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	$i_{L}(t)=I_{0}e^{-{\frac {R}{L}}t}$	$i_{L}(t)=I_{0}(1-e^{-{\frac {R}{L}}t})$	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{L}{R}}

RC Circuits

Discharging

Consider the following circuit:

In the circuit, the capacitor is initially charged and has voltage Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v_C(t)=V_0e^{-\frac{t}{RC}}}

Charging

If the capacitor is initially uncharged and we want to charge it by inserting a voltage source Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_s} in the RC circuit:

The voltage across the capacitor is given by:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=1RC} , and over 99.99% of the final value at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=5RC} .

The magnitudes of the voltage and current of the capacitor in the circuits 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 initially has current $I_{0}=V_{0}/R$ flowing through it; we replace the voltage source with a short circuit at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=0} .

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i_L(t)=I_0e^{-\frac{R}{L}t}}

Charging

If the inductor is initially uncharged and we want to charge it by inserting a voltage source Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_s} in the RL circuit:

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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