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Capacitors II Charging and Discharging. Current The current through a capacitor is given by i C = C(dV C /dt) Capacitors are normally charged or discharged.

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Presentation on theme: "Capacitors II Charging and Discharging. Current The current through a capacitor is given by i C = C(dV C /dt) Capacitors are normally charged or discharged."— Presentation transcript:

1 Capacitors II Charging and Discharging

2 Current The current through a capacitor is given by i C = C(dV C /dt) Capacitors are normally charged or discharged through an RC circuit which contains a voltage source V S, a capacitor C, and a resistor R.

3 Time constant In an RC circuit, the quantity RC is called the time constant Time constant τ = RC If R is expressed in ohms and C in Farads, then the time constant is in seconds. Be careful to convert R to ohms and C to Farads or you will have a significant decimal point error.

4 Discharging a Capacitor When a capacitor C fully charged to the supply voltage V S is discharged through a resistor R, the voltage on the capacitor drops from V S to zero as a function of time according to the equation V C = V S exp(-t/RC) = V S e (-t/RC) V C = V S exp(-t/τ) = V S e (-t/τ)

5 Discharge Current i C = C(dV C /dt) V C = V S exp(-t/τ) = V S e (-t/τ) dV C /dt = V S (-1/τ)exp(-t/τ) i C = C V S (-1/τ)exp(-t/τ) And since τ = RC i C = C V S (-1/RC)exp(-t/τ) i C = (-V S /R) exp(-t/τ)

6 At t = 0, exp(-t/τ) = exp (0) = 1 i C = (-V S /R) which looks like Ohm’s law with the capacitor at V S. At t = ∞, exp(-t/τ) = 1/exp(∞) = 0 So discharge starts at a very rapid rate and slows down asymptotically to zero.

7 Discharge Voltage vs time V (volts) t (msec)

8 Discharge Current vs time I (amps) t (ms)

9 Charging of a Capacitor V c = V s [1 – exp(-t/RC)] V c volts time (ms)

10 Charging Current I c = (V s / R) exp(-t/RC) I c (ma) time (ms)


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