RC Circuits Primer

The questions below are due on Monday October 06, 2025; 10:00:00 PM.

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In this problem, we'll step through the process of solving a couple of different circuits involving a capacitor and a resistor ("RC" circuits), and we'll see some common patterns in the solutions that we can bring to bear on future problems of this sort that we encounter, without needing to rederive them.

1) Capacitors

In lecture next week, we'll start working about circuits involving capacitors. Introducing these components will be a big leap forward for us, because they differ from the other components we've seen so far in at least one key way: they are stateful. That is to say, if we want to know the voltage drop across a capacitor, it's not enough to know the current flowing through the capacitor just at that moment; rather, we need to know the whole history of currents flowing through the device.

In our lumped element abstraction, we'll draw capacitors like so:

And, as with the other components, we'll model a capacitor with a constitutive equation expressed as a constraint on i and v. There are a number of (equivalent) ways that we could write this equation for the capacitor:

v(t) = {1\over C}\int_{-\infty}^t i(\lambda){\rm d}\lambda

v(t) = v(t_0) + {1\over C}\int_{t_0}^t i(\lambda){\rm d}\lambda

i(t) = C{{\rm d}\over {\rm d}t}v(t)

All of these are equally valid, though we'll probably primarily use the last form i = C{{\rm d}v\over {\rm d}t}.

It's important to note that now that we have capacitors in the mix, voltages are currents are functions of time (they won't necessarily be the same from one moment to the next). For simplicity of notation, though, we'll often skip writing the (t) part and just write the variable name; but it's important to remember that we're talking about functions of time.

2) "Natural" Response

Let's start by considering the following circuit, with a resistor and capacitor connected up to each other:

Applying KVL and KCL, we can see that i_{\rm R} = -i_{\rm C} and v_{\rm R} = v_{\rm C}. We'll use that as the basis for solving out this circuit. Let's try to solve for v_{\rm C}(t). We can start with our KCL equation, rearranged slightly:

i_{\rm R}(t) + i_{\rm C}(t) = 0

Then we can substitute in values for each current based on the (shared) voltage:

{v_{\rm C}(t)\over R} + C{{\rm d}\over {\rm d}t}v_{\rm C}(t) = 0

and multiply through by R to get to a more comfortable form:

v_{\rm C}(t) + RC{{\rm d}\over {\rm d}t}v_{\rm C}(t) = 0

One key thing is that, now that we have capacitors in the mix, all of our methods from earlier in the course now give us systems of differential equations to solve, instead of systems of linear equations.

We're not expecting you to have taken a course on differential equations before, but this one has the same form as the ones we walked through last week.

Like we saw then, our first step in solving this will be to take an educated guess at the form of v_{\rm C}(t). In order for that sum to work out to zero, we need v_{\rm C}(t) to look a lot like its time derivative. As such, we'll assume that v_{\rm C}(t) is of the form Ae^{st} for some values of A and s.

Plugging in our assumed answer v_{\rm C}(t) = Ae^{st} to our equation from above, we get:

Ae^{st} + RCsAe^{st} = 0

This equation alone is enough to solve for s:

s = -{1\over RC}

So plugging in, we find that:

v_{\rm C}(t) = Ae^{-t/ RC}

The value RC (which has units of seconds) is important enough that we will give it a name (it's the "time constant" of this circuit) and a symbol (the Greek letter tau: \tau):

v_{\rm C}(t) = Ae^{-t/\tau}

We're going to need to do a little more work to completely characterize our equation, though (KCL isn't enough!). Because of the stateful nature of the capacitor, we'll need to know the value of v_{\rm C}(t) at some point in time in order to solve for values of v_{\rm C} after that point in time (that is, we'll need to know some initial condition in order to fully solve). For now, let's assume that v_{\rm C}(0) is a known value. We can use this to find A:

v_{\rm C}(0) = Ae^{-0/\tau} = A

And we've done it! If we note that this will only hold after our initial condition, we can say that:

v_{\rm C}(t > 0) = v_{\rm C}(0)\times e^{-t/\tau}

Or, slightly more generally, if we know the value of v_{\rm C} at an arbitrary time t_o, we can say that:

v_{\rm C}(t > t_o) = v_{\rm C}(t_o)\times e^{-(t - t_o)/\tau}

If we plot this function starting at time t_o, we see a very characteristic shape:

3) "Step" Response

Now let's consider a slight variant of the circuit from above, now with a voltage source introduced in series with the components:

Let's assume, as an initial condition, that v_{\rm C}(t_o) = 0 and think about how this circuit behaves.

At that instant, what must the current i_{\rm C}(t_o) be? Enter your answer as an expression in terms of V_s, R, C, and/or any other constants you need.

i_{\rm C}(t_o) =~

Note that the derivative of the voltage is proportional to this current. So v_{\rm C} will start to increase from 0. But as v_{\rm C} increases, the voltage drop across the resistor decreases, causing the current to decrease, causing the derivative of v_{\rm C} to decrease, and so on and so on, until, in the limit, things settle out to a point where v_{\rm C} is a constant.

In this limiting case, what is the current i_{\rm C}?
i_{\rm C}(\infty) =~

In this limiting case, what is the voltage v_{\rm C}?
v_{\rm C}(\infty) =~

So now that we know something about the limiting conditions, let's try to solve this out (again under the assumption that v_{\rm C}(0) = 0).

We can start by setting up our equation using KVL:

V_s = v_{\rm C} + v_R

Then can can express v_R in terms of the current running through this loop:

V_s = v_{\rm C} + i_{\rm C}R

And then, since this current i flowing through the resistor is the same as the current flowing through the capacitor, we can use the capacitor's contitutive equation to replace i with C{{\rm d}v_{\rm C}\over {\rm d}t}:

V_s = v_{\rm C} + RC\dot{v}_C

This equation is now in the form that we saw last week. Let's go ahead and solve it, subject to the same initial condition as above (that v_{\rm C}(t=0) is 0).

3.1) Homogeneous Solution

We start by finding the homogeneous part of the solution, which is the answer to 0 = v_{\rm C} + RC\dot{v}_C. Conveniently, that's just the answer to the problem we solved above! So we can say that, for some value of A that will depend on initial conditions:

v_{{\rm C},h} = Ae^{-t\over RC}

3.2) Particular Solution

We also then need to solve for the particular part of our solution, which is any answer to the original differential equation V_s = v_{\rm C} + RC\dot{v}_C. Given the form of this equation, a good guess is to assume that v_{{\rm C},p} is a constant value.

Given the assumption that v_{{\rm C},p) is a constant, what is the appropriate value? Enter your answer as an expression in terms of V_s, R, C, and/or any other constants you need. You may express e^x as e**x or exp(x), and you may express \ln(x) as log(x).

v_{{\rm C}, p}(t) =~

3.3) Complete Solution

Our overall answer for v_{\rm C}(t) will be the sum of the homogeneous part and the particular part of the answer, after subsituting in an appropriate value for A to match the initial condition that v_{\rm C}(0) = 0.

Using this information, solve for v_{\rm C}(t), subject to the constraint that v_{\rm C}(0) = 0. Enter your answer as an expression in terms of V_s, R, C, and/or any other constants you need. You may express e^x as e**x or exp(x), and you may express \ln(x) as log(x).

For t>0, v_{{\rm C}}(t) =~

4) General Answer

This solution generalizes to situations where our initial conditions are not 0. We'll see a more-careful solution to this in class, but here we'll just jump to the conclusion: v_{\rm C} starts at some initial value v_{\rm C}(t_o) and follows a familiar exponential curve that asymptotes out at some final v_{\rm C}(\infty), with the same time constant as our original circuit. So we can say that, for t > t_o:

v_{\rm C}(t > t_o) = \underbrace{v_{\rm C}(t_o)}_\text{initial value} + \left(\underbrace{\left(v_{\rm C}(\infty) - v_{\rm C}(t_o)\right)}_\text{difference between initial and final values}\times \underbrace{\left(1 - e^{-(t - t_o)/\tau}\right)}_\text{upside-down exponential curve}\right)

Plotting this curve, we see a similar kind of shape, but now upside-down: