How Do You Apply Series and Parallel Rules in Circuit Analysis?

[Nikola_Tesla]

Forewarning, the following is a simple grade 11 circuit analysis question that may be quite simple material for others on this thread. I would appreciate any assistance in determininghow this problem may be tackled.

Homework Statement

Homework Equations

Series:
V_T=V₁+V₂+V₃
I_T=I₁=I₂=I₃
R_T=R₁+R₂+R₃

Parallel:
V_T=V₁=V₂=V₃
I_T=I₁+I₂+I₃
1/R_T=1/R₁+1/R₂+1/R₃

The Attempt at a Solution

The difficulty in this question is determining where series and parallel rules apply. The total resistance may first be determined to be 1.2Ω using Ohm's Law. I then believe that it would be in my best interest to determine the combined totals for each line, so the problem may then be treated as parallel. When this is performed, the numbers simply don't add up. Should the voltages of the resistors in each line add up to twelve? Thank you for any assistance in determining what step must be taken in order to solve for the unknowns.

 

[gneill]

Ponder the currents first. What do you know about the current in series and parallel circuits?

Note that there's a current specified for R2... Does that tell you anything about the currents in other components?

 

[Nikola_Tesla]

Yes, I believe that the current would then be 2A in the other resistors in that line, but am unsure as to whether 2 A would also apply to I4 and I5.

 

[Cryptic edge]

Series vs parallel is the easy part, resistor 1,2,3 are all in series, as are 4,5. They are both in parallel to themselves and the battery being the point all parallels have to be equal to.

 

[gneill]

Yes, I believe that the current would then be 2A in the other resistors in that line, but am unsure as to whether 2 A would also apply to I4 and I5.

It will only apply to the resistors in the same branch.

But! What other current do you know?

 

[Nikola_Tesla]

Based on the total current being 10 A, would the second branch currents have to add up to four?

 

[Cryptic edge]

Based on the total current being 10 A, would the second branch currents have to add up to four?

It's been a decade since I've done these, but I believe that's correct.

 

[gneill]

based on the total current being 10 a, would the second branch currents have to add up to four?

2 + 4 = 10 ?

The currents in the components of a given branch do not add; it is the SAME current flowing through all components of a series branch.

 

[Nikola_Tesla]

Ok, so I4 and I5 have to be 8A. Then, how would V2 be determined? Would it be 5.5V based on the total being 12V, or would the second branch voltages also have to be part of the numbers that add up to a total of 12V?

 

[gneill]

Ok, so I4 and I5 have to be 8A. Then, how would V2 be determined? Would it be 5.5V based on the total being 12V, or would the second branch voltages also have to be part of the numbers that add up to a total of 12V?

The branches are in parallel, so the both "see" the same 12 V total. So go ahead and use the fact that the total voltage across the branch is 12 V in order to find V2.

Proceed in the same manner, filling in values as you obtain the required information (a bit like doing a crossword, this!). When you have both the voltage and current for a given resistor, you can work out its resistance.

 

[Nikola_Tesla]

Ok, so I believe that I have it and have made some checks, but wish to double check all of the unknown values if that is okay.

Rt= 1.2 ohms, R1= 1 ohm, R2= 2.75 ohms, R3= 2.25 ohms, R4=0.875 ohms, R5= 0.625 ohms

V2= 5.5V, V4=7V

I1=I2=I3= 2A, I4= 8A, I5= 8A

Thanks you for all your help on such a night.

 

[gneill]

Your results look fine to me

You're very welcome. I'm happy to help.

 

[Nikola_Tesla]

Thanks again for your assistance, it seems like you have helped a lot of people. Have a good night.