Charge and discharge a capacitor

Charge it through a resistor, then let it go: change R and C and watch the time constant.

Charging and discharging a capacitor

Interactive

time0s

voltage V0V

current I0.9mA

charge Q0µC

energy stored0mJ

τ = RC1s

Switch

Battery voltage V₀9 VResistance R10 kΩCapacitance C100 µF

The math, with your numbers

\tau = RC = 10\ \text{k}\Omega \times 100\ \mu\text{F} = 1\ \text{s}

V = V_0\left(1 - e^{-t/RC}\right) = 9\left(1 - e^{-0/1}\right) \approx 0\ \text{V}

Q = CV = 0\ \mu\text{C}, \quad U = \tfrac12 CV^2 = 0\ \text{mJ}

After one τ the capacitor has made 63% of its change; after 5τ, over 99%. Bigger R or bigger C: slower.

Try this

→Charge it fully, then discharge. How long until the voltage falls to about a third (37%)?
→Double R. Does the capacitor end up with less charge, or just take longer?
→Double the battery voltage. By what factor does the stored energy grow?

Want to know why it behaves like this?

The lesson “Capacitors & the RC Time Constant” explains it step by step: How does a camera flash collect energy for seconds and release it in a millisecond?

Read the lesson