Resistor-Capacitor Circuit

A resistor-capacitor circuit is a transient circuit that involves any configuration includes a resistor and a [capacitor](Capacitor%20(Physics).md.

Charging a Capacitor §

Take the

Charging a Capacitor §

Let’s start with a simple circuit involving a Battery, a resistor and a Capacitor:

When a capacitor is introduced, the circuit now has an additional time dependency, because the current flowing through the capacitor (and hence, the circuit), $I$ is dependant on $\frac{dq}{dt}$.

$$\begin{array}{rlrlr}{V}_s-V_{R_1}(t)-V_{C_1}(t)=0& & & & \text{KVL}\\ V_{R_1}(t)=I(t)R_1& & & & \text{Ohm’s Law}\\ C_1\frac{\text{d}{V}_{C_1}}{\text{d}t}=I(t)& & & & \text{Capacitor Equation}\end{array}$$

$$V_s=V_{R_1}(t)+V_{C_1}(t)$$

$$V_s=I(t)R_1+V_{c_1}(t)$$

$$V_s=R_1{C}_1\frac{\text{d}{V}_{C_1}}{\text{d}t}+V_{c_1}(t)\text{Linear First Order ODE}$$

• $f(t)=V_s$, a constant function, is known as the forcing term

The resulting Linear First Order ODE solves to obtain a function in the form:

$$V_C(t)=c_1{e}^{kt}+c_2$$

Analysis using Kirchhoff’s Laws (Physics) §

Being a simple circuit with no junctions, we know the total current is constant, and we can use the conservation of energy:

$$V=\sum IR$$

and again, because there are no junctions:

$$V=I(R_1+R_2)$$

However, for the capacitor, we use the formula for Capacitance and the formula for Current:

$$V_C=\frac{q}{C}I=\frac{\text{d}q}{\text{d}t}$$

$$V=V_C+IR$$

$$V=\frac{q}{C}+\frac{dq}{dt}R$$

$$V-\frac{q}{C}-\frac{dq}{dt}R=0$$

This equation is a Linear First Order ODE:

$$\frac{dq}{dt}+q\times \frac{1}{RC}-\frac{V}{R}=0$$

which evaluates to:

$$V_C(t)=V_s(1-e^{-t/RC})$$

The full derivation, if you're interested:

We can use the integrating factor$I(t)=\exp (\int \frac{1}{RC}dt)$ = $e^{\frac{t}{RC}}$:

$$I(t)\frac{dq}{dt}+\frac{I(t)}{RC}q-\frac{I(t)V}{R}=0$$

$$\frac{d}{dt}(I(t)q)=\frac{I(t)V}{R}$$

$$I(t)q=\int \frac{V}{R}{e}^{t/RC}\text{d}t$$

$$I(t)q=VC{e}^{t/RC}+c$$

$$q=VC(1+\frac{c}{e^{t/RC}})$$

Defining $q=0$ at $t=0$:

$$0=VC(1+c)\to c=-1$$

$$q(t)=VC(1-e^{-t/RC})$$

Since $I=\frac{dq}{dt}$:

$$I(t)=\frac{V}{R}{e}^{-t/RC}$$

We also use tau $\tau$ to be the RC time constant: $\tau =RC$

Which, given constant voltage and resistance across the resistor, looks something like this:

Convergence §

The generally accepted convention in electrical engineering is that $5\tau$ is enough to treat a capacitor as ‘fully charged’, since:

$$V_{C_1}(5\tau )=V_s(1-e)$$

#todo Finish with nice CSS and points?

left=0; right=3;
top=10; bottom=-1;
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y = 10e^{-x}

$$V_c(t)=8e^{-5t}+16$$