# Homework Help: Calculating currents homework question

1. Sep 12, 2010

### jersey

1. The problem statement, all variables and given/known data

I have a closed circuit, with 7 bulbs, all of unequal resistances. I know three currents, 300, 200 and 500. I have 4 unknown currents, those fllowing through A, b, C and D which I need to find.

There is a graphic for this which I don't know how to insert, but the link to go to it is:

http://www.lightandmatter.com/bk4c.pdf [Broken]

GO to the link, scroll to page 123, and the diagram is problem 17.

2. Relevant equations

Loop rule where sum voltages in a closed loop = 0
Junction rule where the sum of currents flowing into and out of a point=0.

3. The attempt at a solution

I started by labelling the known currrents, from the left, G E and F, giving me a name for each one.

I know I need to use the junction rule to calculate the currents and I have this:

Rg+Re+Rf = Ra+Rb+Rc+Rd
1000= Ra+Rb+Rc+Rd

I need to come up with 4 equations for each unknown, so I have:

Ra=1000-Rb-Rc-Rd
Rb=1000-Rc-Rd-Ra
Rc=1000-Rd-Ra-Rb
Rd=1000-Ra-Rb-Rc

then i need to use substitution to solve each one, but the maths is getting lost on me. Can anyone help? Am i in the right direction?

I really appreciate any assistance you can give me to work this out.

Last edited by a moderator: May 4, 2017
2. Sep 12, 2010

### lewando

You can use the "inspection" method (plus the junction rule) to solve this one. For a 3-path junction, if 300mA is going into the junction from one path, and 200mA is coming out of the second path, then what must be going on with the 3rd path?

3. Sep 12, 2010

### jersey

Thank you for your replying. I'm not sure what to do next?

4. Sep 12, 2010

### lewando

The juction rule says that the sum of all currents entering or exiting a junction is zero. Take some time and wrap your mind around that rule (Its important and comes up a lot in circuit analysis).

So, if you have a junction consisting of 3 wires, the currents must add up to zero.

Unless I'm looking at the wrong problem, I see such a junction in the middle of your circuit.

A convention that is helpful is to declare current going into a junction "positive" and current going out "negative"

So, for that center junction we have:

(+300) + (-200) + (X) = 0

Solve for X.

5. Sep 12, 2010

### jersey

Thanks. I'm making it so much harder in my head so it;s good to see it like you put it,

So I have x=-100.

6. Sep 12, 2010

### lewando

Same principle applies. Give it a try, then I'll comment.

7. Sep 12, 2010

### jersey

Thankyou so much!

for the top junction To find A, I have
200+(-500)+A=0
-300=-A

(but this can't be right cause it's the same as one of the given resistances and they're all supposed to be unequal!)

Next, is C.

IF B is 100mA, and I have 500mA then

500 + (100) + c=0
-600=C

How am I going?

8. Sep 12, 2010

### lewando

300 = a

9. Sep 12, 2010

### jersey

hang on for A
200+ (-300)+A=0
-100=-A
A=100

BUt i'm stuck for D

600 + D= 0
D=-600

10. Sep 12, 2010

### lewando

I thought we agreed that "A" (the current going into the upper junction from A) was 300mA.

Not sure where "A=100" is coming from.

To get D, you could either look at the leftmost junction or the rightmost junction.

11. Sep 12, 2010

### jersey

The question states that each of the resistances is unequal. It can't be 300mA as that is the resistance of one of the given bulbs.

For D I have 500+600+D=0
D=1100

So, the 4 unknown currents shown are:
A=?
B=100mA
C=600mA
D=1100mA

HOw does that look?

12. Sep 12, 2010

### jersey

correction - it doesnt say anythging about currents not being the same does it.

So I have:

centre junction:
300 + (-200) + B= 0
B=100 mA , moving away from that junction.

TOP Jnction
(200)+(-500) +A=0
A=300mA moving into the junction

13. Sep 12, 2010

### jersey

to find c
500+100+C=0
C=-600mA

and d?

14. Sep 12, 2010

### lewando

If you know the current going through bulb B (100mA) then you know exactly what the current through bulb C is, yes?.

15. Sep 12, 2010

### lewando

To continue... With your knowledge of the current going through bulb C, you can do the junction current equation for the rightmost junction and get the current running through bulb D, yes?

16. Sep 12, 2010

### jersey

C is 100mA

using this information, i then have

100+100+500+D=0
700+D=0
D=-700

OMG? do I have it? Sorry I'm so slow with this, I really appreciate you sticking with me.

17. Sep 12, 2010

### lewando

Why do "100 + 100" ? Current is like water flowing through a pipe.

If you have 100mA flowing through bulb B, then you have 100mA flowing through bulb C, then you have 100mA flowing into the rightmost junction from that wire.

You are real close.

18. Sep 12, 2010

### jersey

I see! So C is 100mA

To find D
100+500+D=0
600+D=0
D=-600mA
D=600mA moving away from that junction

19. Sep 12, 2010

### lewando

Yes!

To complete your understanding of this, we have been referring to the "rightmost junction", the "middle junction", etc. for the sake of being clear. These are 3-wire junctions, where the interesting activity is happening.

But there are other 2-wire junctions that are less interesting like the one between bulb B and bulb C (shown as a single connecting wire for simplicity). The same junction rule applies, the sum of the currents must equal zero. For this junction:

(+100) + (-100) = 0

Seems trivial, but maybe it will help with your understanding.

20. Sep 12, 2010

### jersey

OK, I haev added that line into my workings and I can see exactly what you mean. You even picke dup the bits that I was not understanding.

I can't thank you enough! Seriously, I totally get it now. Sorry that took me so long ot grasp. Thanks a lot.