Significant figures while finding errors
$begingroup$
The question which arose my confusion:
Two resistors with resistance $R_1 = 100pm 3 Omega$, and $R_2 = 200pm 4 Omega$ are connected in (a) series, (b) in parallel. Find the equivalent resistance of both combinations.
In the end, my answer came out to be (a) $300pm 7 Omega$ and (b) $66.7pm 2 Omega$. But according to my book, my response for (b) is incorrect.
It said that the equivalent resistance should have an error of $pm 1.8$ not my rounded off answer $pm 2$. And it (book) explicitly states that the error in (b) should be "expressed as $pm 1.8$ to keep in conformity with the rules of significant figures." I don't understand how this makes any sense (especially how it conforms to the rules of significant figures). Help anyone?
error-analysis
$endgroup$
add a comment |
$begingroup$
The question which arose my confusion:
Two resistors with resistance $R_1 = 100pm 3 Omega$, and $R_2 = 200pm 4 Omega$ are connected in (a) series, (b) in parallel. Find the equivalent resistance of both combinations.
In the end, my answer came out to be (a) $300pm 7 Omega$ and (b) $66.7pm 2 Omega$. But according to my book, my response for (b) is incorrect.
It said that the equivalent resistance should have an error of $pm 1.8$ not my rounded off answer $pm 2$. And it (book) explicitly states that the error in (b) should be "expressed as $pm 1.8$ to keep in conformity with the rules of significant figures." I don't understand how this makes any sense (especially how it conforms to the rules of significant figures). Help anyone?
error-analysis
$endgroup$
add a comment |
$begingroup$
The question which arose my confusion:
Two resistors with resistance $R_1 = 100pm 3 Omega$, and $R_2 = 200pm 4 Omega$ are connected in (a) series, (b) in parallel. Find the equivalent resistance of both combinations.
In the end, my answer came out to be (a) $300pm 7 Omega$ and (b) $66.7pm 2 Omega$. But according to my book, my response for (b) is incorrect.
It said that the equivalent resistance should have an error of $pm 1.8$ not my rounded off answer $pm 2$. And it (book) explicitly states that the error in (b) should be "expressed as $pm 1.8$ to keep in conformity with the rules of significant figures." I don't understand how this makes any sense (especially how it conforms to the rules of significant figures). Help anyone?
error-analysis
$endgroup$
The question which arose my confusion:
Two resistors with resistance $R_1 = 100pm 3 Omega$, and $R_2 = 200pm 4 Omega$ are connected in (a) series, (b) in parallel. Find the equivalent resistance of both combinations.
In the end, my answer came out to be (a) $300pm 7 Omega$ and (b) $66.7pm 2 Omega$. But according to my book, my response for (b) is incorrect.
It said that the equivalent resistance should have an error of $pm 1.8$ not my rounded off answer $pm 2$. And it (book) explicitly states that the error in (b) should be "expressed as $pm 1.8$ to keep in conformity with the rules of significant figures." I don't understand how this makes any sense (especially how it conforms to the rules of significant figures). Help anyone?
error-analysis
error-analysis
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1597
1597
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2 Answers
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$begingroup$
The easy part is that $66.7 pm 2$ is wrong. It is wrong because unless we know the tenths digit of the error, expressing the main value to the tenth's digit doesn't make sense: we'd be force to drop the result in that digit as soon as we add or subtract. So we should write $67 pm 2$ or write both the main figure and the uncertainty to the tenths column (that is, $66.7 pm 1.8$).
The harder part (and indeed the part with a little wiggle room) is recognizing that both of the inputs are accurate to better than three percent, so they should be treated as having about three digits of precision. However, if you are old enough to recall the slide-rule convention for leading 1s (which requires that $1.00 times 10^2$ is a figure with only two digits of precision), you might feel that that fractional errors of a few percent should imply two digits not three.
Part of the problem is that there is no completely internally consistent way to deal with uncertainty using the crude tool that is significant figures. Working scientists don't follow a checklist on significant figures they just always remember to not write figures that have no meaning. And in that frame of mind I would prefer $66.7 pm 1.8$.
$endgroup$
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$begingroup$
I'm not sure about keeping significant figures, but when you report a number with uncertainty you need to make sure the both values have the same number of decimal places. This is probably what they mean, since if you are going to use $66.7$, then you should report the uncertainty to the same number of decimal places, which goes to $1.8$
From what I have been taught, reporting measurements with uncertainties gets rid of the need for significant figures. Significant figures are more of a "fast and loose" way to report uncertainty in a measurement. For example, if I measure the length of something using a ruler that has ticks every centimeter to be $5.4 rm {cm}$, what I really mean is that I am certain that the length falls between the $5 rm{cm}$ and $6 rm{cm}$ ticks, and I estimate it to be $0.4 rm{cm}$ from the $5 rm{cm}$ tick. The final significant digit tells me which number I am not certain about. This is why if we were to then add multiple length measurements, we need to keep track of significant figures, since this is how we are choosing to keep track of our uncertainty.
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
The easy part is that $66.7 pm 2$ is wrong. It is wrong because unless we know the tenths digit of the error, expressing the main value to the tenth's digit doesn't make sense: we'd be force to drop the result in that digit as soon as we add or subtract. So we should write $67 pm 2$ or write both the main figure and the uncertainty to the tenths column (that is, $66.7 pm 1.8$).
The harder part (and indeed the part with a little wiggle room) is recognizing that both of the inputs are accurate to better than three percent, so they should be treated as having about three digits of precision. However, if you are old enough to recall the slide-rule convention for leading 1s (which requires that $1.00 times 10^2$ is a figure with only two digits of precision), you might feel that that fractional errors of a few percent should imply two digits not three.
Part of the problem is that there is no completely internally consistent way to deal with uncertainty using the crude tool that is significant figures. Working scientists don't follow a checklist on significant figures they just always remember to not write figures that have no meaning. And in that frame of mind I would prefer $66.7 pm 1.8$.
$endgroup$
add a comment |
$begingroup$
The easy part is that $66.7 pm 2$ is wrong. It is wrong because unless we know the tenths digit of the error, expressing the main value to the tenth's digit doesn't make sense: we'd be force to drop the result in that digit as soon as we add or subtract. So we should write $67 pm 2$ or write both the main figure and the uncertainty to the tenths column (that is, $66.7 pm 1.8$).
The harder part (and indeed the part with a little wiggle room) is recognizing that both of the inputs are accurate to better than three percent, so they should be treated as having about three digits of precision. However, if you are old enough to recall the slide-rule convention for leading 1s (which requires that $1.00 times 10^2$ is a figure with only two digits of precision), you might feel that that fractional errors of a few percent should imply two digits not three.
Part of the problem is that there is no completely internally consistent way to deal with uncertainty using the crude tool that is significant figures. Working scientists don't follow a checklist on significant figures they just always remember to not write figures that have no meaning. And in that frame of mind I would prefer $66.7 pm 1.8$.
$endgroup$
add a comment |
$begingroup$
The easy part is that $66.7 pm 2$ is wrong. It is wrong because unless we know the tenths digit of the error, expressing the main value to the tenth's digit doesn't make sense: we'd be force to drop the result in that digit as soon as we add or subtract. So we should write $67 pm 2$ or write both the main figure and the uncertainty to the tenths column (that is, $66.7 pm 1.8$).
The harder part (and indeed the part with a little wiggle room) is recognizing that both of the inputs are accurate to better than three percent, so they should be treated as having about three digits of precision. However, if you are old enough to recall the slide-rule convention for leading 1s (which requires that $1.00 times 10^2$ is a figure with only two digits of precision), you might feel that that fractional errors of a few percent should imply two digits not three.
Part of the problem is that there is no completely internally consistent way to deal with uncertainty using the crude tool that is significant figures. Working scientists don't follow a checklist on significant figures they just always remember to not write figures that have no meaning. And in that frame of mind I would prefer $66.7 pm 1.8$.
$endgroup$
The easy part is that $66.7 pm 2$ is wrong. It is wrong because unless we know the tenths digit of the error, expressing the main value to the tenth's digit doesn't make sense: we'd be force to drop the result in that digit as soon as we add or subtract. So we should write $67 pm 2$ or write both the main figure and the uncertainty to the tenths column (that is, $66.7 pm 1.8$).
The harder part (and indeed the part with a little wiggle room) is recognizing that both of the inputs are accurate to better than three percent, so they should be treated as having about three digits of precision. However, if you are old enough to recall the slide-rule convention for leading 1s (which requires that $1.00 times 10^2$ is a figure with only two digits of precision), you might feel that that fractional errors of a few percent should imply two digits not three.
Part of the problem is that there is no completely internally consistent way to deal with uncertainty using the crude tool that is significant figures. Working scientists don't follow a checklist on significant figures they just always remember to not write figures that have no meaning. And in that frame of mind I would prefer $66.7 pm 1.8$.
answered 1 hour ago
dmckee♦dmckee
74.2k6134267
74.2k6134267
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add a comment |
$begingroup$
I'm not sure about keeping significant figures, but when you report a number with uncertainty you need to make sure the both values have the same number of decimal places. This is probably what they mean, since if you are going to use $66.7$, then you should report the uncertainty to the same number of decimal places, which goes to $1.8$
From what I have been taught, reporting measurements with uncertainties gets rid of the need for significant figures. Significant figures are more of a "fast and loose" way to report uncertainty in a measurement. For example, if I measure the length of something using a ruler that has ticks every centimeter to be $5.4 rm {cm}$, what I really mean is that I am certain that the length falls between the $5 rm{cm}$ and $6 rm{cm}$ ticks, and I estimate it to be $0.4 rm{cm}$ from the $5 rm{cm}$ tick. The final significant digit tells me which number I am not certain about. This is why if we were to then add multiple length measurements, we need to keep track of significant figures, since this is how we are choosing to keep track of our uncertainty.
$endgroup$
add a comment |
$begingroup$
I'm not sure about keeping significant figures, but when you report a number with uncertainty you need to make sure the both values have the same number of decimal places. This is probably what they mean, since if you are going to use $66.7$, then you should report the uncertainty to the same number of decimal places, which goes to $1.8$
From what I have been taught, reporting measurements with uncertainties gets rid of the need for significant figures. Significant figures are more of a "fast and loose" way to report uncertainty in a measurement. For example, if I measure the length of something using a ruler that has ticks every centimeter to be $5.4 rm {cm}$, what I really mean is that I am certain that the length falls between the $5 rm{cm}$ and $6 rm{cm}$ ticks, and I estimate it to be $0.4 rm{cm}$ from the $5 rm{cm}$ tick. The final significant digit tells me which number I am not certain about. This is why if we were to then add multiple length measurements, we need to keep track of significant figures, since this is how we are choosing to keep track of our uncertainty.
$endgroup$
add a comment |
$begingroup$
I'm not sure about keeping significant figures, but when you report a number with uncertainty you need to make sure the both values have the same number of decimal places. This is probably what they mean, since if you are going to use $66.7$, then you should report the uncertainty to the same number of decimal places, which goes to $1.8$
From what I have been taught, reporting measurements with uncertainties gets rid of the need for significant figures. Significant figures are more of a "fast and loose" way to report uncertainty in a measurement. For example, if I measure the length of something using a ruler that has ticks every centimeter to be $5.4 rm {cm}$, what I really mean is that I am certain that the length falls between the $5 rm{cm}$ and $6 rm{cm}$ ticks, and I estimate it to be $0.4 rm{cm}$ from the $5 rm{cm}$ tick. The final significant digit tells me which number I am not certain about. This is why if we were to then add multiple length measurements, we need to keep track of significant figures, since this is how we are choosing to keep track of our uncertainty.
$endgroup$
I'm not sure about keeping significant figures, but when you report a number with uncertainty you need to make sure the both values have the same number of decimal places. This is probably what they mean, since if you are going to use $66.7$, then you should report the uncertainty to the same number of decimal places, which goes to $1.8$
From what I have been taught, reporting measurements with uncertainties gets rid of the need for significant figures. Significant figures are more of a "fast and loose" way to report uncertainty in a measurement. For example, if I measure the length of something using a ruler that has ticks every centimeter to be $5.4 rm {cm}$, what I really mean is that I am certain that the length falls between the $5 rm{cm}$ and $6 rm{cm}$ ticks, and I estimate it to be $0.4 rm{cm}$ from the $5 rm{cm}$ tick. The final significant digit tells me which number I am not certain about. This is why if we were to then add multiple length measurements, we need to keep track of significant figures, since this is how we are choosing to keep track of our uncertainty.
answered 1 hour ago
Aaron StevensAaron Stevens
10.3k31742
10.3k31742
add a comment |
add a comment |
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