In the circuit shown below mathcale 710 v r5 400 omega r3...

In the circuit shown below $\mathcal{E}$ = 71.0 V, $R_5$ = 4.00 $\Omega$, $R_3$ = 2.00 $\Omega$, $R_2$ = 2.20 $\Omega$, $I_5$ = 9.80 A, $I_4$ = 6.50 A, and $I_1$ = 13.3 A. Find the current through $R_2$ and $R_3$, and the values of the resistors $R_1$ and $R_4$. $I_2$ = 6.70 X Based on the known variables, which two junctions should you consider to find the current $I_2$? A $I_3$ = 3.20 A $R_1$ = 5.34 X Which loop will give you an equation with just $R_1$ as the unknown? Did you follow the sign convention for the potential difference across each element in the loop? $\Omega$ $R_4$ = 5.94 X Which loop will give you an equation with just $R_4$ as the unknown? Did you follow the sign convention for the potential difference across each element in the loop? $\Omega$

Transcript

00:02 In this question, we are going to find the resistances and currents given in a complex resistor network in order to find a couple of unknown currents and resistances.

00:18 We're going to end up using kirchhoff's loop and junction rules.

00:23 So the junction rule, of course, basically says that.

00:34 Sum of currents into a junction equals the sum of the currents leaving the junction or going out.

00:48 And the loop rule says that the sum of potential rises minus the sum of potential drops equals zero.

01:12 Or you could also say it that the sum of potential rises in a loop.

01:19 Equals the sum of all the voltage drops in that loop.

01:25 So to start with, we're going to find i2 by looking at the fact that we must have i1 coming up this way and going into contributing to i1.

01:44 And then we are going to have i3 coming up over here on the right side of the circuit.

01:51 And then around and then down.

01:55 So we're really basically saying that, for part a here, that i2 is equal to i1 plus i3.

02:09 Well, that's all well and good, but we don't have i3 yet.

02:14 How are we going to get i3? let's see here.

02:21 So i3 is going to have okay, we'll do it this way.

02:28 So we're going to have i -5 coming across to the right side of the circuit and up.

02:37 All right.

02:38 And then we're going to notice that i -5 is going to split right here into i -4 coming across to the left in this circuit, and it's going to split into i -3, continuing up the right side of the circuit.

03:00 So we are basically able to say that i3 is equal to i5 minus i4.

03:09 So however much current was flowing through the branch with i5, some of it splits off to make i4, whatever's left continues on as i3.

03:18 Well, we can find a value for i3 with the given information because i5 was given as 8 .30amphers.

03:30 And i4 has been given as 6 amperes, so 8 .3 minus 6, and i get a value for i3 of 2 .30 amperes.

03:45 And now that i have this, i can find that i2 is equal to i1, which was given as 12 .1 amperes minus that 2 .30 amperes that we just found.

04:03 Or sorry plus the 2 .30 amperes and i get a value for i2 of 14 .4 amperes excuse me all right and now we can apply the loop rule to find our missing resistances so our loop for r1 is going to start um let me see okay we're going to start at the battery and we're going to come down that center go across to the left, come up through r1.

04:46 So we had a potential rise from the battery e minus the potential drop v1.

04:56 And then if we continue on around our loop, we can then cut down through the center through r2.

05:05 And that brings us back to the negative terminal of the battery.

05:10 So we're going to subtract the voltage across resistor 2, and that's going to be equal to 0.

05:16 Well, how can we find v1? v1, of course, will be equal to the current through resistor 1 multiplied by its resistance.

05:25 V2 will be i2 times r2, which we have, and we have our battery voltage as well, 69 .5 volts, minus, while the current through 1 was given, as 12 .1 amperes multiplied by the unknown r1.

05:46 And then we also have to subtract out i2 times r2...

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