I've redrawn it for you.

Note that the signs across the resistor are opposite to the supply so the voltage will b negative. Don't allow this to confuse you, just put a - sign in front of your answer.

Originally Posted by rayslab

What happened to the 1 Ohm parallel the lower right hand 2 Ohm?
there is a series/parallel combo of 2 + (1//2) = 2 + (2/3) = 2.6666 Ohm, that should parallel your 3 Ohm and the source?

It has a wire across it so it effectively disappears.

Originally Posted by Micheal

Hi Hero999,

The sign is reverse as what you mentioned.

No it's not, the sign on my circuit is the same as the original: the + goes to the - side of the power supply.

But I see the direction of the current across the V3 should be on top + sign.

Please advice why?

Thanks again

What happens if you connect a DVM up with the - input to the positive supply and the + input to the negative supply? It displays a negative voltage.

The signs on the circuit have been put this way to make you think.

Originally Posted by rayslab

2) Add series of left hand 2 Ohm and combo 2/3 Ohm (1 parallel 2) = 2.666 Ohm

You're not listening, the left hand 2Ω resistor has a piece of wire across it so its resistance is 0Ω.

The voltage across the 2Ω resistor will be 0V.

Look at my previous attachment.

so V(1) = (-) 0.706Volts

That's wrong.

The resistances are all in parallel, I'm not going to say this again.

The resistance of the 3 resistors in parallel is 0.5454 ohms. Since the values are given to one decimal place, (actually no decimal places) the answer should also be limited to the same accuracy, namely 0.5 ohms
The voltage V will be: I(amps) x R (ohms) = 2 x 0.5 = 1v

It would be more accurate just to solve using fractions:

1) A one Ohm resistor and a two Ohm in parallel are:

R1 x R2 / (R1 + R2)

which is:

(1 x 2) / (1 + 2) = 2/3

2) A 2/3 Ohm resistor in parallel with a 3 Ohm resistor is: (3 x 2/3) / (3 + 2/3) = (2) / (11/3) = 6/11 Ohms (I wouldn't round that off to 0.5)

Originally Posted by colin55

Firstly the values of resistance are given in full units. The final result is 0.54 ohms.
You cannot specify the result to an accuray of hundredths when the inital values are not stated to that accuracy.
A value of 0.54 is rounded to 0.5 That's how the final "technical" result was obtained.

Ugh, that reminds me of chemistry class, the sig fig police. lol, I always lost half a point on an exam for getting the sig figs wrong.

Here's my opinion:
I think you should as a general rule give the answer to one more significant figure than the numbers given to you.

If in doubt display the answer to three significant figures then round it.

Here's how I'd solve the problem:

Rounded to 2 significant figures: 1.1V

Originally Posted by colin55

Firstly the values of resistance are given in full units. The final result is 0.54 ohms.
You cannot specify the result to an accuray of hundredths when the inital values are not stated to that accuracy.
A value of 0.54 is rounded to 0.5 That's how the final "technical" result was obtained.

I did NOT specify the accuracy to hundredths, I specified it as a fraction: 5/11 is a fraction which is the EXACT answer to the question and you can carry the fraction out to any decimal you want or leave it as is. There is no reason an answer to a problem must be given in decimal form. In this case, it would imply to an old instructor like myself that I was reading a paper from a "calculator cripple" which is somebody who didn't know basic math skills because they always rely on a calculator. Giving the answer as 5/11 shows the person actually understands the techniques required to solve the problem and does not assume accuracy beyond the stated problem parameters.

Originally Posted by Hero999

Here's my opinion:
I think you should as a general rule give the answer to one more significant figure than the numbers given to you.

If in doubt display the answer to three significant figures then round it.

Here's how I'd solve the problem:

Rounded to 2 significant figures: 1.1V

Just out of curiousity: where does it say the answer must be decimalized or rounded off? The exact answer for the problem is 12/11V. Since the initial values are whole integers, you can carry them through the entire calculation as fractions and give the exact answer. When you round things off, you make assumptions and throw away information. It is ALWAYS best to display the exact answer (if there is one) and let the next user decide where to round it off. Once you round it off, all info past the last digit is lost.

Originally Posted by bountyhunter

Since the initial values are whole integers, you can carry them through the entire calculation as fractions and give the exact answer. When you round things off, you make assumptions and throw away information. It is ALWAYS best to display the exact answer (if there is one) and let the next user decide where to round it off. Once you round it off, all info past the last digit is lost.

I agree but some schools and colleges don't see it that way, from their perspective you can't give figures beyond the number of decimal places given in the question. I know that's complete rubbish but I don't make the rules.

Originally Posted by colin54

A value of 0.54 is rounded to 0.5

You mead 0.55

Originally Posted by Hero999

I agree but some schools and colleges don't see it that way, from their perspective you can't give figures beyond the number of decimal places given in the question. I know that's complete rubbish but I don't make the rules.

But, I didn't do that. Like I said: no law says it has to be converted to decimal. I did it in fractions keeping all values in whole integer expressions or ratios.