## RC circuits

Resistors are simple circuit elements.  A constant current flows when a resistor is connected across the terminals of a battery.  If a capacitor is added to the circuit, the situation changes.  If a parallel-plate capacitor C is connected in series with a resistor R, and the two ends of the chain are connected to a battery as shown in the diagram to the right, then the battery will pump electrons from the plate connected to the negative terminal to the plate connected to the positive terminal.  The magnitude of the net charge on each plate will increase with time as

Q = Q0(1 - e-t/τ),

where τ = RC is the time constant of the RC (resistor-capacitor) circuit.
At t = 0, Q = 0, and when t becomes very large compared to τ, then e-t/τ << 0, and Q = Q0 = CV0
For the voltage across the capacitor we have V = Q/C, so

V = V0(1 - e-t/τ),

where V0 is the battery voltage.
The current flowing in the circuit as the capacitor is being charged is

I = ΔQ/Δt= I0e-t/τ = (V0/R)e-t/τ.

The time it takes to charge the capacitor so that it holds half of the maximum charge for a given voltage is t½ = τ ln2.
This is also the time it takes for the voltage across the capacitor to be equal to ½ of the battery voltage and the time it takes for the current to decrease to ½ of its maximum value.

If, after the capacitor has been fully charged, it is disconnected from the battery and the switch is closed, i.e, the RC circuit is shorted out, then the charge on the capacitor will decrease as Q = Q0e-t/τ, the voltage will decrease as V = V0e-t/τ, and the current flowing in the circuit will be I = ΔQ/Δt = -I0e-t/τ = -(V0/R)e-t/τ.  The minus sign indicates that the current that discharges the capacitor flows in the opposite direction as the current that charges the capacitor.

#### Problem:

The intermittent wiper system in a car is based on an RC time constant and utilizes a 0.40 μF capacitor.  What resistance R is needed for a 2 s time constant?

Solution
The time constant of an RC circuit is τ = RC.
R = τ/C = (2 s)/(4*10-7) F = 5*106 Ω = 5 MΩ.