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emf x emf s = IR x IR s = R x R s . size 12{ { {"emf" rSub { size 8{x} } } over {"emf" rSub { size 8{s} } } } = { { ital "IR" rSub { size 8{x} } } over { ital "IR" rSub { size 8{s} } } } = { {R rSub { size 8{x} } } over {R rSub { size 8{s} } } } } {}

Solving for emf x size 12{"emf" rSub { size 8{x} } } {} gives

emf x = emf s R x R s . size 12{"emf" rSub { size 8{x} } ="emf" rSub { size 8{s} } { {R rSub { size 8{x} } } over {R rSub { size 8{s} } } } } {}
Two circuits are shown. The first circuit has a cell of e m f script E and internal resistance r connected in series to a resistor R. The second diagram shows the same circuit with the addition of a galvanometer and unknown voltage source connected with a variable contact that can be adjusted up and down the length of the resistor R.
The potentiometer, a null measurement device. (a) A voltage source connected to a long wire resistor passes a constant current I size 12{I} {} through it. (b) An unknown emf (labeled script E x in the figure) is connected as shown, and the point of contact along R size 12{R} {} is adjusted until the galvanometer reads zero. The segment of wire has a resistance R x size 12{R rSub { size 8{x} } } {} and script E x = IR x size 12{E rSub { size 8{x} } = ital "IR" rSub { size 8{x} } } {} , where I size 12{I} {} is unaffected by the connection since no current flows through the galvanometer. The unknown emf is thus proportional to the resistance of the wire segment.

Because a long uniform wire is used for R size 12{R} {} , the ratio of resistances R x / R s size 12{R rSub { size 8{x} } /R rSub { size 8{s} } } {} is the same as the ratio of the lengths of wire that zero the galvanometer for each emf. The three quantities on the right-hand side of the equation are now known or measured, and emf x size 12{"emf" rSub { size 8{x} } } {} can be calculated. The uncertainty in this calculation can be considerably smaller than when using a voltmeter directly, but it is not zero. There is always some uncertainty in the ratio of resistances R x / R s size 12{R rSub { size 8{x} } /R rSub { size 8{s} } } {} and in the standard emf s size 12{"emf" rSub { size 8{s} } } {} . Furthermore, it is not possible to tell when the galvanometer reads exactly zero, which introduces error into both R x size 12{R rSub { size 8{x} } } {} and R s size 12{R rSub { size 8{s} } } {} , and may also affect the current I size 12{I} {} .

Resistance measurements and the wheatstone bridge

There is a variety of so-called ohmmeters that purport to measure resistance. What the most common ohmmeters actually do is to apply a voltage to a resistance, measure the current, and calculate the resistance using Ohm’s law. Their readout is this calculated resistance. Two configurations for ohmmeters using standard voltmeters and ammeters are shown in [link] . Such configurations are limited in accuracy, because the meters alter both the voltage applied to the resistor and the current that flows through it.

The diagram shows two circuits. The first one has a cell of e m f script E and internal resistance r connected in series to an ammeter A and a resistor R. The second circuit is the same as the first, but in addition there is a voltmeter connected across the voltage source E.
Two methods for measuring resistance with standard meters. (a) Assuming a known voltage for the source, an ammeter measures current, and resistance is calculated as R = V I size 12{R= { {V} over {I} } } {} . (b) Since the terminal voltage V size 12{V} {} varies with current, it is better to measure it. V size 12{V} {} is most accurately known when I size 12{I} {} is small, but I size 12{I} {} itself is most accurately known when it is large.

The Wheatstone bridge    is a null measurement device for calculating resistance by balancing potential drops in a circuit. (See [link] .) The device is called a bridge because the galvanometer forms a bridge between two branches. A variety of bridge devices are used to make null measurements in circuits.

Resistors R 1 size 12{R rSub { size 8{1} } } {} and R 2 size 12{R rSub { size 8{2} } } {} are precisely known, while the arrow through R 3 size 12{R rSub { size 8{3} } } {} indicates that it is a variable resistance. The value of R 3 size 12{R rSub { size 8{3} } } {} can be precisely read. With the unknown resistance R x size 12{R rSub { size 8{x} } } {} in the circuit, R 3 size 12{R rSub { size 8{3} } } {} is adjusted until the galvanometer reads zero. The potential difference between points b and d is then zero, meaning that b and d are at the same potential. With no current running through the galvanometer, it has no effect on the rest of the circuit. So the branches abc and adc are in parallel, and each branch has the full voltage of the source. That is, the IR size 12{ ital "IR"} {} drops along abc and adc are the same. Since b and d are at the same potential, the IR size 12{ ital "IR"} {} drop along ad must equal the IR size 12{ ital "IR"} {} drop along ab. Thus,

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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on number 2 question How did you got 2x +2
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Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
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Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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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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