Question

The resistors have the values Ra=2.00 ohm, Rb=9.00 ohm, Rc=1.00 ohm, Rd=1.00 ohm, Re=6.00 ohm, and the batteries, which have negiligible internal resistance, have emf's of Eu=3.50 V, Ev=9.50 V, Ew=1.50 V.
Calculate the power dissipated in Re

Answer 1

Calculate the current through Rd. Use a + sign for current flowing to the left in Rd.

Calculate the power delivered by Ev.

Answer 2

I know I need to add the left and right loops to get the currents, but I'm having trouble setting up the equations

Answer 3

how do name the current in loop 1 and loop 2? I1 and I2 for instance?

Answer 4

yeah, so I1 would be current through Re to the right, I2 would be current through Rc tupward and I3 would be current through Rd to the left

Answer 5

No need to use three currents. Imagine you have I1, clockwise, passing through Ew, Ev, Rc,Re and Rb
And I2 passing clockwise through Eu, Ra, Rd , Rc and Ev

Answer 6

Ok not sure how to set that up then

Answer 7

Is the equation for loop1 (forget about I2 now)
I1(Rc+Re+Rb)+Ew+Ev=0?

Answer 8

That looks correct to me. So would the second loop be I2(Ra+Rd+Rc)+Eu+Ev=0 or would it be different because its in a different direction?

Answer 9

YEs except for Ev, whose sign has to be "minus" because I2 is hitting the battery by the negative part

Answer 10

ok, so from those I can find the currents. To get the power dissipated there is it just current^2 * resistance? also not sure about power delivered

Answer 11

No, no, bear with me, we have not finished yet the equations!

Answer 12

ah ok

Answer 13

In the equation that I wrote, remember I told you to discard I2?

Answer 14

yep

Answer 15

What is the effect of I2 in Loop1? Isn't it -I2xRc

Answer 16

yep but not sure what that gives you

Answer 17

Just put -I2xRc in my equation and you have the right equation for loop1. Now you do the same for loop2. See the effect of I1 in loop2

Answer 18

so you subtract that form loop one and add it to loop 2?

Answer 19

The effect of I1 in Loop2 is -I1 x Rc. You get it? (I1 goes opposite to I2, therefore its effect on common parts to both loops will be negative)

Answer 20

Put -I1 x Rc in your equation for loop2

Answer 21

So it's opposite?

Answer 22

Not sure where you're going with this

Answer 23

Yep. Now you can tell me the complete equations for loop1 and 2

Answer 24

I1(Rc+Re+Rb)+Ew+Ev-I2xRc=0 for loop 1?

Answer 25

Yes, I give you the second. Check with mine:
I1x(Rc+Re+Rb)-I2xRc=-Ew-Ev
-I2xRc+I2x(Ra+Rd+Rc)=-Eu+Ev

Answer 26

yep, looks correct. So from the currents then how do you get the power dissipated and delivered?

Answer 27

The power is RI^2 and the current that passes through Re is I1, then you have it

Answer 28

From the first equation I get I2 =2, does that seem correct?

Answer 29

Woops, I mean second equation

Answer 30

Since there is only one variable

Answer 31

No, there are two variables. It is an equations system:


Loop1: I1x(Rc+Re+Rb)-I2xRc=-Ew-Ev


Loop2: -I2xRc+I2x(Ra+Rd+Rc)=-Eu+Ev

Answer 32

Solve the system

Answer 33

I don't see 2 variables in the second equation. Only I2

Answer 34

Mistake of mine: The effect of I1 is -Rc x I1, then the second equation is:
-I1xRc+I2x(Ra+Rd+Rc)=-Eu+Ev

Answer 35

ah ok

Answer 36

Sorry for that

Answer 37

no problem, give me a sec to work it out

Answer 38

I end up with a negative number for I1, but that should be ok right?

Answer 39

I get I1=-38/63

Answer 40

yep same here

Answer 41

And I2=85/63

Answer 42

yep

Answer 43

You got it. What is the current through Re?

Answer 44

-38/63

Answer 45

That is, I1, then We=I1^2 x Re

Answer 46

gives me 2888/1323 Jor 2.18 J

Answer 47

We=(38/63)^2 x 6 in Watts

Answer 48

oh woops yeah watts lol

Answer 49

If I asked to calculate the power in Rc, what current would you use?

Answer 50

I'm not sure. both?

Answer 51

The current through Rd should just be I2, right?

Answer 52

Yep, I1-I2 or I2-I1, is the same because for the power you use power two

Answer 53

and through Rd is I2, true

Answer 54

Great! so now for the power delivered by Ev?

Answer 55

should be related to the current through Rc

Answer 56

Maybe the way to solve is a bit fuzzy but see that it is like writing the equations for each loop but substracting the effect of the current of adjacent loops. like this you may solve any circuit no matter how complex.

Answer 57

That makes sense. I think I need more practice

Answer 58

Yes, but take into account that if I1-I2 is negative you can say that the battery is delivering power, otherwise we are charging it

Answer 59

so Ev should just be (I1-I2)^2 * Rc

Answer 60

No. It is (I2-I1)xEv (the battery delivers power for a current that is flowing out from the battery, that is whay I said I1-I2 has to be negative)
Then I2-I1 has to be positive and the power is (I2-I1)xEv

Answer 61

It gives me a number around 19 Watts

Answer 62

Ah woops, was thinking about the wrong thing. Thank you, that worked and maeks more sense to me now. Time to go practice some more

Answer 63

I leave you, I am going to hit the hay. Enjoy circuits and electronics. It is really amusing