Calculating voltages help

[Sam Clay]

Hi,

So I just started a course at uni and there is a fair amount of electronics which I have no background in really. I did a lab the other day where we built the circuit shown in the diagram here. https://drive.google.com/file/d/0B8Ni7zBwY_YMaFdKSVhFZU5SR2s/view?usp=sharing

I can't work out why the voltage across where the volt meter is attached is 3.33, and how to prove it mathematically.

Any help would be greatly appreciated.
Thanks

 

[(*steve*)]

Have you covered KVL and KCL? If so, you can use them to determine the voltages at the nodes where the voltmeter is connected (compared to some reference node). What is the difference between them?

If you've not covered that, but understand ohms law and formulas for resistors in series and parallel, you can achieve the same result.

 

[Sam Clay]

Have you covered KVL and KCL? If so, you can use them to determine the voltages at the nodes where the voltmeter is connected (compared to some reference node). What is the difference between them?

If you've not covered that, but understand ohms law and formulas for resistors in series and parallel, you can achieve the same result.

Hi, I haven't covered that yet I'm afraid, but understand Ohm's law and calculating resistance in series and parallel.

 

[Laplace]

Sometimes it helps to redraw the circuit in a more familiar orientation.....


If you have learned about voltage dividers, now may be the time to divide some voltage!

 

[chopnhack]

@Laplace - That was clever!! - shall we call it commutative property of voltage in parallel

Sam, I am still struggling with the math, much to everyone's chagrin around here, but this site might help you understand the calculations behind this circuit.

 

[jbelectric777]

Sam, Series/parallel resistance isn't tough at all, maybe this will help: Series just add the sum which in your diagram series are 90Ω then 60Ω now we did series now we parallel those two:
1/90 =.011 and 1/60 = .016 the sum is .011111 + .0166666 = .0277778 then then sum of all divided by 1 to get closest resistance value or 1/35.99997Ω (Ohms law: E/R=V or 10/3Ω = 3.33) It looks like on the last stage they rounded 5 or less to nearest tenth and used 3Ω
or maybe so the student could understand the digit 3 rounded is better than your calculators finishing with all these decimals and just showed the close as possible end product. Ask the moderators if that seems to be the case. Jim B

 

[(*steve*)]

The answer to this question doesn't actually require any resistance in parallel calculation.

 

[(*steve*)]

Series just add the sum which in your diagram series are 90Ω then 60Ω now we did series now we parallel those two:
1/90 =.011 and 1/60 = .016 the sum is .011111 + .0166666 = .0277778 then then sum of all divided by 1 to get closest resistance value or 1/35.99997Ω (Ohms law: E/R=V or 10/3Ω = 3.33) It looks like on the last stage they rounded 5 or less to nearest tenth and used 3Ω

I don't follow this at all. I can see how you get a total equivalent resistance of 36 ohms, but I'm not sure that this helps you at all in answering the question.

Then for the last step... where did 3 ohms come from?

The solution involves finding the voltage at the points labelled + and - by Laplace, and then finding the difference between them (because that's what is being measured).

However, I think at this point @Sam Clay has probably just found another forum that has given him the answer and thus won't be returning here to try to solve it himself.