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Question Number 33272 by NECx last updated on 14/Apr/18

The resistance R of an   unknown resistor is found by  measuring the potential  difference V across the  resistor and the current I through  it and using the equation R=(V/I).  The voltmeter reading has a 3%  uncertainty and the ammeter  reading has a 2% uncertainty.  what is the uncertainty in the  calculated resistance?

$${The}\:\boldsymbol{{resistance}}\:\boldsymbol{{R}}\:\boldsymbol{{of}}\:\boldsymbol{{an}}\: \\ $$$$\boldsymbol{{unknown}}\:\boldsymbol{{resistor}}\:\boldsymbol{{is}}\:\boldsymbol{{found}}\:\boldsymbol{{by}} \\ $$$$\boldsymbol{{measuring}}\:\boldsymbol{{the}}\:\boldsymbol{{potential}} \\ $$$$\boldsymbol{{difference}}\:\boldsymbol{{V}}\:\boldsymbol{{across}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{resistor}}\:\boldsymbol{{and}}\:\boldsymbol{{the}}\:\boldsymbol{{current}}\:\boldsymbol{{I}}\:\boldsymbol{{through}} \\ $$$$\boldsymbol{{it}}\:\boldsymbol{{and}}\:\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}}\:\boldsymbol{{R}}=\frac{{V}}{{I}}. \\ $$$$\boldsymbol{{The}}\:\boldsymbol{{voltmeter}}\:\boldsymbol{{reading}}\:\boldsymbol{{has}}\:\boldsymbol{{a}}\:\mathrm{3\%} \\ $$$$\boldsymbol{{uncertainty}}\:\boldsymbol{{and}}\:\boldsymbol{{the}}\:\boldsymbol{{ammeter}} \\ $$$$\boldsymbol{{reading}}\:\boldsymbol{{has}}\:\boldsymbol{{a}}\:\mathrm{2\%}\:\boldsymbol{{uncertainty}}. \\ $$$$\boldsymbol{{what}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{uncertainty}}\:\boldsymbol{{in}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{calculated}}\:\boldsymbol{{resistance}}? \\ $$

Commented by NECx last updated on 14/Apr/18

please help

$${please}\:{help} \\ $$

Answered by MJS last updated on 14/Apr/18

R=((1.03V)/(.98I))≈1.05(V/I)  R=((.97V)/(1.02I))≈.95(V/I)  so the uncertainty in R is 5%    with both division and multiplication  the resulting uncertainty is the  sum of the single uncertainties

$${R}=\frac{\mathrm{1}.\mathrm{03}{V}}{.\mathrm{98}{I}}\approx\mathrm{1}.\mathrm{05}\frac{{V}}{{I}} \\ $$$${R}=\frac{.\mathrm{97}{V}}{\mathrm{1}.\mathrm{02}{I}}\approx.\mathrm{95}\frac{{V}}{{I}} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{uncertainty}\:\mathrm{in}\:{R}\:\mathrm{is}\:\mathrm{5\%} \\ $$$$ \\ $$$$\mathrm{with}\:\mathrm{both}\:\mathrm{division}\:\mathrm{and}\:\mathrm{multiplication} \\ $$$$\mathrm{the}\:\mathrm{resulting}\:\mathrm{uncertainty}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{single}\:\mathrm{uncertainties} \\ $$

Commented by NECx last updated on 14/Apr/18

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

Commented by NECx last updated on 14/Apr/18

This what I did tho I dont know  what went wrong    R=(V/I) → (dR/R)=(dV/V) − (dI/I)    dR=R((dV/V)−(dI/I))  dV=3%V  dI=2%I  dR=R(3%−2%)    dR=1%R    please whats wrong

$${This}\:{what}\:{I}\:{did}\:{tho}\:{I}\:{dont}\:{know} \\ $$$${what}\:{went}\:{wrong} \\ $$$$ \\ $$$${R}=\frac{{V}}{{I}}\:\rightarrow\:\frac{{dR}}{{R}}=\frac{{dV}}{{V}}\:−\:\frac{{dI}}{{I}} \\ $$$$ \\ $$$${dR}={R}\left(\frac{{dV}}{{V}}−\frac{{dI}}{{I}}\right) \\ $$$${dV}=\mathrm{3\%}{V}\:\:{dI}=\mathrm{2\%}{I} \\ $$$${dR}={R}\left(\mathrm{3\%}−\mathrm{2\%}\right) \\ $$$$ \\ $$$${dR}=\mathrm{1\%}{R} \\ $$$$ \\ $$$${please}\:{whats}\:{wrong} \\ $$

Commented by MJS last updated on 14/Apr/18

it′s not +3% and +2%,   it′s ±3% and ±2% ⇒  ⇒ dV=3% or dV=−3%         dI=2% or dI=−2%  with my method:  it could be ((1.03V)/(1.02I)) or ((.97V)/(.98I)) but  the maximum is higher, as I  showed above

$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:+\mathrm{3\%}\:\mathrm{and}\:+\mathrm{2\%},\: \\ $$$$\mathrm{it}'\mathrm{s}\:\pm\mathrm{3\%}\:\mathrm{and}\:\pm\mathrm{2\%}\:\Rightarrow \\ $$$$\Rightarrow\:{dV}=\mathrm{3\%}\:\mathrm{or}\:{dV}=−\mathrm{3\%} \\ $$$$\:\:\:\:\:\:\:{dI}=\mathrm{2\%}\:\mathrm{or}\:{dI}=−\mathrm{2\%} \\ $$$$\mathrm{with}\:\mathrm{my}\:\mathrm{method}: \\ $$$$\mathrm{it}\:\mathrm{could}\:\mathrm{be}\:\frac{\mathrm{1}.\mathrm{03}{V}}{\mathrm{1}.\mathrm{02}{I}}\:\mathrm{or}\:\frac{.\mathrm{97}{V}}{.\mathrm{98}{I}}\:\mathrm{but} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{higher},\:\mathrm{as}\:\mathrm{I} \\ $$$$\mathrm{showed}\:\mathrm{above} \\ $$$$ \\ $$

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