RC and RL Exponential Responses

RC Circuits

Discharging

Consider the following circuit:

In the circuit, the capacitor is initally charged and has voltage $V_{0}$ aross it, and the switch is initially open. At time $t=0$, we close the circuit and the capacitor will discharage through the resistor. The voltage across a capacitor discharging through a resistor as a function of time is given as:

$v_{C}(t)=V_{0}e^{-{\frac {t}{RC}}}$

Charging

If the capacitor is initially uncharged and we want to charge it by inserting a voltage source $V_{s}$ in the RC cicuit:

The voltage across the capacitor is given by:

$v_{C}(t)=V_{s}(1-e^{-{\frac {t}{RC}}})$

The term RC is the resistance of the resistor multiplied by the capacitance of the capacitor, and known as the time constant, which is a unit of time. The value of the function will be 63% of the final value at $t=1RC$, and over 99.99% of the final value at $t=5RC$.

The magnitudes of the voltage and current of the capacitor in the circuit above are shown in the graphs below:

Capacitor

	Voltage	Current
Charge	File:RC charge voltage.jpg	File:RC charge current.jpg
Discharge	File:RC discharge voltage.jpg	File:RC discharge current.jpg

RL Circuits

Discharging

In the following circuit, the inductor initally has current $I_{0}=V_{0}/R$ flowing through it; we replace the voltage source with a short circuit at $t=0$.

The current flowing through the inductor at time t is given by:

$i_{L}(t)=I_{0}e^{-{\frac {R}{L}}t}$

Charging

If the inductor is initially uncharged and we want to charge it by inserting a voltage source $V_{s}$ in the RL cicuit:

The current through the inductor is given by:

$i_{L}(t)=I_{0}(1-e^{-{\frac {R}{L}}t})$

The time constant for the RL circuit is equal to $L/R$.

The magnitudes of the voltage and current of the inductor for the circuits above are given by the graphs below:

Inductor

	Voltage	Current
Charge	File:RL charge voltage.jpg	File:RL charge current.jpg
Discharge	File:RL discharge voltage.jpg	File:RL discharge current.jpg