# Need to calculate current flowing through parallel resistors.

1. Aug 28, 2011

### rdhc1330

1. The problem statement, all variables and given/known data
"What is current flowing through R4?"

Hope you're able to see the diagram I attached, any help would be awesome, I'm really quite confused.

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2. Aug 28, 2011

### Zryn

Could you post any attempts you have made at answering the question?

Are you familiar with any appropriate laws, how resistances add when combined in series or parallel and voltage or current dividers?

3. Aug 29, 2011

### rdhc1330

Well we have learned quite a bit. I know that in series the current is the same all the way around the circuit, while in parallel it is a portion. Rt = R1 + R2.... etc so I = V/R, 10/ (Rn/Rt)

4. Aug 29, 2011

### Zryn

Sounds good.

This is the case for resistors in series. How are the resistors in this circuit arranged? Are there any resistances in series? Do you know how resistors will add in parallel as well?

I = V/R is Ohms Law rearranged, but what does the 10/ (Rn/Rt) relate to?

Instead of jumping in to the calculations, can you 'see' (describe) the method by which you would obtain the current through R4?

5. Aug 29, 2011

### rdhc1330

The 10 is voltage and Rn/Rt was supposed to be Rtotal but I wrote it wrong.
I know for Rtotal it must be 1/(1/R1 + 1/R2... ) etc.

I use LTSpice usually for doing this, I did physics a long time ago! But I know I would have to Figure out the overall total and then take the piece that R4 is?

6. Aug 29, 2011

### Zryn

The voltage source in the diagram is labelled as v1 = 12V, not 10, and I wanted you to write out the equations that got to Rn/Rt so we could see where you were going wrong, whether it was with your method or your math .

There are several methods you could use to determine the current in R4, but here are two:

a) Combine all the resistors in to one resistance, calculate the series current, and then use a current divider to determine the current in R4

b) Combine each set of parallel resistors into one resistance, calculate the voltage divider and then use Ohms Law to determine the current in R4

Which looks preferable to you? (Ideally you could do both to assist your knowledge and confirm your answers)

7. Aug 29, 2011

### rdhc1330

Oh sorry! I didn't realise I had written 12 not 10V. The most preferable would probably be (b), I have already written the equation above so would just need to add them all together for Rtotal?

8. Aug 29, 2011

### Zryn

Using Rt = 1/(1/R1 + 1/R2 + ... + 1/Rn) will allow you to find the total resistance of resistors in parallel, but there are multiple different sets of parallel resistors in this diagram so be careful with the math. Also, if you were to add them all together, you would be proceding down the path as mentioning in a), rather than b).

Lets go with b) instead. Firstly, can you see the multiple sets of parallel resistance? Which ones are they? What values do you get if you find the total resistance for each different set?

When you have these values, redraw the circuit with each parallel set's total resistance represented by a single resistor, and then find the voltage across each one using either Ohm's Law or the voltage divider equation.

Lastly, use your knowledge of how the voltage across each total resistance in your new diagram relates to each parallel resistor set in the original diagram, how voltages in parallel operate and Ohm's Law, and let us know the answers you get.

9. Aug 31, 2011

### rdhc1330

I would say (R1 + R2) and (R3 + R4 + R5 + R6) are the multiple sets?

10. Aug 31, 2011

### Zryn

Yep, thats a good start.

Another convention to write combinations of resistors is using the '||' symbol for parallel combinations and the '+' symbol for series combinations.

You could therefore say that the total resistance of the circuit is Rt = R1||R2 + R3||R4||R5||R6.

11. Aug 31, 2011

### rdhc1330

awesome, that helped, so I have now figured out that across R1 (R1 + R2) that it is 17.24Ohms and R2 (R3 + R4 + R5 + R6) is 18.32Ohms, then used the voltage divider equation Vin * R1/R1+R2 to find that R1 = 5.82V and R2 = 6.18V. Now I'm confused about what to do next?

12. Aug 31, 2011

### rdhc1330

I'm going to take a stab at this and say =

80/300 * 18.32ohms = 4.89ohms (resistance segment of R4)
then using ohms law --> 6.18/4.89 = 1.26A ? Am I on the right track?

13. Aug 31, 2011

### Zryn

Can you show your math for this calculation? I got something ever so slightly different.

Yep, this looks good.

As my result for R1||R2 was slightly different to yours I have gotten slightly different results, but the method is good.

Not this:

If you have a total circuit resistance of 17.24R + 18.32R = 35.56R, and a voltage source of 12V, then by Ohms Law the circuit current is I = V/R = 12/35.56 = 0.337A, and thus the current through R4 must not be 1.26A. You can always check your work using methods like this, and everything should match up, or something has gone wrong somewhere.

What you have done so far is determine that across the original set of R3||R4||R5||R6 (and its equivalent 18.32R combination) there is a voltage of 6.18V. Using the above method there is also a current of 0.337A flowing into, splitting between the resistors, and then re-combining and flowing out of this set.

Given resistors in parallel, what do you know about the voltage across them and the current through them?

14. Aug 31, 2011

### rdhc1330

Isn't the voltage drop the same across all resistors in parallel? While the current is split

also sorry: Rt(R1) = 1/ (1/30 + 1/40)

15. Aug 31, 2011

### Zryn

Yep!

So given that you know the voltage across the resistors, you know that since they are in parallel that the same voltage will be across each individual resistor and you know the value of each individual resistance, you can now use Ohms Law to calculate the current through each one.

To check if you have the correct values as well, you can then add these currents up, and the result should equal the previously mentioned total circuit series current.

Alternatively, knowing the series current through the resistors and their resistance value you can use the current divider equation to calculate the current through each individual resistance.

To check if you have the correct values, you can multiply these currents by their associate resistances and you should get the previously mentioned parallel voltage.

That math looks good, when I put it into the calculator though I get 17.14R instead of your result of 17.24R. It's a very minor error, but as it propagates down through your equations it may result in the final answer being very wrong.

For this reason it's good to write out your math to show that you know what you are doing, and in the instance that you get the wrong number somewhere your markers will likely give you marks for knowing what you are doing, whereas were you to put down no math explaining your method, and then the final incorrect answer, you may get few marks at all.

16. Aug 31, 2011

### rdhc1330

alright I'll check the math on that and then let you know what I get for the final answer for the current through R4.... Hold tight

17. Aug 31, 2011

### rdhc1330

ok so I got (35.56/80) * 0.337 = 0.15A

18. Aug 31, 2011

### Zryn

Using the same method, can you find the current through R3, R5 & R6 as well?

Also, can you write out your algebraic equation for the current divider, like you did for the voltage divider earlier as Vin * R1/(R1+R2) ?

19. Aug 31, 2011

### rdhc1330

Ir4 = (Rt/R4) * It

Ir3 = (35.48/90) * 0.338 = 0.133A
Ir5 = 0.171A
Ir6 = 0.200A

20. Aug 31, 2011

### Zryn

The series current everywhere in the circuit is 0.337A, as calculated previously. When this current enters the parallel set of R3||R4||R5||R6 it will split according to the inverse resistance values (the path of least resistance will be the largest current) and then it will recombine when it exits the parallel set.

This means that the sum of the currents through these 4 resistors must be equal to the series current of the circuit.

Is = Ir3 + Ir4 + Ir5 + Ir6

0.337 = 0.133 + 0.15 + 0.171 + 0.200

0.337 = 0.654

Something is wrong.

Alternatively, you know that the voltage across each of these resistors is 6.2V. Therefore by Ohms Law:

V = Ir4 * R4

6.2 = 0.15 * 80

6.2 = 12

Something is definitely wrong.

The current divider equation is actually:

Ir = Rt / (Rt + Rn) * I

Where 'Rt' is the total resistance of the resistors which are *not* the resistor in question which you are trying to find the current through, 'Rn' is the resistor in question which you are trying to find the current through and 'I' is the source current.