Resistor-Capacitor (RC) Circuits
Introduced| English | Meaning |
|---|---|
| time constant/taɪm ˈkɒnstənt/ | The characteristic time for an exponential process to change by a factor of e; in decay, the quantity falls to one over e of its starting value. |
| exponential/ˌekspəˈnenʃl/ | Describing a relationship in which a variable occurs in an exponent. |
Why does a camera flash take a moment to recharge?
- Press the shutter, and the flash needs a second before it can fire again.
- A capacitor is refilling with charge through a resistor.
- The resistor limits the current, so the filling is gradual, not instant.
- A resistor and capacitor together make an RC circuit — the physics of timing.
Charging rises toward a limit
- Close the switch and charge flows onto the capacitor: $Q(t) = CV\left(1 - e^{-t/RC}\right)$.
- It climbs fast at first, then slows as the capacitor fills.
- It approaches, but never quite reaches, the final charge $CV$.
- The current does the opposite: large at first, fading to zero.
Charge through a resistor
Watch the capacitor fill gradually and the charging current fade toward zero.
The time constant τ = RC
- The product $\tau = RC$ is the time constant 时间常数 — the natural timescale.
- In one $\tau$, the charge reaches about $63\%$ of its final value.
- After about $5\tau$, it is essentially fully charged.
- Bigger $R$ or bigger $C$ means slower charging.
The time constant of an RC circuit is:
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$\tau = RC$ sets the charging and discharging timescale.
A $5\ \Omega$ resistor charges a $4\ \text{F}$ capacitor. Find $\tau$ (in s).
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$\tau = RC = 5 \times 4 = 20\ \text{s}$.
After one time constant, a charging capacitor is fully charged.
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False. After one $\tau$ the charge reaches $1 - e^{-1} \approx 63\%$ of its final value; it approaches full charge only after several time constants.
Discharging fades away
- Disconnect the source and let the capacitor drive the resistor.
- The charge decays: $Q(t) = Q_0\, e^{-t/RC}$ — an exponential 指数 fall.
- The current and voltage decay with the same time constant $\tau = RC$.
- Each $\tau$ the charge drops to about $37\%$ of what it was.
A discharging capacitor loses charge as an ____ decay, $Q = Q_0 e^{-t/RC}$.
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Discharge is exponential: $Q = Q_0 e^{-t/RC}$.
The capacitor's two extremes
- At the first instant ($t = 0$): the empty capacitor acts like a plain wire.
- After a long time ($t \to \infty$): the full capacitor acts like an open gap.
- So current starts high and ends at zero as the capacitor fills.
- Those two limits let you check any RC answer quickly.
At the first instant of charging ($t = 0$), an uncharged capacitor acts like:
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Empty, it offers no back-voltage, so it behaves like a wire (max current).
Select all true statements about RC circuits.
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Exponential charging, τ = RC, open gap when full. Charging is gradual, never instant.
A $2\ \Omega$ resistor charges a $3\ \text{F}$ capacitor. Find the time constant.
- $\tau = RC = 2 \times 3 = 6\ \text{s}$.
- After $6\ \text{s}$ the charge is about $63\%$ of its final value.
At $t = 0$ an uncharged capacitor behaves like a wire (max current), and at $t \to \infty$ a charged one behaves like an open gap (zero current). Swapping these two limits is the classic RC mistake — check which instant the question asks about.
In an RC circuit, charging follows $Q = CV(1 - e^{-t/RC})$ and discharging $Q = Q_0 e^{-t/RC}$, both with time constant $\tau = RC$ (about $63\%$ per $\tau$). A capacitor starts like a wire ($t=0$) and ends like an open gap ($t\to\infty$).
Write from a time constant of 16 seconds
- A charging circuit has R = 8 ohms and C = 2 farad.
- Find the time constant. Say what fraction of the final charge that one time constant means.
A resistor of 8 ohms charges a capacitor of 2 farad. What is the time constant, and is the capacitor fully charged after one time constant?
8 times 2 is 16 seconds. The exponential 1 minus 1/e is about 63 percent.
What is the time constant, and what does it mean?
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RC is 16 seconds, and one time constant is not the end of charging.