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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.

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?

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?

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

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.

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 } {}

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 } {} ?

(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 } {}

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 } {}

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
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Sherica
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Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, College physics -- hlca 1104. OpenStax CNX. May 18, 2013 Download for free at http://legacy.cnx.org/content/col11525/1.1
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