Hi to all
I'm new at this (retired pharmacist with a interest in electronics), I'm been reading and studying some texts
I've come up with one problem (of many) that I can't figure out how to get all the answers .
The answers to the problem is as follows:
a) 16 amps (Rt=5 ohms)- I figured this one
b) I3=I9=4 amps,
c) I8= 1 amp
d) Vab= 14 volts.

Please see attached network...

Anyone willing to help would be most appreciated( not so easy c-out a teacher)

Thanks.............Buddy

 

Let me help you to put the diagram right. It would help make people to see it much easily.

 

Since you got that Req = 5Ω, then you should be able to answer the following fairly easily:

Q1) How much of the 16A flows to the right when it hits the bottom node from the battery?
Q2) What is the resistance of the path consisting of R{3,4,5,7,8}?
Q3) What is the resistance of the path through R{6,9}?
Q4) With these in mind, what is the current in each of these two paths?
Q5) Given I3, how much of that current goes through R4?
Q6) How does the portion of I3 that doesn't go through R4 split between R5 and R{7,8}?

Given I8 and I9, you should be able to find Va (relative to the bottom node) and Vb (again, relative to the bottom node) and then take the difference since Vab = Va - Vb.

BTW: The answers given are correct.

 

Seems to me all your answers are correct at least at the integer parts. May be you could try something more difficult.

Allen

 

Seems to me all your answers are correct at least at the integer parts. May be you could try something more difficult.

Allen

I think the answers were given and he is having trouble figuring out how to get all but the first one.

 

Ah, I see. Should have figured that out. Thanks.

Allen

 

Let me help you to put the diagram right. It would help make people to see it much easily.View attachment 80220

Thanks , I thought I did.

Since you got that Req = 5Ω, then you should be able to answer the following fairly easily:

Q1) How much of the 16A flows to the right when it hits the bottom node from the battery?
Q2) What is the resistance of the path consisting of R{3,4,5,7,8}?
Q3) What is the resistance of the path through R{6,9}?
Q4) With these in mind, what is the current in each of these two paths?
Q5) Given I3, how much of that current goes through R4?
Q6) How does the portion of I3 that doesn't go through R4 split between R5 and R{7,8}?

Given I8 and I9, you should be able to find Va (relative to the bottom node) and Vb (again, relative to the bottom node) and then take the difference since Vab = Va - Vb.

BTW: The answers given are correct.(I know it's from the text)
WBahn- I really appreciate your help , I get most of the theory but my answers aren't coming up correct, here we go:
Q1- 16 A flows to the right ?( it hasn't hit any elements yet)
Q2 R4||5= 2.67 ohms, R7+R8(in series)= 8 ohms, then R4||5 and || R7,8 = 2.00 ohms, (now in series) with R3, which = 10 ohms(path I3)
Q3 R6,9(in series) = 10 ohms(path I9)
Q4 1st path V=IR , 80= I3(10), I3= 8 amps( ?), 2nd path V=IR, 80 = I9(10), I9= 8 amps( ?) (should I have used the Current divider law)
Q5 KCL applies but #'s aren't working
Q6 I get that BUT my amps are wrong for the individual currents
Thanks , What am I doing wrong?
Regards...............KJ

 

With regards to your answer to Q1, if all of the current flows to the right, then that means that NONE of it flows to the left and goes through the 10Ω resistor on that side. Does that make any sense?

How much current flows in that 10Ω resistor? How much voltage does it have across it?

 

the 10 ohm resistor has 8 amps flowing though it. 80 V , since it's across the potental difference of the battery(80 v)

 

the 10 ohm resistor has 8 amps flowing though it. 80 V , since it's across the potental difference of the battery(80 v)

Okay, so if 16A is coming downward from the battery and 8A of that goes off to the left, how much goes off to the right?

 

8 amps, I'm concerned about the other answers also.

Thanks

 

8 amps, I'm concerned about the other answers also.

Thanks

As you should be.

Actually, your answers to Q2 and Q3 are fine, although the way you got Q2 wasn't optimal, but that's not a biggie. But you should get in the habit of looking for the easy way (and it takes practice and experience to spot it quickly). R7+R8 is 8Ω which is in parallel with R5, which is another 8Ω, so the combination of these is 4Ω and this is then in parallel with R4, another 4Ω, so the total of the three parallel branches is 2Ω. Added to R3=8Ω in series and we have 10Ω.

For Q4 you should have used the current divider law because you have so much current dividing between two equal resistance branches before coming back together, so it will split in two. However, because of the wrong answer to Q1, you would have gotten the same wrong answer you got using Ohm's Law (which you used incorrectly). So let's turn attention to that. You've made one of the classic mistakes when applying Ohm's Law in that you grabbed a voltage (80V) and you grabbed a resistance (10Ω) and threw them at a formula to get 8A. But Ohm's Law related the voltage ACROSS a resistor to the current THROUGH that resistor and the resistance of THAT resistor. The 80V is NOT the voltage that appears across that 10Ω of resistance! You've also got that 5Ω resistor in that path. To find the resistance that corresponds to the 80V you need to combine the two 10Ω paths from Q2 and Q3 to get 5Ω which is then in series with the 5Ω from R2 to get 10Ω. Now you can apply Ohm's Law to get 8A for the entire right side of the circuit (which matches what we already knew from Q1) and it is THIS current that splits evenly between the two branches described in Q2 and Q3.

So see what you can do from this point.

 

I read it over quickly, I'll get back to you with a response by this evening(i'm in NY).
However, thank you for sticking with me, i'm really intent on teaching myself as much as I can(all the math is no problem it's the concept
of electronics that is challenging)
Regards.............Buddy

 

Hi,
Q4 I3=(I)(R9')/(R3'+R9')=(8) (10)/ (10+10)= 4 amps, since I9=I3= 4amps
Q5 R7,8= 8 ohms, then R7,8||R5 which =4 ohms, and R4 and R5= 4 ohms therefore the 4 amps is split equally into 2 amps each(R4 gets 2 amps)
Q6 R7,8 gets 1 amp ( and R5= 1amp)
Vab = Va-Vb, Va=(R8)(I8)= 2 ohms x 1amp= 2 Volts, Vb= I9 x R9= 4 amps x 4 ohms= 16 volts , Va-Vb= 2 volts-16volts= -14 volts(polarity in opposite direction)
THANK YOU!!!!!!!!!!!!!!!!!!!,
However(one last question- not really?) had the R1(10 ohms resistor) been on the other aide of the battery would it have made a difference?
The polarity of the Vab came out negative meaning that the Vb was more negative than Va ?
Once again THANK YOU FOR ALL YOUR HELP..........KJ

 

Hi,
Q4 I3=(I)(R9')/(R3'+R9')=(8) (10)/ (10+10)= 4 amps, since I9=I3= 4amps
Q5 R7,8= 8 ohms, then R7,8||R5 which =4 ohms, and R4 and R5= 4 ohms therefore the 4 amps is split equally into 2 amps each(R4 gets 2 amps)
Q6 R7,8 gets 1 amp ( and R5= 1amp)
Vab = Va-Vb, Va=(R8)(I8)= 2 ohms x 1amp= 2 Volts, Vb= I9 x R9= 4 amps x 4 ohms= 16 volts , Va-Vb= 2 volts-16volts= -14 volts(polarity in opposite direction)
THANK YOU!!!!!!!!!!!!!!!!!!!,
However(one last question- not really?) had the R1(10 ohms resistor) been on the other aide of the battery would it have made a difference?
The polarity of the Vab came out negative meaning that the Vb was more negative than Va ?
Once again THANK YOU FOR ALL YOUR HELP..........KJ

You are very close, but not quite there. You are getting a bit sloppy with the polarity of your voltage drops.

You need to sanity check your work. If the bottom node is taken at ground (the most obvious choice), does it make sense that Va or Vb are positive?

Let's call the bottom node G (we can arbitrarily define it to be 0V, but that doesn't matter).

Because we the current I8 flowing upward in through R8 from Node G to Node A, we know that (Vg-Va) = I8·R8, similarly we have (Vg-Vb) = I9·R9.

We want

Vab = Va - Vb
Va = Vg - I8·R8
Vb = Vg - I9·R9

Vab = (Vg - I8·R8) - (Vg - I9·R9)
Vab = I9·R9 - I8·R8
Vab = (4A)(4Ω) - (1A)(2Ω) = 16V - 2V = 14V

 

OK.....KJ

 

Would it have made a difference if the R1 (10 ohm) was on the other(right)side of the battery ?

 

Would it have made a difference if the R1 (10 ohm) was on the other(right)side of the battery ?

Nope. You would have had all 16A going to the right, but at some point 8A of it would have gone up through R1. That might have happened before the branch with Node A, between Node A and Node B, or after the branch with Node B. But none of that would have mattered as far as how much current went up each of those two branches.

Now, in the real world all of those wires have some resistance and if the resistance is high enough and the current is high enough, then the placement of the branches does have an effect because the voltage drops along the wires depends on how many branches have yet to be supplied.

 

Thanks, you really explained alot of questions I had with this one problem.

.......KJ