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I 1 R 1 = I 2 R 3 . size 12{I rSub { size 8{1} } R rSub { size 8{1} } =I rSub { size 8{2} } R rSub { size 8{3} } } {}

Again, since b and d are at the same potential, the IR size 12{ ital "IR"} {} drop along dc must equal the IR size 12{ ital "IR"} {} drop along bc. Thus,

I 1 R 2 = I 2 R x . size 12{I rSub { size 8{1} } R rSub { size 8{2} } =I rSub { size 8{2} } R rSub { size 8{x} } } {}

Taking the ratio of these last two expressions gives

I 1 R 1 I 1 R 2 = I 2 R 3 I 2 R x . size 12{ { {I rSub { size 8{1} } R rSub { size 8{1} } } over {I rSub { size 8{1} } R rSub { size 8{2} } } } = { {I rSub { size 8{2} } R rSub { size 8{3} } } over {I rSub { size 8{2} } R rSub { size 8{x} } } } } {}

Canceling the currents and solving for R x yields

R x = R 3 R 2 R 1 . size 12{R rSub { size 8{x} } =R rSub { size 8{3} } { {R rSub { size 8{2} } } over {R rSub { size 8{1} } } } } {}
This complex circuit diagram shows a galvanometer connected in the center arm of a Wheatstone bridge arrangement. All the other four arms have a resistor. The bridge is connected to a cell of e m f script E and internal resistance r.
The Wheatstone bridge is used to calculate unknown resistances. The variable resistance R 3 size 12{R rSub { size 8{3} } } {} is adjusted until the galvanometer reads zero with the switch closed. This simplifies the circuit, allowing R x size 12{R rSub { size 8{x} } } {} to be calculated based on the IR size 12{ ital "IR"} {} drops as discussed in the text.

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 size 12{R rSub { size 8{1} } } {} , R 2 size 12{R rSub { size 8{2} } } {} , and R 3 size 12{R rSub { size 8{3} } } {} , which contribute to the uncertainty in R x size 12{R rSub { size 8{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.

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Section summary

  • Null measurement techniques achieve greater accuracy by balancing a circuit so that no current flows through the measuring device.
  • One such device, for determining voltage, is a potentiometer.
  • Another null measurement device, for determining resistance, is the Wheatstone bridge.
  • Other physical quantities can also be measured with null measurement techniques.

Conceptual questions

Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of null measurements?

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If a potentiometer is used to measure cell emfs on the order of a few volts, why is it most accurate for the standard emf s size 12{"emf" rSub { size 8{s} } } {} to be the same order of magnitude and the resistances to be in the range of a few ohms?

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Problem exercises

What is the emf x size 12{"emf" rSub { size 8{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 x = 5 . 000 Ω size 12{R rSub { size 8{x} } =5 "." "000" %OMEGA } {} and R s = 2 . 500 Ω size 12{R rSub { size 8{s} } =2 "." "500" %OMEGA } {} ?

24.0 V

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Calculate the emf x size 12{"emf" rSub { size 8{x} } } {} of a dry cell for which a potentiometer is balanced when R x = 1 . 200 Ω size 12{R rSub { size 8{x} } =1 "." "200" %OMEGA } {} , while an alkaline standard cell with an emf of 1.600 V requires R s = 1 . 247 Ω size 12{R rSub { size 8{s} } =1 "." "247" %OMEGA } {} to balance the potentiometer.

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When an unknown resistance R x size 12{R rSub { size 8{x} } } {} is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting R 3 size 12{R rSub { size 8{3} } } {} to be 2500 Ω size 12{"2500" %OMEGA } {} . What is R x size 12{R rSub { size 8{x} } } {} if R 2 R 1 = 0 . 625 size 12{ { {R rSub { size 8{2} } } over {R rSub { size 8{1} } } } =0 "." "625"} {} ?

1 . 56 k Ω size 12{1 "." "56 k" %OMEGA } {}

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To what value must you adjust R 3 size 12{R rSub { size 8{3} } } {} to balance a Wheatstone bridge, if the unknown resistance R x size 12{R rSub { size 8{x} } } {} is 100 Ω size 12{"100" %OMEGA } {} , R 1 size 12{R rSub { size 8{1} } } {} is 50 . 0 Ω size 12{"50" "." 0 %OMEGA } {} , and R 2 size 12{R rSub { size 8{2} } } {} is 175 Ω size 12{"175" %OMEGA } {} ?

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(a) What is the unknown emf x size 12{"emf" rSub { size 8{x} } } {} in a potentiometer that balances when R x size 12{R rSub { size 8{x} } } {} is 10 . 0 Ω size 12{"10" "." 0 %OMEGA } {} , and balances when R s size 12{R rSub { size 8{s} } } {} is 15 . 0 Ω size 12{"15" "." 0 %OMEGA } {} for a standard 3.000-V emf? (b) The same emf x size 12{"emf" rSub { size 8{x} } } {} is placed in the same potentiometer, which now balances when R s size 12{R rSub { size 8{s} } } {} is 15 . 0 Ω size 12{"15" "." 0 %OMEGA } {} for a standard emf of 3.100 V. At what resistance R x size 12{R rSub { size 8{x} } } {} will the potentiometer balance?

(a) 2.00 V

(b) 9 . 68 Ω size 12{9 "." "68 " %OMEGA } {}

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Suppose you want to measure resistances in the range from 10 . 0 Ω size 12{"10" "." 0 %OMEGA } {} to 10 . 0 kΩ size 12{"10" "." 0" k" %OMEGA } {} using a Wheatstone bridge that has R 2 R 1 = 2 . 000 size 12{ { {R rSub { size 8{2} } } over {R rSub { size 8{1} } } } =2 "." "000"} {} . Over what range should R 3 size 12{R rSub { size 8{3} } } {} be adjustable?

Range = 5 . 00 Ω to 5 . 00 k Ω size 12{"Range=5" "." "00 " %OMEGA " to "5 "." "00"" k" %OMEGA } {}
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Questions & Answers

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Temiloluwa
the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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