V 1 A R1 C R2 S A 2 Circuit 4: Figure 8. Circuit 4 1....

V 1 A R1 C R2 S A 2 Circuit 4: Figure 8. Circuit 4 1. Explain the values of the currents found in the data qualitatively... that is, why are the values as they are (do not compute them) 2. Using Kirchhoff's Rules, set up equations to analyze the circuit shown. a. Determine the current through the ammeter connected to battery #1. b. Determine the sum of the potential difference across the capacitor and the switch. R1 and R2 are equal to 33 ohms. V1=7.5v V2=10v. C=25000 uF What happens to 1 2a and b if the switch closes?

Transcript

00:01 Okay, so we're doing chapter 26, problem 49.

00:04 So we have this circuit here, which has all, all the resistors have the same resistance, big r.

00:10 And it says that time equals zero, the capacitor c is uncharged and the switch is closed.

00:17 And it says that time t equals zero, the three currents can be determining by, determined by analyzing a simpler but equivalent circuit, identify this, and use it to find the values of i, 2, and 3.

00:29 So if the capacitor is uncharged, there's no voltage dropped against it, which means it's just like the capacitor doesn't exist.

00:38 So for part a, the simplest circuit looks like this.

00:44 It's just the same thing without the capacitor.

00:55 So it will look like this with r, r, r, and r.

00:58 This is emf.

01:00 So we have i1 going this way, i2 down here.

01:06 I3 over here.

01:09 So now from this simple circuit, we can figure this out.

01:12 So first of all, we should know that we can figure out the req for this.

01:17 So we have one resistor here and then two more added in parallel program.

01:23 And these two parallel ones are in series with the first one.

01:28 So that means this is r plus the parallel branch.

01:35 And these are in series.

01:36 So we add these up and this becomes, three halves are.

01:43 So now using the loop or cure troughs looperol, we know that i1 is given by emf over iq.

01:53 So in this case, this is just two emf over three are.

02:04 So this is i1.

02:07 And now we also know that i1 is going to equal i2 plus i3 because if we look at this junction right here, we have i1 coming in and i2 and three, split apart and then coming back together here.

02:22 So from that, we should also know that i2 equals i3 because both branches have the same resistance.

02:33 So based on that, we know that i1 equals 2 i2 or i2 equals emf over 3r, which also equals i3.

02:47 Cool...

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