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The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of $ C $ farads $ (F), $ and a resistor with a resistance of $ R $ ohms $ (\Omega). $ The voltage drop across the capacitor is $ Q/C, $ where $ Q $ is the charge (in coulombs), so in this case Kirchhoff's Law gives

$ RI + \frac {Q}{C} = E(t) $

But $ I = dQ/dt, $ so we have

$ R \frac {dQ}{dt} + \frac {1}{C} Q = E(t) $

Suppose the resistance is 5 $ \Omega, $ the capacitance is 0.05 F, and a battery gives a constant voltage of 60V.

(a) Draw a direction field for this differential equation.

(b) What is the limiting value of the charge?

(c) Is there an equilibrium solution?

(d) If the initial charge is $ Q(0) = 0 C, $ use the direction field to sketch the solution curve.

(e) If the initial charge is $ Q(0) = 0 C, $ use Euler's method with step size 0.1 to estimate the charge after half a second.

a. $=12$

b. 3

c. $Q=3$ is an equilibrium solution.

d.

e. $Q(0.5) \approx 2.77 \mathrm{C}$

Differential Equations

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Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

we're getting the diagram of a circuit and were asked to solve the differential equation associated with this circuit as well as approximate solutions. This differential equation in part A. So the circuit were given as four distinct parts. We have e, which is our battery, which supplies voltage. We're told that bolted, just supplied constantly at 60 volts, see is our capacitor, which we're told has a good passions of 0.5 Henry's and our is a resistor, which we're told has resistance of five homes. Finally, we have a switch determines whether or not the circuit is open or closed like your chops long. You know that the change in voltage so the voltage drops some together equals the voltage coming out of the battery. So we have that the voltage drop across the resistor is our I nobles. Drop across the capacitor is Q. Where Q is the charge over? See, it's alright plus Q. Over C is equal e of t. But by definition, I the current is the same as the change in the charge with time or D. Q T t. So get substitute that into the equation to get our de que Deep T plus one of her seat times. Q equals E. T. And we've simplify this even further by plugging in our values for R C and P A. T to get five. Thank you, Bt Plus one ever see one over 10.5 This is the same is one of a 10.5 20 que is equal to 60. So we see we have this ordinary differential equation, which has degree of one. It's a linear differential equation. In part A were asked to draw the slope field of this equation so we can simplify this first by reading this as five de que DT is equal to 60 minus 20 que so that de que t t. She's also Q prime. If we think that Q Prime is the derivative respect, time is equal to 60. Divided by five is well minus 20. About it by five. Is for this world minus four. Q. You can actually make this even simpler for writing this four times three minus cute. So we now have Here are differential equation, you know, some that's more familiar to us. So if we take C to be some real number, which is B u t t. Then we have that the slope To have this value, it must be that C is equal to four times three minus que so in particular do you see that? Que must be equal to see over four minus three, native or three minus See here before where the sea is not the capacities This is just some real number c So this key was actually a function of time. So we have Here is a Siris of horizontal lines in a graph of Q of T vs t, the points on which all have the same slope. So we see that in particular, if we have C is equal to zero, then Q is equal to three. So you have three all the points on three. When have Slope zero we know that it's C is going to be greater than zero. It follows that que has he got a three minus sear before? Is going to be less Stan three. So we see that as we go further and further below three, our slopes are going to get steeper and steeper. So particular that's a Q equals zero or Q equals negative one. We have C equals 16 Q equals zero. We have C equals 12 so it gets quite steep quite quickly. It's going to be pretty hard to draw. Actually, in this I'm gonna redraw some of this. It's those I really don't need to draw the second quadrant either, because we're only concerned about positive T generally, so we'll move the y axis over as well. This will be three. This will be to so be one. It's very like this is Q T now, as I discussed, as we get further and further below the line, Q T equals three slips are going to get steeper and steeper. So if he was equal to say to when we have SI is equal to four already very steep slopes here, this is just an approximation what this should look like until we're essentially almost vertical as we crossed the X axis and we see that if C is going to be less than zero, so we have negative slopes, he was going to be greater than three, so in particular that he was equal to four. Then we'll have a slope of negative force already pretty steep. You see that as we get further and further away from the line. Q X equals three Q T equals three that you simply get steeper and steeper. Tell where nearly vertical places one part of the slope field. Probably most important part to see. And that's our insert apart. They in part B, we were asked, Find the limiting value off the charge. So what is the value of que as t approaches? Affinity? Well, this doesn't actually depend on our initial value. So no matter what we pick for, why a zero? We see that we follow these slopes as tangents. You will always end up following the line at Q of T equals three as an acid tip, so we see that no matter which solution Q of T, no matter the initial values, you see women as T approaches Infinity on q of t is three, and there's a more analytical way to see this, so limit as T approaches Infinity Q. Of t. You can see just from the graph, this is equal to three. Alternatively, I wouldn't know where to find this is you're assuming that we reached a steady state, which implies that change in the chart of time eventually is going to be zero. So we have the limit of this. As T approaches infinity, it's going to be zero and therefore we have that this the limit of this which is equal to the limits as T approaches positive infinity on four times three minus que, which using our properties of limits begin. Simply write this says 12 minus four times the limit As T approaches infinity q of t. We can solve this to get with the women as T approaches infinity, our q of t is equal to three. And this is because again, we assumed a steady state means that as t approaches infinity derivative people zero his limit zero in Parsi were asked to find an approximation. So you're asked first if there is an equilibrium solution. Remember that in equilibrium solution is when we had some constant C in the rial members such that potentially, um, change. Their function is equal to this constant. Well, this implies that four times three minus que is equal to see. Therefore, this implies that que is equal to three minus sear before this is what we calculated earlier. And you know this isn't equilibrium solution if SI is equal to zero. So we have an equilibrium solution of Q of t equals three in Part D. You're given an initial charge. We were asked to sketch the solution curve, using the direction field so back to were given that the initial charge Q of zero is equal to zero. And if I graph this in red our slope field, see that we have a point at 00 And if I follow the slope field, we quickly increase on these tangents until start leveling out at around Q of T equals three you follow. That is an asset. Looks a little bit like some kind of exponential growth in Part E were given initial charges again Q of zero equals zero and were asked to use Oilers method with a certain steps sized estimate charge at a certain time. So you're given once again that Q of zero equals zero and you're us to use a step size each equals 0.1 and to find the charge approximately after half a second. So from what we're given, you know that X not is equal to zero, and why not? Is why of X not, which is still equal to zero. From our step size, we get X one secret 0.1. We're gonna expect five steps. Why? One is equal to y 00 plus step size 0.1 times are function evaluated at 00 So what was our function? Well, we go back, we have that de que d t is equal to four times three minus cute. That's our function. So evaluated 00 because this is simply going to be 12 this is equal to 1.2 second step X two is equal 2.1 plus 0.1 just point to And why to, as he put a why one, which is 1.2 plus our step size times value our function at the point 0.11 point two This is four times three minus 1.2. This is equal to 1.92 Third step x three is equal 2.3 And why three people to why, to which is 1.92 plus our step size times value of the function at point point to 1.92 which is four times three minus one point nine to which is equal to two point 35 to our fourth step. X four is equal 2.4 and why four is able to why three child is represented as why three for brevity, plus our step value times function evaluated at the point point three y three, which is going to be four times three minus. Why three. This is equal to two point 6112 and finally, in step five, we have X five is equal to 25 we have. It's an approximation. Why have 0.5 is about equal to why five, which is equal to y for plus 0.1 times or times three minus y for I mean, this is equal to two points. 7667 to this is our approximation to Y 0.5, which can also be written more briefly, is two points. 77 Since this is in charge, we're looking at cool ums