# Homework Help: Terminals within a parallel circuit question

1. Feb 8, 2013

### Kevin2341

1. The problem statement, all variables and given/known data
Here's the circuit in question
http://s52.beta.photobucket.com/user/I_eat_corn/media/circuit1.png.html

I need to calculate the voltage at terminals A, B, C, and D. (A and B are finished and checked)
2. Relevant equations

Ohms Law, E=IR, I=E\R, R=E\I, P=VI or RI^2

3. The attempt at a solution

Knowing that Vs = 10V, and Req = 21.5909Ω, I = 463.2mA

I calculated A to be 10V (0.463mA x 21.5909Ω = 10V)
B is 0.463mA x 11.5909Ω = 5.368V

C is where things are getting tricky. I am stumbling with what exactly I'm supposed to do here as I've never encountered a circuit like this before (I'm obviously an amateur).

I THINK that my calculations are going to be relative to how parallel branches R2 and (R3+R4) meet up and then pass through R5 to ground, considering how my reference point is ground.

I have punched the circuit into my computer and ran a simulation, so I know what the answers are, I just cannot figure out how to get them. I double checked my points A and B and proved my answers to be correct, but I'm struggling with coming up with an answer for C. I think after C, D will be easy considering it will be an extension so to speak of C.

Voltage @ C: 4.105V, Amperage: 84.21mA
Voltage @ D: 3.474V, Amperage: 84.21mA
Voltage after R2: 3.474, Amperage is 378.9mA

As for my own personal attempts:
- I tried calculating voltage by removing R1 and R3 from the Req, so I have (1\5 + 1\7.5)^-1+7.5. That gave me 10.5ohms as my Req-R1-R3. I multiplied that by Voltage at B (5.368)
-I tried just 5.368\((1\r4)^-1 + 7.5) that gave me an answer... closer... but not correct nevertheless...

I'm pretty clueless now, I've made other attempts but cannot remember them off the top of my head.

2. Feb 8, 2013

### The Electrician

Do you understand why the current through R5 is exactly the same as the current through R1?

Since you therefore know the current through R5, you can calculate the voltage across R5 and that is the voltage at D.

Now since you know the voltage at B and at D, you know the voltage across the series combination of R3 and R4. If you know the voltage across that series combination, you can calculate the current through the series combination of R3 and R4. The current through R3 is the same as the current through R4. Knowing the current through R4, you can calculate the voltage across R4.

Knowing the voltage across R4 and the voltage at D, you can calculate the voltage at C.

3. Feb 9, 2013

### Kevin2341

From my prior problems and what I learned from my trials and errors, and I'm going to attempt to throw out some lingo which I'll probably say incorrectly but I hope you know what I mean, voltage and resistance are proportional to amperage. If the voltage source is 10v, and the equivalent resistance is 10 (lets say we have 10 1ohm resistors in series), as the first resistor, I = 10v\10ohms = 1 amp, at resistor two, 1 amp times 9 ohms is 9 volts, third resistor: 1 amp times 8 ohms is 8 volts. So on, so forth, so basically as you get a smaller voltage in proportion to a smaller resistance, the amperage stays the same.

That's why the current is the same throughout the entire circuit (in a really simplified way).

Where I'm having trouble is coming up with the correct numbers for the amperage split between the two parallels. I think once I figure that out I will make a breakthrough on it and finish the problem. I was getting something along the lines of like 82% of the current go through the 5 ohm resistor on the left hand parallel branch, and 18% of current through the right hand. But I cannot figure out why it is so much, all of my number crunching isn't giving me the answer.

4. Feb 9, 2013

### The Electrician

Let's go a step at a time. I asked you if you understand why the current in R5 is the same as the current in R1, and you didn't answer me. The current in R1 is the same as the current in R5 because R1 and R5 are in series with each other and with the 10 volt source.

You know that the current in R1 is .463 amps, and therefore the current in R5 is also .463 amps. You know what the resistance of R5 is, so now calculate the voltage across R5 and then tell me what the voltage at D is.

5. Feb 9, 2013

### Kevin2341

Well at point D, the only resistor left in the circuit will be R5, so I can go to say that 7.5ohms times .463amps = 3.474 volts.

My circuit maker program is showing it at 3.474V. So we now know what D is. That's good. But where do I go with C now? Every attempt I make at finding voltage and amperage there is coming out incorrect.

6. Feb 9, 2013

### The Electrician

You know the voltage at B and at D. Subtract D from B and you will have the voltage across the R3 and R4 series combination. Since the R3 and R4 series combination is equivalent to a single 22.5Ω resistor, we can calculate the current in that equivalent resistance since we know the voltage across it (B-D volts).

What value do you get for that current?

That same current is the current in both R3 and R4. Now, knowing the current in R4 we can calculate the voltage across R4. What value do you get for that?

Now add the voltage across R4 to the voltage at D and you will have the voltage at C. What do you get for that?

7. Feb 10, 2013

### Kevin2341

Step by step with what you gave me:
Vb: 5.368
Vd: 3.474

Subtract Vd - Vb = 3.474-.5368 = -1.894 (not sure why you said Vd-Vb, shouldn't it be the other way around so that we get positive numbers?)

R3+R4 = 22.5 ohms.

I = V\R, I = 1.894\22.5 ≈ 0.084178 ≈ 0.08421 [CircuitMaker reading] --> (84.21mA)

V = 0.084178x15 = 1.26267

5.368-1.26267 = 4.105V (which matches with CircuitMaker), That's my C reading.

THEN 0.084178x22.5 = 1.894V (Which is my voltage drop from B to D!)

Thank you SO much! I never even imagine trying something like this, but it makes sense now!

8. Feb 12, 2013

### The Electrician

Yep, you got it.

Look again at what I said, which was " Subtract D from B and you will have the voltage across the R3 and R4 series combination."

When I said "Subtract D from B", that means Vb-Vd. Be careful. One of the most common mistakes EE students make when analyzing networks is to get a sign wrong. I can't tell you how often I did that in the beginning of my studies.