EE281 - Electric Circuits

RC Circuits

First-Order Circuits

Consider a circuit with resistor and capacitor as shown below.

At t=0, the capacitor voltage is \(V_o\). Applying KCL at the node:

\(i_c + i_R = 0\)

\(C \dfrac{dv}{dt} + \dfrac{v}{R} = 0\)

\(\dfrac{dv}{dt} = - \dfrac{v}{RC} \)

which is a first order differential equation.

\(\dfrac{dv}{v} = - \dfrac{1}{RC}dt\)

Integrating both sides:

\( ln (v) = -\dfrac{t}{RC} + ln (A)\)

or:

\( ln (\dfrac{v}{A}) = -\dfrac{t}{RC}\)

taking exponential power on each sides:

\( v(t) = Ae^{-\dfrac{t}{RC}}\)

As we know the initial voltage is \(V_0\). Thus:

\( v(t) = V_oe^{-\dfrac{t}{RC}}\)

Thus, the voltage response of a RC circuit is an exponential decay of voltage. The time constant of the circuit is given as:

\(\tau=RC\)

The unit of the time constant is in seconds.

Thus, the equation above becomes:

\( v(t) = V_oe^{-\dfrac{t}{\tau}}\)

, which is presented in the figure below:

![](../images/RC_tau.png)

Exercise:

If the initial voltage of the capacitor is 60 V, then determine Vc, Vx, io.

60, 20, -5