The figure shows a circuit containing an electromotive...

The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of $ C $ farads $ (F), $ and a resistor with resistance of $ R $ ohms $ (\Omega). $ The voltage drop across 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 $ (see Examples 3.7.3), 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, a battery gives a constant voltage of 60 V, and the initial charge is $ Q(0) = 0 C. $ Find the charge and the current at time $ t. $

The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of $ C $ farads $ (F), $ and a resistor with resistance of $ R $ ohms $ (\Omega). $ The voltage drop across 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 $ (see Examples 3.7.3), 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, a battery gives a constant voltage of 60 V, and the initial charge is $ Q(0) = 0 C. $ Find the charge and the current at time $ t. $

Transcript

00:01 All right, so we've been given the data of the problem, r, dq, dt, plus 1 over c times q equals e of t.

00:13 And we've also been given, we'll use this later, that q of zero equals zero.

00:20 All right, so again, r is the resistance, dq, dt is the current, c is the capacitance, q is the charge, and e of t is the voltage supply to the battery.

00:30 We were given numbers for all of those things.

00:32 So let's plug them in.

00:35 Dq d t times 5, since that was the resistance, plus 1 over 0 .05 times q equals 60.

00:45 And we know that this is a differential equations problem, and that we want, because we're going to be using the method of integrating factors, we want the, excuse me, the constant multiplying dqdt to be 1.

01:02 We have to divide through by 5, but before we do that, it would be easier to do so if we just remembered that 1 over 0 .05 is 20.

01:11 So let me rewrite that.

01:13 So 5 dq d t plus 20 times q equals 60.

01:21 And when we divide through by 5, it just gives us dq d t plus 4 times q equals 12.

01:29 And because we're using the method of integrating factors, i'm going to call it i.

01:35 I of t is e to the integral of whatever is multiplying q, which in this case is 4, dt, and the integral of 4 with respect to t is just 4t.

01:48 And since it's the method of integrating factors, we don't have to worry about plus c over here.

01:54 Okay? okay.

01:55 So now that we have that, we can multiply both sides of the differential equation by e to the 4t, which i will do on this new slide.

02:05 So we've got e to the 4t times dq d t plus 4e to the 4t times q times q equals 12, e to the 4t.

02:20 And we want to do something with this, but it's much easier to do so if we recognize that the left -hand side is just the derivative with respect to t of e to the 4 -t times q.

02:40 And i'll explain what that means in a second.

02:43 So if you take the derivative of e to the 4 -t times q, you know you have to use the product rule.

02:49 And the derivative of q is dqdt.

02:52 That's this half right here.

02:53 And the derivative of e to the 4t is 4 times e to the 4t times q.

03:03 So that's your left -hand side, that's your right -hand side, and we have a comparatively complicated derivative on the left -hand side.

03:12 And since we wanted to find q and dqdt, it would be much easier if we could simplify that.

03:17 Now, the way to do it here is by taking the integral of both sides, which would get rid of this ddt in the front.

03:24 So integrating both sides.

03:34 Sides, we can just look at what's on the inside of the derivative here, so that would mean e to the 4t times q, and then the integral of 12e to the 4t dt.

03:48 Now the integral of this, just to save time, use u substitution, where u equals 4t, and that integral comes out to e to the 4, 3 times e to the 4t plus c.

04:05 All right.

04:08 And you can do that out...

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