# Understanding The Art of Electronics mathematical explanation for capacitors

The author mentions that the reader should not worry if they cannot keep up with the mathematics, I however feel uncomfortable doing so. The issue is the fact that the author does not explain what the mathematical variables are for so I would like someone who understood it to clarify. The give the following circuit simulate this circuit – Schematic created using CircuitLab

It mentions that at time = 0 the equation for the current is $$I = C{dv\over{dt}} = {V_i - V \over {R}}$$ The author does not specify what Vi or V are. However, I can conclude that you would need to subtract the voltage supplied by the battery by the voltage of the capacitor. Vi is the source voltage, V is the voltage on capacitor. Afterwards they define the V (capacitor voltage) as $$V=V_i+Ae^{-t\over{RC}}$$ What confused me here is what the term Vi was doing in the expression for the initial voltage across the capacitor or does Vi stand for something else? also what does A signify? As the author then states that A is determined by initial conditions V=0 ,t=0. A will be equal to -Vi so then $$I={V_i -(V_i + (-V_i * e^{-t/RC})\over{R}} = {V_i * -e^{-t/RC}\over{R}}$$ at this point I have lost track of the intuition behind it and if someone could walk me through what the author means, or really just the intuition behind the mathematics I would appreciate it.

• dV is the change in voltage, so Vi is in the initial voltage and V is the final voltage at the time you care about. 'A' is just a placeholder constant which he then derives.
– ACD
Dec 30 '14 at 17:43
• So there is no term for the voltage being supplied? Dec 30 '14 at 17:59
• I am currently reading and it states V= Vi * (1-e^-t/RC) if Vi represented initial voltage on the capacitor, if it were zero then V would always be zero, it makes more sense if Vi was the supply voltage since the left term is approaching 1 which is multiplied by Vi (supply voltage) to give a V that is approaching the supply voltage Vi Dec 30 '14 at 18:08
• I think your problem is that the diagram notation and the mathematical notation have not been matched up by the original author or picked up by a proof reader. C refers to C1, R refers to R1, Vi (not even shown on the diagram) is the initial voltage across the capacitor and V1 is the source voltage V. Dec 30 '14 at 18:41
• In my copy of this (excellent) book - paperback, 1983 - what you have shown as voltage source V1 is instead illustrated as a battery labelled Vi Dec 30 '14 at 18:54

First, you should be aware that the treatment of circuit theory in The Art of Electronics is very brief. An actual circuit analysis textbook would cover time constants in much more detail and provide more examples.

Your circuit appears to be Figure 1.31 from section 1.13. You left off the switch and the initial conditions, and mislabeled the voltage source. The battery is $V_i$, not $V_1$. Here's a corrected version: simulate this circuit – Schematic created using CircuitLab

It is implied (but not stated) that the capacitor is discharged at t = 0, so that V starts out at zero volts. With the switch open, there is no DC path from V to ground, so we have to make an assumption like this.

Once the switch is closed, current can start to flow. The mathematical way of approaching this is to write an equation using Kirchhoff's Current Law (KCL). Because of the capacitor, this will be a first-order differential equation:

$$current\ out\ of\ V_i = current\ into\ C$$ $$\frac{V_i - V}{R} = C\frac{dV}{dt}$$

(A differential equation is an equation involving a rate of change. Here, $\frac{dV}{dt}$ is the rate of change of V with respect to time.) You can then solve this to get an equation of the form:

$$V = V_i + Ae^{-t/RC}$$

where e is the base of the natural logarithm (~2.718) and A is an unknown constant. You can solve for the constant using your initial condition of $V_{t=0} = 0$.

An alternate way of looking at it is to say that a capacitor acts like an open circuit at DC, and like a short circuit when the voltages in the circuit are changing rapidly. At the instant we close the switch, we have a rapid change -- $V_i$ is suddenly applied all at once. The capacitor acts like a short to ground, so the current is $V_i/R$. After a long time, the voltage has stabilized and we effectively have a DC circuit. The capacitor acts like an open circuit, so $V = V_i$ and no current flows.

The transition between these two states in an exponential decay. This means that the equation for V will have a term like $e^{-t/\tau}$, where $\tau$ (tau), called the "time constant", determines the rate of decay. At t = 0, this exponential term is equal to 1. As t -> infinity, the exponential term decays to 0. We can use this to get an equation for V:

$$V = (final\ condition) - (difference\ between\ final\ and\ initial\ conditions) * (exponential\ term)$$

At t = 0, when the exponential term is equal to 1, this gives us:

$$V = (final\ condition) - (difference\ between\ final\ and\ initial\ conditions) = (initial\ condition)$$

At t = infinity, when the exponential term is equal to 0, it gives us:

$$V = (final\ condition) - 0 = (final\ condition)$$

In this circuit, our initial condition is $V = 0$. Our final condition is $V = V_i$. The difference between them is $V_i - 0 = V_i$. We would normally have to solve the differential equation to get the time constant $\tau$, but what the book is telling us is that for an R-C circuit, $\tau = RC$. Now we can write the final equation:

$$V = V_i - V_ie^{-t/RC}$$

When the initial condition is zero (as it is here), we can write the equation as:

$$V = (final\ condition) * (1 - (exponential\ term))$$

which gives:

$$V = V_i(1 - e^{-t/RC})$$

And that's exactly what's in the book.