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18.1 Electrical Conduction
18.2 Ohms law and Resistivity
18.3 Variation of resistance with temperature
18.4 Electromotive force (emf), internal
resistance and potential difference
18.5 Electrical energy and power
18.6 Resistors in series and parallel18.7 Kirchhoffs Laws
18.8 Potential divider
18.9 Potentiometer and Wheatstone Bridge
TOPIC 18 : ELECTRIC CURRENT AND
DIRECT-CURRENT CIRCUITS
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18.1 Electrical Conduction
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Conductors contain many free electrons and move randomly.
If a continuous wire is connected to the terminal of a battery, the
potential difference between the terminals of the battery sets up an
electric field inside the wire and parallel to it, directed from the positive
toward the negative terminal.
Thus free electrons are attracted into the positive terminal (are
forced to drift in one direction).
This direction is in the direction opposite to the field, E. The velocity of these free electrons is called drift velocity.
battery
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18.1 Electrical Conduction
Thus free electrons at one end of the wire are attracted into
the positive terminal, and at the same time, electrons leave
the negative terminal of the battery and enter the wire at theother end.
There is a continuous flow of electrons through the wire that
begins as soon as the wire is connected to both terminals.
However, when the conventions of positive and negativecharge were advised two centuries ago, it was assumed that
positive charge flowed in a wire.
For nearly all purposes, positive charge flowing in one
direction is exactly equivalent to negative charge flowing in
the opposite direction.
Today we still use the historical convention of positive current
when discussing the direction of a current. So when we speak
of the current in a circuit, we mean the direction positive
charge would flow. 5
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SUBTOPIC :
LEARNING OUTCOMES :
a) State and use Ohms Law.
b) Define and use resistivity formulae,
At the end of this lesson, the students should
be able to :
RA
l
18.2 Ohms Law and Resistivity
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18.2 Resistivity and Ohms Law
Ohms Law
Ohms law states that the potential difference across a
conductor, Vis directly proportional to the current,I
through it, if its physical conditions and the temperature
are constant.
V I
constant
Ohm's Law
V
I
V R V IRI
18.2 Ohms Law and Resistivity
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18.2 Resistivity and Ohms Law
V
I
constant
V
I
VRI
Ohmic conductorsare conductors which obey Ohmslaw. Examples: pure metals. (Figure A)
Non-ohmic conductorsdo not obey Ohms law.
Example: junction diode. (Figure B)
Figure A Figure B
V
I
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Resistivity
Resistivity is a measure of a materials ability to oppose
the flow of electric current through the material.
Resistivity is defined as the resistance of a sample of thematerial of cross-sectional area 1 m2 and of length 1m.
It is a constant value.
Its formulae is given by .
Its unit is m.
Its value depends on the material.
All conductors have smaller resistivity. Insulators have larger resistivity.
RA
l
where l= length of the conductor (m)A = area of cross-section of the conductor (m-2)
18.2 Resistivity and Ohms Law
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5.2 Resistivity and Ohms Law
Example 18.2
A wire (length=2.0 m, diameter=1.0 mm) has a resistance of
0.45 . What is the resistivity of the material used to make the
wire ?
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5.2 Resistivity and Ohms Law
Example 18.3
What voltage will be measured across a 1000- resistor in a
circuit if we determine that there is a current of 2.50 mA flowing
through it ?
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SUBTOPIC :
LEARNING OUTCOMES :
a) Explain the effect of temperature on electrical
resistance in metals and superconductors.
b) Define and use temperature coefficient of resistivity,.
c) Apply resistanceR= Ro [1 + (T-To)].
At the end of this lesson, the students should
be able to :
18.3 Variation of Resistance with
Temperature
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More collisons occur between free electrons and ions.
18.3 Variation of Resistance with Temperature
These electrons are slowed down thus increases theresistance.
The resistance of a metal can be represented by theequation below
R=Ro[1+(T)] ,R-Ro= RR=Ro+T
where R = the resistance at temperature T,
Ro= the resistance at temperature To = 20o C or 0oC,
= the temperature coefficient of resistance (o
C-1
)
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18.3 Variation of Resistance with Temperature
is a constant value and it is depends on the material.
o
RR
T
Temperature coefficient of resistance , is defined as the
fractional change in resistance per Celsius degree.
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Example 18.4
A platinum wire has a resistance of 0.50 at 0oC. It is
placed in a water bath, where its resistance rises to a
final value of 0.60 . What is the temperature of the
bath ? ( = 3.93 x 10-3oC -1)
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18.3 Variation of Resistance with Temperature
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Example 18.5
A narrow rod of pure iron has a resistance of 0.10 at 20oC.
What is its resistance at 50 oC ? ( = 5.0 x 10-3oC -1)
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18.3 Variation of resistance with temperature
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18.3 Variation of Resistance with TemperatureSuperconductor
As the temperature decreases, the resistance at first
decreases smoothly.
(T, R)
At a certain critical temperature Tc (4.2 K for mercury)
the resistance suddenly drops to zero.
R
T
metal
R
T
superconductor
Graph of resistance Ragainst temperature T
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SUBTOPIC :
LEARNING OUTCOMES :
a) Define emf , .
b) Explain the difference between emf of a battery and
potential difference across the battery terminals.
c) Apply formulae V= Ir.
At the end of this lesson, the students should
be able to :
18.4 Electromotive Force (emf), Internal
Resistance and Potential Difference
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18.4 Electromotive Force (emf), Internal
Resistance And Potential Difference
The e.m.f of a source is the p.d across the term inals o f
the source in open circui t(no current is flowing,I= 0).
The e.m.f of a source is the wo rk done per uni t charge.
The e.m.f of a source is defined as the electr ical energythat generated by a source so that the charges canflow
from one terminal to another terminalof the source
through any res is tor.
The e.m.f of a battery is the po tential di f ference across
i ts terminalwhen it is not connected to a circuit.
What is electromotive force,emf ( or)?
SI unit : Volt (V)
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18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
What is in ternal resistance?
In a cell or battery, the negative ions are attracted by
anode and the positive ions are attracted by the cathode.
The flow of these ions produces current.
However the collisions between the ions and therecombination of opposite ions reduce the flow ofcurrent. This resistance in the cell is called internalresistance, r.
18 4 El i F ( f) I l R i d P i l Diff
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Suppose a battery of emf, and internal resistance r
is connected to an external resistor,R.
Total resistance in the circuit is (R + r).
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18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
18 4 El i F ( f) I l R i d P i l Diff
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The e.m.fof this battery is given as
( )
ab
I R rIr IR
Ir V
ab dcV V
Vab = Vb
Va = terminal voltage (potential difference acrossthe battery terminals)
R = external resistance
r= internal resistance25
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
18 4 El t ti F ( f) I t l R i t d P t ti l Diff
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In a circuit diagram, this symbol
represents a res is torin a circuit that dissipateselectrical
energy.
A straight line represents a conducting wire
with negl igib le resistance.
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18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
18 4 El t ti F ( f) I t l R i t d P t ti l Diff
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ab
ab
Ir V
V Ir
Notes:
a) Vab< when the battery of emf is connected to theexternal circuit with resistanceR.
b) Vab> when the battery of emf is being charged byother battery.
c) Vab= when the battery of emf has no internalresistance (r=0) and connected to the external circuit
with resistanceR..
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terminal voltage (potential
difference across terminals)emf
potential difference
across internal resistance
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
18 4 El t ti F ( f) I t l R i t d P t ti l Diff
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Example 18.6
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A battery with a terminal voltage of 11.5 V when delivering0.50 A has an internal resistance of 0.10 . What is its emf?
18.4 Electromotive Force (emf), Internal Resistance and Potential Difference
18 4 Electromotive Force (Emf) Internal Resistance and Potential Difference
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Example 18.7
29
The battery in a circuit has an emf of 9.0 V. It is attached to aresistor and an ammeter that shows a current of 0.10 A. If a
voltmeter across the batterys terminals reads 8.9 V, what is
its internal resistance ?
18.4 Electromotive Force (Emf), Internal Resistance and Potential Difference
SUBTOPIC
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SUBTOPIC :
LEARNING OUTCOMES :
a) Apply formulaP=IVand electrical energy, W=VIt.
At the end of this lesson, the students should
be able to :
18.5 Electrical Energy and Power
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The electrical (potential) energy, Wis the energy gained by
the charge Q from a voltage source (battery) having a
terminal voltage V.
W= QV (the work done by the source on the charge)
But Q=It, then W= VI t
Unit : Joule (J)
The rate of energy delivered to the external circuit by the
battery is called the electric powergiven by,
Unit : watt ( 1 W = 1J/s) 32
@
,W QV
P Q I t t t
P I V P I
18.5 Electrical Energy and Power
18 5 Electrical Energy and Power
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The energy dissipated per second in an electric device
(rate of energy dissipated) is given as
for any deviceW VIt
P VIt t
A passive resistor is a resistor which converts all the
electrical energy into heat. For example, a metal wire.
22
but
or
P VI V IR
V
P I R P R
only for resistor
18.5 Electrical Energy and Power
18 5 El t i l E d P
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Calculate the resistance of a 40 W automobile headlight
designed for 12 V?
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Example 18.818.5 Electrical Energy and Power
18 5 Electrical Energy and Power
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The current through a refrigerator of resistance 12 is 13
A. What is the power consumed by the refrigerator?
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Example 18.918.5 Electrical Energy and Power
18 5 Electrical Energy and Power
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An electric iron with a 15-ohm heating element operates at
120 V. How many joules of energy does the iron convert toheat in 1.0 h ?
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Example 18.1018.5 Electrical Energy and Power
SUBTOPIC
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SUBTOPIC :
LEARNING OUTCOMES :
a) Deduce and calculate effective resistance of resistors
in series and parallel.
At the end of this lesson, the students should
be able to :
18.6 Resistors in series and parallel
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18.6 Resistors in series and parallel
1R 2R 3R
V
1V 2V 3V
I I
Resistors in Series
The properties of resistors in series are given below.o The same currentI flows through each resistor
where
321 IIII
battery , r=0
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18.6 Resistors in series and parallel
o The sum of the voltages around a circuit loop (thatis, the gains and losses with + and - ,respectively) is
zero.
(Assuming that the connecting wires have no resistance)
1 2 3
0ii
ii
V V
V V
V V V V
total potential difference
;22
IRV ;33
IRV 1 1
bur ;V I R
321eq IRIRIRIR eq
IRV
321eq RRRR
where resistance)(effectiveequivalent:eqR
18.6 Resistors in series and parallel
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Resistors in Parallel
1R
3R
V1V
2V
3V
I I
2R2I
1I
3I The properties of resistors inparallel are given below.o There is the same potential
difference, Vacross eachresistor where
321 VVVV
o Charge is conserved, thereforethe total currentIin the circuitis given by
321 IIII
18.6 Resistors in series and parallel
18.6 Resistors in series and parallel
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;22 R
V
I ;33 RV
I 1 1but
;
V
I R
321eq R
V
R
V
R
V
R
V
eqR
V
I
321eq R
1
R
1
R
1
R
1
18.6 Resistors in series and parallel
18.6 Resistors in series and parallel
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Example 18.10
.04.02
V06 .
12
Calculate :
a. the total resistance of the circuit.
b. the total current in the circuit.c. the potential difference across 4.0 resistor.
18.6 Resistors in series and parallel
18.6 Resistors in series and parallel
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Solution 18.10
.04.02
V06 .
12
p
18.6 Resistors in series and parallel
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Solution 18.10
b. Total current,
c. The potential difference acrossR1=2.0 is
Therefore the potential difference acrossR3=4.0
is given by
p
18.6 Resistors in series and parallel
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Example 18.11
For the circuits shown below, calculate the equivalent
resistance between points x and y.
.03
.01.01
x
y
.02
.02
.018
.016
.08
yx .09
.016
.06
.020
(0.79 ) (8.0 )
p
SUBTOPIC :
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SUBTOPIC :
LEARNING OUTCOMES :
a) State and use Kirchhoffs Laws.
At the end of this lesson, the students should
be able to :
18.7 Kirchhoffs Laws
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18.7 Kirchhoffs Laws
Kirchhoffs first law (junction/current law)
It states that the algebraic sum of the currents at any
junction of a circuit is zero,
0
or
i n out
I
I I
18.7 Kirchhoffs Laws.
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Kirchhoffs second law (loop/voltage law)
I R
It states that the algebraic sum of the voltages across
all of the elements of any closed loop is zero.or
It states thatin any closed loop, the algebraic sum of
e.m.fs is equal to the algebraic sum of the products
of current and resistance.
Sign convention
IR
+IR
across resistor
-IR
+
-
across battery
18.7 Kirchhoffs Laws
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Example 18.12
Using Kirchhoffs rules, find the current in each resistor.
R1 = 10 R2 = 20
210 V
1 20 V
18.7 Kirchhoff s Laws
18.7 Kirchhoffs Laws
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Solution 18.12
R1 = 10 R2 = 20
210 V
1 20 V
Step
1. Draw current. (arbitrary)
2. Draw loop. (arbitrary)
3. Apply Kirchhoffs laws.
0 ,I I R
I
18.7 Kirchhoffs Laws
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Example 18.13
Apply Kirchhoffs rules to the circuit in figure below and find
the current in each resistor.
R1 = 3.0
R2 = 3.0
R3 = 3.0
R4 = 3.0
2 2 0 V.
1 2 0 V.
18.7 Kirchhoffs Laws
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Solution 18.13
R1 = 3.0
R2 = 4.0
R3 = 5.0
R4 = 2.0
2 3 0 V.
1 6 0 V.
I
2nd KL,
( )I R
18.7 Kirchhoffs Laws
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Example 18.14
Find the current in each resistor in the circuit shown below.
R2 = 4.0
R3 = 4.0
R1 = 4.0 3 5 0 V.
1 10 V
2 5 0 V.
I1=3.75 A up, I2= 1.25 A left, I3= 1.25 A right
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18.7 Kirchhoffs Laws
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Example 18.16
Calculate the currentsI1,I2 andI3. Neglect the internalresistance in each battery.
1R1
.10R3
V151
.50R2
V102
V033 .
1I
2I
3I
;. A6917I1
;. A6214I2
A073I3
.
E l 18 1718.7 Kirchhoffs Laws
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Example 18.17
Given 1=8 V, R2=2 , R3=3 ,
R1 =1 andI=3 A. Ignore the internal resistance in
each battery.
Calculate
a. the currentsI1and I2.b. the e.m.f. 2.
Ans. : 1 A, 4 A , 17 V
3R
1
2R2
1I
2I
I
1R
SUBTOPIC :
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SUBTOPIC :
LEARNING OUTCOMES :
a) Explain the principle of a potential divider.
b) Apply equation of potential divider
At the end of this lesson, the students should
be able to :
18.8 Potential divider
11
1 2
.R
V V
R R
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18.8 Potential divider
A potential divider is used to tap a fraction of the voltage
supplied by a source of e.m.f.
Potential divider circuit
11
1 2
RV V
R R
Potential difference across l1 orR1 is
V
I
2V1V
1RI
2ReqRVI
Two resistors are connected in series.
The current flows in each
resistor is the same ;
1 2,
eqR R R
21
RR
VI
11 IRV
18.8 Potential divider.
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V
I
2l1lI
ba c
2V1V
Vll
lV
21
11
lR
A
Potential difference across l1 is
11
1 2
RV V
R R
ResistanceR1 andR2 are replaced by a uniformhomogeneous wire as shown in figure below.
E l 18 1818.8 Potential divider
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Example 18.18
Resistors of 3.0 and 6.0 are connected in series to a
12.0 V battery of negligible internal resistance. What are the
potential difference across the (a) 3.0 and (b) 6.0
resistors ?
4.0 V, 8.0 V
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18 9 Potentiometer and Wheatstone Bridge
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18.9 Potentiometer and Wheatstone Bridge
A potentiometer is mainly used to measure potentiometer.
It consists of a uniform wire. Basically a potentiometer circuit consists of a uniform wire
AB of length 100.0cm, connected in series to a driver cell
with emfVof negligible internal resistance.
V
I I
BA C
xV
I
G
+ -
I
(Unknown Voltage)
Jockey
(Driver cell -accumulator)
Potentiometer
18.9 Potentiometer and Wheatstone Bridge
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The potentiometer is balanced when the jockey (sliding
contact) is at such a position on wire AB that there is no
current through the galvanometer. Thus
When the potentiometer in balanced, the unknown
voltage (potential difference being measured) is equal tothe voltage across AC.
Galvanometer reading = 0
ACx VV
Potentiometer can be used to :i) Measure an unknown e.m.f. of a cell.ii) Compare the e.m.f.s of two cells.
iii) Measure the internal resistance of a cell.
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18.9 Potentiometer and Wheatstone Bridge
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i) Measure an unknown e.m.f. of a cell.
V
I I
BAC
I
G
+ -
I
(Unknown cell)
Jockey
(Driver cell -accumulator) When the potentiometer is
balanced, IG = 0
VAC=
VAC= IRAC1
AC
AC
lR
A
2 , ABAB
lR
A
3
2 , ...43
AC
A C A B
AB
lR R
l
...5AB
VI
R
4 and 5 into 1,AC
AC AB
AB AB
AC
AB
lVV R
R l
lV
l
Balance length = lAC
18.9 Potentiometer and Wheatstone Bridge
ii) C th f f t ll
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ii) Compare the e.m.f.s of two cells.
V
I I
BAI
G
I
(2)
(1)
2
1
S
CJ
D
When the potentiometer is
balanced, IG = 0
Balance length,
lAC= l1 for1 and lCD= l2 for2
AC
AB
l Vl
From , thus
1 2
2 1
22 1
1
...1 and ...2
2 1 ,
AC CD
AB AB
CD
AC
l lV V
l l
l
l
l
l
1l2l
Wheatstone Bridge18.9 Potentiometer and Wheatstone Bridge
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Wheatstone Bridge
BA G
C
D
1R 2R
3R XR
I
2I
I
1I
2I
1I
0
It is used to measure the unknown resistance of the resistor.
Figure below shows the Wheatstone bridge circuit consists of a cell
of e.m.f. (accumulator), a galvanometer , known resistances (R1,R2
andR3) and unknown resistanceRx.
The Wheatstone bridge is said to be balanced when no current
flows through the galvanometer. Hence
Then ,
Therefore
Since
1CBAC III
2DBAD III
and
Potential at C = Potential at D
ADAC VV and BDBC VV
IRV thus
3211 RIRI X221 RIRI and
X2
32
21
11
RI
RI
RI
RI 3
1
2X R
R
RR
18.9 Potentiometer and Wheatstone Bridge
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BA G
C
D
1R 2R
3R XR
I
2I
I
1I
2I
1I
03
1
2
X RR
R
R
XR
A
GB
R
J
Thick copper
strip
(Unknown
resistance)
Jockey
(resistance box)
Accumulator
Wire of uniform
resistance
0
1I 1I
2I I
I
1
2
XR l
R l
2
3 1
XR R
R R
18.9 Potentiometer and Wheatstone Bridge
Example 18 19
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Example 18.19
An unknown length of platinum wire 0.920
mm in diameter is placed as the unknown
resistance in a Wheatstone bridge as
shown in figure below.
Resistors R1 and R2 have resistance of
38.0 and 46.0 respectively.
Balance is achieved when the switch
closed and R3 is 3.48 . Find the
length of the platinum wire if its
resistivity is 10.6 x 10-8 m.