Circuit 1 with Resistors and a Battery A circuit is...

Circuit 1 with Resistors and a Battery A circuit is constructed with five resistors and a battery as shown. The battery voltage is V = 12 V. The values for the resistors are: R1 = 75 ?, R2 = 148 ?, R3 = 167 ?, and R4 = 121 ?. The value for Rx is unknown, but it is known that I4, the current that flows through resistor R4, is zero. 1) What is I1, the magnitude of the current that flows through the resistor R1? 2) What is V2, the magnitude of the voltage across the resistor R2? 3) What is I2, the magnitude of the current that flows through the resistor R2? 4) What is Rx, the value of the unknown resistor Rx? 5) What is V1, the magnitude of the voltage across the resistor R1? 6) If the value of the resistor R2 were doubled, how would the value of the resistor R3 have to change in order to keep the current through R4 equal to zero? R3 would need to be increased R3 would need to be decreased R3 would not need to be changed There is no change that could be made to R3 to keep the current through R4 equal to zero.

Transcript

00:02 Okay, so for this problem, we're asked in the first part to rank the values of the potential difference in all of the resistors.

00:10 So it's actually quite easy if we understand how this works and we actually don't have to do any calculations.

00:18 Okay, so i already simplified, okay, the values of the resistors.

00:24 So these are to the marking in red.

00:27 Okay, is this sum of this? r2 resistor and i'm going to put it here and these are three resistors, these are in series.

00:37 So 6r equals just r2 plus r3, okay? and this one, since this are in r4 and this 6r resistor are in parallel, then when we find, if we remember how to sum, of resistors in parallel, then we can actually see that if you make these calculations, the result is this two are.

01:09 Okay? so i did this to simplify and for you to understand how the thinking is going to be.

01:19 So we need to find, and i'm making this in purple, how the total potential difference it's going to divide between this.

01:29 So i know that when i have a resistance in series, the potential difference divides, okay? and when i have them in parallel, it's the same.

01:40 So i'm going to look in this first.

01:44 I know that this is in series, so the potential is going to divide between r1 and these 2r.

01:52 Okay, so how this is going to be? i know that r1 equals r.

01:56 And the other resistance is 2r.

02:00 So i notice that when the potential difference divides, since this is half as much as this, okay, these r1 is going to need just half of the voltage difference than these 2r.

02:20 Okay, remember how this is, how the relation is.

02:25 Okay, so this is proportional.

02:27 So this, so if i have twice as much, i need just twice as much as the potential difference, okay? so i know that v1, that is how i'm going to call the potential difference, is going to be one -third, okay, of the total potential difference, and v of 2r, i'm just going to put it like this, is going to be two -thirds of the total potential difference, and this is actually going to be the final.

02:57 Value, okay, in terms of the total potential difference.

03:02 And this is for here.

03:03 And remember when i add up v1 plus v2r, okay, this has to be just a, and this is satisfying, okay? so what happens when i get through here, so for the second part, okay? so i know that these two are resistant.

03:28 Is actually the parallel sum of r4 and r6r.

03:37 That's the ones i have here.

03:42 Okay, so, but i know that since i it's a parallel configuration, the value of the potential difference is going to be the same.

03:53 Okay, so i know that v4 is actually going to be the same that the value i have here.

03:59 So it's going to be two -thirds of e, and this is going to be, or more like is the final value.

04:08 And i know that v of 6r, that is how i'm calling this, is 2 thirds also of the value.

04:21 And i still know that this 6r resistor is actually the sum.

04:30 Of r2.

04:32 Actually, i have it here, no? it's r2 plus r3.

04:36 And since these two are in series, okay? the potential difference or the voltage is going to divide.

04:45 Okay, so if i notice again how the values of r2 and r3 are, this is two times r and this is four times r.

04:56 So i notice that since this is, since this is twice as much as this, i will need two -thirds of the total value of the voltage than here.

05:12 Okay? it's the same thinking with it before.

05:15 So, following this, i know that v2 is going, is one -third of the total value of the voltage for 6r.

05:28 And i know that 6 is this, so it's going to be 1 third plus 2 thirds of e, and i will find that v2 is 2 over 9 times e.

05:44 And if i do the same thinking for v3, this is 4 over 9, and this is the final value for both of this.

06:00 So for ranking, okay, if i notice these values, i will see that v4 it's the biggest one, then it's b3, then it's b2, b1, i'm sorry, and then it's v2.

06:16 So this is how we rank these potential differences.

06:21 Okay, and the next part, we're asked to find how this is, how the potential differences are terms of the total potential difference, but we already did.

06:34 No, it's actually these values that i calculated before.

06:41 So i'm going to erase this, okay? and because we are going to do the same thinking, but now for current...

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