Voltage Drop

[bountyhunter]

I showed up to reply to a post in this thread that was sent to my email (which contained the wrong answer) but obviously it has been deleted from the thread because it's not here.

 

[colin55]

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

 

[bountyhunter]

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)

 

[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.

 

[Mikebits]

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.

 

[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:
[latex]
R = \frac{1}{ \frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}} = \frac{1}{ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}}= 0.545\Omega \\
V = IR = 2 \times 0.545 \times 2= 1.09V[/latex]
Rounded to 2 significant figures: 1.1V

 

[bountyhunter]

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.

 

[bountyhunter]

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:
[latex]
R = \frac{1}{ \frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}} = \frac{1}{ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}}= 0.545\Omega \\
V = IR = 2 \times 0.545 \times 2= 1.09V[/latex]
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.

 

[Hero999]

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.

A value of 0.54 is rounded to 0.5

You mead 0.55

 

[bountyhunter]

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.

 

[Hero999]

It depends on how he's expected to work it out, they might want him to use fractions so he'll loose a mark if he does it using a calculator.