Series-Parallel

[Ohm1]

Here is a good one. I actually go this one wrong. My answer was 7amps. Yet, I'm an expert on it now, so obviously I have the correct answer in my head. If you want to take a shot at it, go for it. My special saying: There is no shame in this learning game. Meaning: If you've been doing electrical for 4 or more years, then don't feel like you have to be ashamed of getting the answer wrong. Many loose it, when they don't use it!




What is the total (I) in the circuit?

[CraigFL]

6 Amps?

Craig

[joed]

I get 6 amps as well.

[Ohm1]

I hope we aren't going to some resource (books, online, etc etc) for our answers--that's cheating. I haven't worked a problem like this in over 4 years, yet I tried it with no resources, and got pretty close to the correct answer<<<<This is not to say the above is correct. :D

I'll await others to respond.

[CR500]

R3,4,5,6,7 = 2/3 ohm

Total Resistance = 6 2/3 ohms

I = 6 amps

[Mr T]

I got 6A too. It's been a while since ive done this too. Just dont be breaking out thevenin / norton equations! :pcguru:

[Fischer]

Not being an electrician, I thought it was modern art. Bauhaus I think.

Anyone care to explain how you figure this one out?

[Ohm1]

Not being an electrician, I thought it was modern art. Bauhaus I think.

Anyone care to explain how you figure this one out?


I will try to explain it the best I can.

The above drawing presents a series-parallel circuit. To find the total current (I) in the circuit, we should first reduce the circuit to only one resistance--a resistance that is similar to the total value of all the series and parallel resistances. If you look at the drawing above, you will find R6 and R7 in one branch. R3 in another. R4 and R5 in the last of three. Theses three will need to be reduced. We reduce them by identifying the kind of circuit that exist in each branch. R6,7,3,4 and 5 are in series. [Note: To calculate a resistance in series, we simply add the numbers together.]. So R6=2 and R7=4-- 2+4=6. Now we go to the next series branch that needs to be added: R4=2 and R5=1 --2+1=3. Then the last is simply R3=1. Now we have three individual resistors in the circuit: 6/3/and 1. We now need to calculate the branches (6/3/and1) using the reciprocal or the product over sum method for calculating parallel resistances [Note: We will use the reciprocal method to solve this one).


1/6=.17
1/3=.33
1/1=1
Total of three branches= 1.5 So: 1/1.5=.66
The above calculation has reduced the three branches, making the above drawing a series circuit. As we said above, to calculate the resistance in a series circuit, we simply add the resistances. So: .66+1+3+2=6.66

Then we use ohms law to conclude:
I=E/R
I=40/6.66
I=6 Amps


Hope that helps!!

[Mr T]

To clarify further, the formula for parallel resistiance is:

1
-------------------
1 1 1
--- + --- + ---
R1 R2 R3

[Ohm1]

To clarify further, the formula for parallel resistiance is:

1
-------------------
1 1 1
--- + --- + ---
R1 R2 R3


I'm sure you know that this isn't the only formula to calculate parallel resistances.

[MJ CORMIER]

I've Got 6.5
MIKE:willy:

[Fischer]

I will try to explain it the best I can.

The above drawing presents a series-parallel circuit. To find the total current (I) in the circuit, we should first reduce the circuit to only one resistance--a resistance that is similar to the total value of all the series and parallel resistances. If you look at the drawing above, you will find R6 and R7 in one branch. R3 in another. R4 and R5 in the last of three. Theses three will need to be reduced. We reduce them by identifying the kind of circuit that exist in each branch. R6,7,3,4 and 5 are in series. [Note: To calculate a resistance in series, we simply add the numbers together.]. So R6=2 and R7=4-- 2+4=6. Now we go to the next series branch that needs to be added: R4=2 and R5=1 --2+1=3. Then the last is simply R3=1. Now we have three individual resistors in the circuit: 6/3/and 1. We now need to calculate the branches (6/3/and1) using the reciprocal or the product over sum method for calculating parallel resistances [Note: We will use the reciprocal method to solve this one).


1/6=.17
1/3=.33
1/1=1
Total of three branches= 1.5 So: 1/1.5=.66
The above calculation has reduced the three branches, making the above drawing a series circuit. As we said above, to calculate the resistance in a series circuit, we simply add the resistances. So: .66+1+3+2=6.66

Then we use ohms law to conclude:
I=E/R
I=40/6.66
I=6 Amps


Hope that helps!!

Thankfully, I will probably never need to implement this formula, but it's great to learn new things. I can actually follow the formula, but don't test me on it.

Thanks for the info.

[junkcollector]

I remember I was first exposed to this stuff back in high school. I remember using another formula though, "product over sum" as its called. R1 X R2 / R1 + R2 etc. If you had more than two, you would just to R1 and R2, then the total to R3, and so on. If you have all the same value resistance, you can just divide the value in ohms by the number of resistance. The "one over one over R" formula mentioned above is the most valuable because it can be used for all types of parallel circuits. (easy if you have a calculator...)

The trick is to solve the parallel segment into one equivalent resistor, so you end up with one simple series circuit.

This is kid stuff, AC is more fun!!!

[Mr T]

I'm sure you know that this isn't the only formula to calculate parallel resistances.

Its one of the easier for someone who is learning and for when there are more then 2 resistances in parallel.

[Ohm1]

Its one of the easier for someone who is learning and for when there are more then 2 resistances in parallel.
I agree!!!!!

[Ohm1]

This is kid stuff, AC is more fun!!! What you said is Interesting!!

[Mr T]

I remember some of my early electronics classes . The course started with a 45 minute lecture / hands on with a ohmmeter on how dangerous electricity is and how little it takes to kill.

We did problems that had a few more resistors in it and he would pick one and say to figure out the current flowing through it or the voltage at a certain point of the circuit. Sometimes it would be a wattage calculation. It was about a roomy 30 person classroom that had a chalkboard (remember those?) across 2 walls. The circuit was shown on a overhead slide and the workout started at one end of the room and went across both chalkboards then he had to erase the front board and it filled up 2/3 of the front board again. It took about 40 minutes to solve. I'm guessing over 50' of chalkboard was used. Our exams consisted of a few of these questions. So on that note I do agree.. kid stuff.

[CR500]

... "product over sum" as its called. R1 X R2 / R1 + R2 etc.

JC,

The equation you provided isn't another formula, it just hasn't been reduced. If you combine like terms (divide by R1 and R2) the result will be the same as Mr T's.

[junkcollector]

Perhaps I should have said, "another method".

[Ohm1]

Bang on

I got exactly the same answer after a few attempts of trying to remember the recipricol method. Got any more?
Yes, I do!
I will post another later tonight!

[Ohm1]

What is the total current (I) in this circuit?

[joed]

We need a voltage(E) to go with the total R of 31 ohms to figure the current.

[Ohm1]

We need a voltage(E) to go with the total R of 31 ohms to figure the current.
Sorry. lets make it 30 volts.

[joed]

Then .97 amps flows.

[Ohm1]

Yes

0.97 amps rounded off from 0.967.

I will think of some myself when I get the time.

Regards

If you do, start a new thread.

I will post my answer tonight or tomorrow.

[Ohm1]

What is the total current (I) in this circuit?
(30)(20)/30+20=R
600/50=12 ohms
12+9+10=31 ohm
E/R=I
30/31=.96 or .97 amps