# Homework Help: Potential Difference across a resistance

1. Nov 1, 2014

### Differentiate1

1. The problem statement, all variables and given/known data

a) Find Req of the circuit
b) Find the potential difference across resistance R3

2. Relevant equations
Resistors in parallel = (1/R1 + 1/R2 + 1/R3 + ...)^-1
Resistors in series = R1 + R2 + R3 + ...

3. The attempt at a solution

a)
R3 and R4 are in parallel --> [ (1/40Ω) + (1/50Ω) ]^-1 = 200/9 Ω
R2 and R_34 are then in series --> 50Ω + (200/9)Ω = 650/9 Ω
R1 and R_234 are then in parallel --> [ (1/40) + (650/9) ]^-1 = 25.7Ω

b)
This is the part I need help with as I'm clueless on even how to start. Any guide is appreciated.

2. Nov 1, 2014

### Orodruin

Staff Emeritus
How is the potential split between two resistances in series?

3. Nov 1, 2014

### Differentiate1

In series, the potential is equal to the sum of the individual voltages I recall. So Vtot = V1 + V2 + ...

4. Nov 1, 2014

### Orodruin

Staff Emeritus
Yes, and how would you find the individual voltages?

5. Nov 1, 2014

### Differentiate1

Using Ohm's Law, V = IR --> I would say find the current first. So I = V/R.
Vtotal = 10 V here --> I = 10V/Rtotal = 10/26
V1 = I(R1)
V2 = I(R2)
V4 = I(R4)
Then, V3 = I(R3) = 15.38 V

The answer however is 5.5 V.

6. Nov 1, 2014

### Orodruin

Staff Emeritus
The current you have computed is the total current through the equivalent circuit, not the current through R3.

7. Nov 1, 2014

### Differentiate1

I am unsure of how to find the current through R3, the individual resistor.

Last edited: Nov 1, 2014
8. Nov 1, 2014

### Orodruin

Staff Emeritus
You do not need to. There are several ways of approaching this and the one I had in mind does not rely upon it. Let us try another one: Can you find a relation between the potential across R2 and that across R3?

9. Nov 1, 2014

### Differentiate1

Well since R2 and R3 are in series, V23 = V2 + V3

10. Nov 1, 2014

### Orodruin

Staff Emeritus
And what does it have to equal to?

11. Nov 1, 2014

### Differentiate1

Since this is a closed loop, V23 would have to equal 0.

12. Nov 1, 2014

### Orodruin

Staff Emeritus
You can use Kirchoff's laws for sure. However, you are not closing a loop by going across R2 and R3. In order to close it, what other component do you need to cross? (There are two choices, one is simpler than the other...)

13. Nov 1, 2014

### Differentiate1

Would need to cross R1 to close it?

14. Nov 1, 2014

### Orodruin

Staff Emeritus
Yes, or ... (You picked the slightly more complicated choice)

Alternatively, what is the potential difference across R1?

15. Nov 1, 2014

### Differentiate1

I = 10/26
So V1 = (10/26)(40) = 15.38 V

16. Nov 1, 2014

### Orodruin

Staff Emeritus
No. All the current does not go through R1 either. You do not need to know the current to know the voltage across R1, it is enough to use Kirchoff's voltage law.

17. Nov 1, 2014

### Differentiate1

So would I set V1 + V2 + V3 - 10 = 0 ?

18. Nov 1, 2014

### Orodruin

Staff Emeritus
No, you are now confusing two of the loops for KVL. Just use one loop at a time.

19. Nov 1, 2014

### Differentiate1

For R1: V1 would have to be 10V. Need to clarify that before moving on.

20. Nov 1, 2014

### Orodruin

Staff Emeritus
Correct. So what does this tell you about V2 and V3?

21. Nov 1, 2014

### Differentiate1

V2 + V3 would have to equal 0?

22. Nov 1, 2014

### Orodruin

Staff Emeritus
No. What loop are you getting this from?

23. Nov 1, 2014

### Differentiate1

Just redid it.
V1 - I2(R2) - I3(R3) = 0
--> I2(R2) - I3(R3) = 10 V
This next step correct?

24. Nov 1, 2014

### Orodruin

Staff Emeritus
You have made a mistake in solving for V2 + V3. I also strongly suggest you work with the voltages instead of the currents, since you are looking for a voltage.

Once you have solved it correctly, ask yourself what the current through R2 is and how you can use this to find V2 and thus V3.

I am now going to bed so this is my last post for today, it is 4 am here.

25. Nov 1, 2014

### Differentiate1

Alright thanks again. I'll see what I can work with.