# Finding the current in the circuit.

Discussion in 'Homework Help' started by cdummie, May 23, 2015.

1. ### cdummie Thread Starter Member

Feb 6, 2015
104
1
In electrical circuit shown in the picture, there are resistances R2=R3=R4=R5=300Ω. R1 and R6 are unknown. If E1=30V and E2=E3=0V current I2=100mA and I3=40mA. Find the current I1 if E1=0V, E2=15V and E3=60V.

Now, i think this could be solved using superposition theorem and linearity theorem(i think that's the correct name). I mean i could put all of the resistors in passive network, but i don't know what to do next, since i can't find any solved examples similar to this.

2. ### Liqauttro New Member

May 23, 2015
17
1
Hello to you cdummie,
Well ,this circuit can be solved by a method called "Mesh Analysis"
In general, you should apply a few steps:
1)For example in this circuit you should place 4 current loops for each inner "closed circuit"(in general-- number of currents=number of loops) - one current for each loop
2)Choose a random direction of each one of them(the loops directions here are "independent",unlike the usual Kirchoff Currents Law)
3)Apply Kirchoff Voltage Law for each one of these loops
4)Express the voltage across each one of the resistors by Ohm Law ,pay attention that the total voltage across R3 should be a superposition of the
KVL of 2 loops (Since it is "common" for both "indepemdent " loops, a similar procces should be applied on R4 and also for R5 ,for the same reasons)
5)Now you should have 4 equations (number of loops= number of equation)
6)Solve them ,now pay attention the results of the currents for instance can be negative , for example I1=-0.6A ,The mening is - the direction you chose for I1 loop is opposite than in
what actually happens in "reality" , thats it(but its magnitutde remains 0.6A)

I truly hope I made it clear enough , If you have any questions or any attempts for solving this question , I can gladly help you

3. ### cdummie Thread Starter Member

Feb 6, 2015
104
1

I've tried it that way, but since i don't have all values i can't get the solution.

May 23, 2015
17
1

5. ### cdummie Thread Starter Member

Feb 6, 2015
104
1
I1 is upper left current loop, I2 is upper right current loop, I3 is down left current loop and I4 is down right current loop.

Now, i don't know what to do next since i don't have R6 and R1, and i have E1, E2 and E3 for a specific values of currents labeled in my first post.

6. ### Liqauttro New Member

May 23, 2015
17
1
First , it seems that the equations don't match the figure since for instance I1 loop goes through R3,R1,E2 --I2 goes through E2,R4.R2
My idea will be to make 2 more equations involving the missing resistances R1,R6
such as :
1)Vr2+E3+Vr6+Vr1+E1=0
2)Vr1+E1+Vr6+Vr5+E2=0

7. ### cdummie Thread Starter Member

Feb 6, 2015
104
1

But still i don't know what to do with E1, E2 and E3.

8. ### Liqauttro New Member

May 23, 2015
17
1
In which meaning ?What is your exact problem?

9. ### cdummie Thread Starter Member

Feb 6, 2015
104
1
Well, i don't know which values for E1, E2 and E3 to use, since there are two cases, in first case E1=30V, E2=E3=0V, then I2=100mA, I3=40mA (not the loop currents I2 and I3 but currents labeled in my first post), yet if E1=0V, E2=15V and E3=60V then i have to find I1.

10. ### Liqauttro New Member

May 23, 2015
17
1
I guess you should do mesh analysis twice
The first one with E1=30V, E2=E3=0V,I The second with E1=0V, E2=15V and E3=60V
I guess the first one should help you finding the missing resistances ,and then continue with the second analysis with your new knownw R1 and R6- if not , try writing the equations for "both circuits"
And then try to derive something from a combination of all equations by comparing one to another

11. ### Liqauttro New Member

May 23, 2015
17
1
I will try to make things a bit clearer:

Well let's list the things we know(in Red) and don't know(Black) in the first circuit ( with E1=30V, E2=E3=0V),
I1
I2
I3

I4
E1
E2
E3

R1
R2
R3
R4
R5

R6

As you can see 4 things are unknown-- so from mesh analysis ( with E1=30V, E2=E3=0V), you get 4 equations ,you should find these 4
and then do a second mesh anlysis with R1 and R6 (in case of: E1=0V, E2=15V and E3=60V) which you just found wih another 4 equations to find the new I1

cdummie likes this.
12. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
You have two unknown resistances and so you need two more equations. They've provided you with two additional pieces of information to provide those equations, namely the currents I2 and I3. So you need to use those to write constrain equations involving your mesh currents. That will let you solve for R1 and R6. Now you just have a normal network to solve for the second part of the problem.

I'm more concerned with the statement "since I can't find any solved examples similar to this." That's not the way to approach circuit analysis. You want to learn and understand the fundamentals so that you can apply that understanding to problems that you, and possibly no one else, has ever seen before.

cdummie likes this.
13. ### cdummie Thread Starter Member

Feb 6, 2015
104
1
I'll try to implement what Liqauttro suggested, it makes sense, but i could never think of solving it that way, the problem is that i solved only problems when i know all values, and when something like this appears it just confuses me (because i'm not that good at solving circuits), but is there a way to solve this using linearity theorem along with superposition (maybe it could be faster, and it could prove that current i get using mesh analysis is correct)? Anyway, thanks both of you guys, you really helped me.

14. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Pretend you don't know ANY of the values (which is really the way you should approach most problems, for a number of reasons).

If you use mesh current analysis then you will have four unknown mesh currents along with three unknown voltages and six unknown resistances, for a total of thirteen unknowns. So you need thirteen equations to solve the system. You have:

4 from the mesh analysis.
4 from the explicit values of the resistors.
3 from the explicit values of the voltage sources.
2 from the explicit values of two of the currents.

You thus have thirteen equations.

Try working a simpler problem that is along the same vein.

Forget that R1 is unknown. First come up with the mesh equations for IA and IB in terms of {Vs, R1, R2, R3}. You have two equations and three unknowns {IA, IB, R1}. Now write IA and IB in terms of I1. You know have three equations and three unknowns.