at

t = 0

"To find when charging is 90% done, I set

e^{- t / τ} = 0.90

Wrong quantity. Charging fraction is

1 - e^{- t / τ}

, so 90% done means

1 - e^{- t / τ} = 0.90

, i.e.

e^{- t / τ} = 0.10

. Setting

e^{- t / τ} = 0.90

finds when it's only 10% done.

"Since

τ = R C

has units of

Ω \cdot F

, and those aren't seconds,

τ

isn't really a time."

Units do collapse to seconds:

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

. The mismatch is only surface-level notation.

"During discharge

d Q / d t = + Q / (R C)

, so charge grows."

Sign error. KVL with no source gives

R \frac{d Q}{d t} = - \frac{Q}{C}

, so

\frac{d Q}{d t} = - \frac{Q}{R C} < 0

— the charge decays. The minus sign encodes "losing charge."

"A capacitor blocks DC, so no current ever flows when I connect the battery."

It blocks DC only in the final steady state. During the transient (roughly the first few

τ

), current does flow to deliver the charge; it dies to zero only once

V_{C} = E

"Because current stops at the end, the resistor was useless — remove it and it charges instantly."

With

R \to 0

τ = R C \to 0

, so the current spike

E / R \to \infty

. Nothing charges "instantly" in reality; the resistor is what makes the current finite and the process well-behaved.

Why questions

Why does the charging curve bend over (concave down) instead of being a straight line?

Because the rate

\frac{d Q}{d t} = \frac{1}{R C} (Q_{m a x} - Q)

is proportional to how far you still have to go. As

Q

approaches

Q_{m a x}

that gap shrinks, so the slope continuously flattens.

Why do charging and discharging share the exact same

τ

The time constant depends only on the components in the loop,

R

and

C

, not on the presence of a battery. Both processes are governed by the same

R C

product, so they run on the same clock.

Why is the current identical in magnitude everywhere in a series RC circuit at any instant?

Series means one path with no junctions, and current is the flow of charge — charge cannot pile up in the wire, so the same

I = d Q / d t

passes through

R

and onto the capacitor plate simultaneously. This is Kirchhoff's companion, current continuity.

Why does the same mathematics describe Newton's Law of Cooling and radioactive decay?

All three obey "rate of change proportional to how far from equilibrium," which forces a solution built from

e^{- t / τ}

. See Exponential Decay and Differential Equations — the physics differs but the equation is one template.

Why is

τ = L / R

for an LR circuit but

τ = R C

here?

In an RC circuit larger

R

slows things (less current), so

R

multiplies. In an LR circuit larger

R

speeds the decay of current, so

R

divides. The role of

R

flips because it limits current in one and dissipates stored current-energy in the other.

Why does a bigger

R

make charging slower even though it lowers the current the whole time?

Slower charging means it takes longer to deliver the same total charge

Q_{m a x} = E C

. A big

R

throttles the delivery rate at every instant, so accumulating that fixed charge simply takes more time.

Edge cases

What happens as

R \to 0

(ideal wire, no resistor)?

τ = R C \to 0

, so charging is effectively instantaneous and the initial current

E / R \to \infty

— an idealized infinite spike. Real circuits always have some resistance that tames this.

What happens as

R \to \infty

(broken wire / open switch)?

τ = R C \to \infty

, so the current

E / R \to 0

and no charge ever flows — the capacitor stays at whatever voltage it started with, forever. Infinite resistance is just an open circuit.

What happens as

C \to 0

(vanishingly small capacitor)?

τ \to 0

and

Q_{m a x} = E C \to 0

: it "charges" instantly to essentially no charge. A zero-capacitance element behaves like an open gap that never stores anything.

What happens as

C \to \infty

(enormous capacitor)?

τ \to \infty

, so charging never practically finishes; over any finite time the voltage barely rises. A huge capacitor looks like a short (a plain wire holding

V_{C} \approx 0

) for a long time.

If you connect an already-charged capacitor to a battery of the same voltage

E

, what current flows?

Zero at all times. The net driving voltage

E - V_{C} = 0

from the start, so

I = 0

— the circuit is born in its own steady state and nothing changes.

t \to \infty

during charging, what are

I

V_{C}

, and

V_{R}

I \to 0

V_{C} \to E

, and

V_{R} = I R \to 0

. The capacitor holds the entire emf and acts like an open switch; no current means no resistor drop.

What if you discharge a capacitor through zero resistance?

τ \to 0

gives an ideal instantaneous dump and infinite momentary current. Physically this is a spark/short; real resistance and inductance keep it finite.

If two identical resistors are in series in the charging loop, how does

τ

change?

The effective resistance doubles to

2 R

, so

τ = 2 R C

— charging takes twice as long. Only the net series resistance in the loop matters, per Ohm's Law and Resistance.

Recall One-line self-test

Charge rises to 63%, decays to 37%, and after $5 τ$ it's practically done — which of these three did you get wrong most often above? Answer ::: Most people trip on "reaches 63% vs falls to 37%" — they are complementary, not the same number.

Connections

Parent: RC circuits & τ = RC
Capacitors and Capacitance · Ohm's Law and Resistance · Kirchhoff's Voltage Law
Exponential Decay and Differential Equations · Newton's Law of Cooling
Energy Stored in a Capacitor · LR Circuits