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Again, since b and d are at the same potential, the $\text{IR}$ drop along dc must equal the $\text{IR}$ drop along bc. Thus,
Taking the ratio of these last two expressions gives
Canceling the currents and solving for R _{x} yields
This equation is used to calculate the unknown resistance when current through the galvanometer is zero. This method can be very accurate (often to four significant digits), but it is limited by two factors. First, it is not possible to get the current through the galvanometer to be exactly zero. Second, there are always uncertainties in ${R}_{1}$ , ${R}_{2}$ , and ${R}_{3}$ , which contribute to the uncertainty in ${R}_{\mathrm{x}}$ .
Identify other factors that might limit the accuracy of null measurements. Would the use of a digital device that is more sensitive than a galvanometer improve the accuracy of null measurements?
One factor would be resistance in the wires and connections in a null measurement. These are impossible to make zero, and they can change over time. Another factor would be temperature variations in resistance, which can be reduced but not completely eliminated by choice of material. Digital devices sensitive to smaller currents than analog devices do improve the accuracy of null measurements because they allow you to get the current closer to zero.
Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of null measurements?
If a potentiometer is used to measure cell emfs on the order of a few volts, why is it most accurate for the standard ${\text{emf}}_{\text{s}}$ to be the same order of magnitude and the resistances to be in the range of a few ohms?
What is the ${\text{emf}}_{\text{x}}$ of a cell being measured in a potentiometer, if the standard cell’s emf is 12.0 V and the potentiometer balances for ${R}_{\text{x}}=5\text{.}\text{000}\phantom{\rule{0.15em}{0ex}}\Omega $ and ${R}_{\text{s}}=2\text{.}\text{500}\phantom{\rule{0.15em}{0ex}}\Omega $ ?
24.0 V
Calculate the ${\text{emf}}_{\text{x}}$ of a dry cell for which a potentiometer is balanced when ${R}_{\text{x}}=1\text{.}\text{200}\phantom{\rule{0.25em}{0ex}}\Omega $ , while an alkaline standard cell with an emf of 1.600 V requires ${R}_{\text{s}}=1\text{.}\text{247}\phantom{\rule{0.25em}{0ex}}\Omega $ to balance the potentiometer.
When an unknown resistance ${R}_{\text{x}}$ is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting ${R}_{3}$ to be $\text{2500}\phantom{\rule{0.25em}{0ex}}\Omega $ . What is ${R}_{\text{x}}$ if $\frac{{R}_{2}}{{R}_{1}}=0\text{.}\text{625}$ ?
$1\text{.}\text{56 k}\Omega $
To what value must you adjust ${R}_{3}$ to balance a Wheatstone bridge, if the unknown resistance ${R}_{\text{x}}$ is $\text{100}\phantom{\rule{0.15em}{0ex}}\Omega $ , ${R}_{1}$ is $\text{50}\text{.}0\phantom{\rule{0.15em}{0ex}}\Omega $ , and ${R}_{2}$ is $\text{175}\phantom{\rule{0.15em}{0ex}}\Omega $ ?
(a) What is the unknown ${\text{emf}}_{\text{x}}$ in a potentiometer that balances when ${R}_{\text{x}}$ is $\text{10}\text{.}0\phantom{\rule{0.15em}{0ex}}\Omega $ , and balances when ${R}_{\text{s}}$ is $\text{15}\text{.}0\phantom{\rule{0.15em}{0ex}}\Omega $ for a standard 3.000-V emf? (b) The same ${\text{emf}}_{\text{x}}$ is placed in the same potentiometer, which now balances when ${R}_{\text{s}}$ is $\text{15}\text{.}0\phantom{\rule{0.15em}{0ex}}\Omega $ for a standard emf of 3.100 V. At what resistance ${R}_{\text{x}}$ will the potentiometer balance?
(a) 2.00 V
(b) $9\text{.}\text{68}\phantom{\rule{0.25em}{0ex}}\Omega $
Suppose you want to measure resistances in the range from $\text{10}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega $ to $\text{10}\text{.}\mathrm{0\; k\Omega}$ using a Wheatstone bridge that has $\frac{{R}_{2}}{{R}_{1}}=2\text{.}\text{000}$ . Over what range should ${R}_{3}$ be adjustable?
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