calculating currents homework question...

[jersey001]

HI,

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:

**broken link removed**


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


Can anyone help me?

 

[MrAl]

Hello there,


The way to solve that problem is to realize the node rule about currents that you mentioned:
"The sum of all the currents entering the node equals the sum of all the currents leaving the node."
Or:
"The algebraic sum of all the currents entering the node equals zero".

For example:
If we have a node with three branches the sum of currents into and out of that node must equal zero, so I1+I2+I3=0.
This also means that if we know any two of those three currents we can calculate the third just by solving for that one current. If we knew I1 and I2 we could calculate I3 by solving that equation for I3:
I1+I2+I3=0
solve for I3 by subtracting both I1 and I2 from both sides to get:
I3=-I1-I2.

It's also a convention to call currents entering the node positive and currents leaving the node negative, so if I1 and I2 were entering and I3 was leaving we would get:
(-I3)=-(+I1)-(+I2)
or
-I3=-I1-I2
or
I3=I1+I2

and it makes sense that at least one current would be leaving the node.

Numerical example for a three branch node:
We have 1 amp entering the node in one branch and 3 amps leaving the node in the second branch. What current is flowing in the third branch?
Since
I1+I2+I3=0

and we have 1 amp entering and 3 amps leaving with a third unknown, we have:
1-3+I3=0 and the unknown I3 is positive because 3>1, and simplifying:
-2+I3=0

solving for I3:
-2=-I3

or
2=I3

or
I3=2

so the remaining current is 2 amps.

If we instead had four branches we would have had to form I1+I2+I3+I4=0 and go from there.