Resistor-Capacitor Circuit

A resistor-capacitor circuit is a transient circuit that involves any configuration includes a resistor and a capacitor ^head %%🖋 Edit in Excalidraw, and the light exported image%%

Formulae §

Charging Capacitor in RC Circuit

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

$$I_C(t)=\frac{V_s}{R}{e}^{-t/RC}=I_s{e}^{-t/\tau }$$

Note

• $V_C$ = Capacitor voltage (in $\text{V}$)
• $I_C$ = Capacitor current (in $\text{A}$)
• $V_s$ = Source voltage (voltage series to capacitor and resistor) (in $\text{V}$)
• $I_s$ = Source current (current flowing through resistor and capacitor) (in $\text{A}$)
• $C$ = Capacitance of the capacitor (in Farads, $\text{F}$)
• $R$ = Resistance of the resistor (in Ohms, $\Omega$)
• $\tau$ = Time constant = $RC$

Time Constant §

The time constant, $\tau$ is given as the product of the capacitance and resistance:

$$\tau =RC$$

At time $t=\tau$, we have:

$$V_C(\tau )=V_s(1-e^{-1})\approx 0.632V_s$$

Which means the voltage through the capacitor is roughly 63% of the source voltage at this time, which we can use to estimate tau.

Convergence §

$$V_C(5\tau )=V_s(1-e^{-5})\approx 0.993V_s\approx V_s$$

Graph §

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Derivation §

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 $\frac{dq}{dt}$.

$$\begin{array}{rlrlr}{V}_s-V_R(t)-V_C(t)=0& & & & \text{KVL}\\ V_R(t)=I(t)R& & & & \text{Ohm’s Law}\\ C\frac{\text{d}{V}_C}{\text{d}t}=I(t)& & & & \text{Capacitor Equation}\end{array}$$

$$V_s=V_R(t)+V_C(t)$$

$$V_s=I(t)R+V_C(t)$$

$$V_s=RC\frac{\text{d}{V}_C}{\text{d}t}+V_C(t)\text{Linear First Order ODE}$$

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

Solving the ODE §

The equation is an inhomogeneous linear first order ODE. It can be solved in the following steps:

1. Find the homogeneous solution $V_{C_h}(t)$. This usually involves an integrating factor, but the form is always:

$$V_{C_h}(t)=c_1{e}^{k\cdot t}$$

where $k$ and $c_1$ are constants. NOTE: $c_1\ne 0$

2. The particular solution, $V_{C_p}(t)$ is of the same function family as the forcing term. Here it is a constant:

$$V_{C_p}(t)=c_2$$

3. Hence, the solution is of the form:

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

4. Obtain $\frac{\text{d}{V}_c}{\text{d}t}$:

$$\frac{\text{d}{V}_C}{\text{d}t}=k\cdot c_1\cdot e^{kt}$$

5. Plug both back into the original ODE:

$$c_1{e}^{kt}+c_2+RC(k{c}_1\cdot e^{kt})=V_s$$

$$c_1{e}^{kt}(1+kRC)+c_2=V_s+0\cdot e^{kt}$$

6. Equate coefficients:

$$\{\begin{array}{ll}{c}_1(1+kRC)=0,& (1)\\ c_2=V_s,& (2)\end{array}$$

7. For $(1)$, $c_1$ is defined to be not equal to $0$, hence by the Null Factor Theorem:

$$1+kRC=0\to k=\frac{-1}{RC}$$

8. Plug into solution:

$$V_C(t)=c_1{e}^{-t/RC}+V_s$$

9. Use an initial condition $v_0$(the voltage the capacitor starts when $t=0$) to find $c_1$:

$$V_C(0)=0$$

$$\to 0=c_1+V_s(e^x\ne 0)$$

$$\to c_1=-V_s$$

10. Finally:

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

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

Current Formula §

Using the characteristic capacitor equation:

$$\begin{array}{rl}{I}_C(t)& =C\frac{\text{d}{V}_C}{\text{d}t}\\ & =C\cdot \frac{\text{d}}{\text{d}t}(V_s-V_s\cdot e^{-t/RC})\\ & =C\cdot (\frac{V_s}{RC}\cdot e^{-t/RC})\\ & =\cancel{C}\cdot (\frac{V_s}{R\cancel{C}}\cdot e^{-t/RC})\\ & =\frac{V_s}{R}\cdot e^{-t/RC}=I_s\cdot e^{-t/RC}\end{array}$$

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:

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

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 $t=0$:

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

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

Dividing by $C$:

V(t) = V_{s}(1-e^{-t/RC})

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

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