Meaning and mechanism
The capacitor stores electric charge, and the resistor limits the current. As the capacitor charges, the voltage across the resistor and the charging current decrease.
When discharging, the stored charge moves through a resistor, resulting in a decrease in voltage. As the remaining voltage decreases, the discharge current also decreases.
Relationships and conditions
τ=RC; charging follows V₀(1−e⁻ᵗ/τ), and discharging follows V₀e⁻ᵗ/τ. After one τ, the charging voltage is about 63.2% of its final value and the discharge voltage about 36.8% of its initial value.
We assume an ideal resistor and capacitor, along with a constant power source. Actual leakage, internal resistance, and the load of the instruments can alter the observed curve.
What to observe in the simulation
At t = τ, charging reaches about 63.2% of the source voltage and discharging about 36.8%.
The source, resistor and capacitor are ideal. Leakage and parasitic effects are omitted. Discharging starts from V₀. The scene uses an analytical solution; engine checks use backward Euler. Use this simulation; do not reproduce it with real power supplies or wall outlets.
Key relationships and symbols
τ = RC · Vcharge = V₀(1 − exp(−t/τ)) · Vdischarge = V₀exp(−t/τ)
| Symbol | Meaning | Unit |
|---|---|---|
| R | Series resistance | Ω |
| C | Capacitance | F |
| τ | Time constant RC | s |
Worked example
With R=10 kΩ and C=100 μF, what is the time constant after changing only R to 20 kΩ?
2 s
Common misconceptions
- The concept of a “one-hour constant” does not mean that charging or discharging is completely finished.