RC and RL Exponential Responses

Summary of Equations

Exponential responses of capacitors and inductors

	Discharging	Charging	Time Constant
Capacitor	$v_{C}(t)=V_{0}e^{-{\frac {t}{RC}}}$	$v_{C}(t)=V_{0}(1-e^{-{\frac {t}{RC}}})$	$RC\,$
Inductor	$i_{L}(t)={\frac {V_{0}}{R}}e^{-{\frac {R}{L}}t}$	$i_{L}(t)={\frac {V_{0}}{R}}(1-e^{-{\frac {R}{L}}t})$	${\frac {L}{R}}$

RC Circuits

Discharging

Consider the following circuit:

In the circuit, the capacitor is initially 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 discharge 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 circuit:

Current will flow into the capacitor and accumulate a charge there. As the charge increases, the voltage rises, and eventually the voltage of the capacitor will equal the voltage of the source, and current will stop flowing. The voltage across the capacitor is given by:

$v_{C}(t)=V_{0}(1-e^{-{\frac {t}{RC}}})$

where $V_{0}=V_{s}$.

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 voltage and current of the capacitor in the circuits above are shown in the graphs below, from t=0 to t=5RC. Note the polaritiy—the voltage is the voltage measured at the "+" terminal of the capacitor relative to the ground (0V). A positive current flows into the capacitor from this terminal; a negative current flows out of this terminal.:

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

Note that the current through the capacitor can change instantly at t=0, but the voltage changes slowly.

RL Circuits

Discharging

In the following circuit, the inductor initially has current $I_{0}=V_{s}/R$ 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:

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

Charging

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

The inductor initially has a very high resistance, as energy is going into building up a magnetic field. Once the magnetic field is up and no longer changing, the inductor will act like a short circuit. The current at steady state will be equal to $V_{0}/R$ and in this case, $V_{0}=V_{s}$. Since the inductor is acting like a short circuit at steady state, the voltage across the inductor will be 0. The current through the inductor is given by:

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

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

The voltage and current of the inductor for the circuits above are given by the graphs below, from t=0 to t=5L/R. The voltage is measured at the "+" terminal of the inductor, relative to the ground. A positive current flows into the inductor from this terminal; a negative current flows out of this terminal.:

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

Note that the voltage across the inductor can change instantly at t=0, but the current changes slowly.