This document discusses electric current and direct current circuits. It begins by providing an analogy comparing voltage, resistance, and current to gravity, friction, and the flow of water in a river. It then defines electric current microscopically as the drift velocity of electrons in a metal when an electric field is applied. It provides the formula for current as the rate of flow of charge. Several examples are worked through applying this formula. The document continues by defining Ohm's law and resistivity. It explains how resistance depends on the properties and dimensions of the conductor. Finally, it discusses how resistance varies with temperature and defines the temperature coefficient of resistivity.

1. 18. Electric Current & Direct Current Circuits
 SF026 2015/ 16

2. ANALOGY: Electric circuits
Voltage: A force
that pushes the
current through the
circuit (in this picture
it would be
equivalent to gravity)
Resistance:
Friction that
impedes flow of
current through
the circuit (rocks
in the river)
Current:
The actual
“substance”
that is
flowing
through the
wires of the
circuit
(electrons!)

3. 18.1 Electrical Conduction
(a)Describe microscopic model of current
(b)Define and use electric current formulae,
dt
dQ
I 

4. Microscopic Model of Current
In wire (metal)
the charge carrier
is free electrons
Without battery,
V = 0  No
electric field, E
Free e
undergoes
random motion

5. Microscopic Model of Current
When a potential
difference is
applied across
the metal, an
electric field, E
is set up
This field exerts
an electric force
on the freely
moving electron
The freely moving electrons
tend to drift with constant
average velocity (drift
velocity, vd) along the metal
in a direction opposite that of
the E
http://www.schoolphysics.co.uk/age16-
19/Electricity%20and%20magnetism/Current%20electricity/text/Electric_current/index.html

6. Electric Current, I
 Consider a simple closed circuit consists of
wires, a battery and a light bulb as shown
 From the diagram,
 Direction of electric field or electric
current: (+)ve to (–)ve
 Direction of electron flows: (–)ve to (+)ve
 Electric current, I is defined as the rate of flow of
charge
Mathematically,
 One Ampere is the flow of one C through an area in one
second
t
Q
I 
dt
dQ
I  Instantaneous
current
Average
current
It is a base and scalar quantity
S.I. unit: ampere (A).
1
sC1
second1
coulomb1
ampere1 


7. Example 18.1
(a) There is a current of 0.5
A in a flashlight bulb for
2 min. How much charge
passes through the bulb
during this time?
Solution:
(b) A silver wire carries a
current of 3.0 A.
Determine
(i) the number of electrons
per second pass
through the wire,
(ii) the amount of charge
flows through a cross-
sectional area of the
wire in 55 s.
t
Q
I  C60
119
selectrons1088.1 

t
N
C165Q

8. 18.2 Ohm’s Law & Resistivity
(a)State and use Ohm’s law
(b)Define and use resistivity, l
RA
ρ 

9. Ohm’s Law
 States that the potential
difference across a
conductor is
proportional to the
current flowing through
it if its temperature is
constant
where T is constant
 Materials that obey Ohm’s law
are materials that have constant
resistance over a wide range of
voltage  ohmic conductor
 Materials that do not obey the
Ohm’s law  non-ohmic
conductors
IV 
IRV 
 Then
R = resistance
(V)V
(A)I0
Gradient, m
= R
Ohmic conductor
(metal)
http://www.schoolphysics.co.uk/animations/Ohms_law/index.html
file:///C:/DOCUME~1/KMS/LOCALS~1/Temp/phet-ohms-law/ohms-law_en.html

10. Ohm’s Law – Non-ohmic conductor
V
I
0
Semiconductor Carbon
V
I
0
Electrolyte
V
I
0

11. Resistance, R – electric property which is
impedes or limits current in an electrical circuit
 is defined as a ratio of the potential difference across
an electrical component to the current passing
through it
Mathematically,
 Scalar quantity, unit: ohm ( ) or V A1
I
V
R 
Resistance
Type of
material
Length, l
Cross
sectional
area, A
Temperature
 In general, the
resistance of
a conductor
increases
with
temperature
V: potential difference (voltage),
I: current

12. Resistivity &
Conductivity
Resistivity,  (specific
resistance)
 is defined as the resistance of
a unit cross-sectional area
per unit length of the
material
Mathematically,
 Scalar quantity
 Unit: ohm meter ( m)
 It is a measure of a material’s
ability to oppose the flow of
an electric current
 Resistivity depends on the
type of the material &
temperature
l
RA
ρ 
ρ
σ
1

Conductivity, 
 is defined as the reciprocal of
the resistivity of a material
Mathematically,
 Scalar quantity,
unit: 1 m1
Material Resistivity,  ( m)
Silver 1.59  108
Copper 1.68  108
Aluminum 2.82  108
Gold 2.44  108
Glass 10101014
 A good electric conductors 
have a very low resistivities,
 good insulators  have very
high resistivities
file:///C:/DOCUME~1/KMS/LOCALS~1/Temp/phet-resistance-in-a-wire/resistance-in-a-wire_en.html

13. Example 18.2
A constantan wire of
length 1.0m and cross
sectional area of 0.5 mm2
has a resistivity of 4.9 x
10–7 Ωm. Find the
resistance of the wire.

14. Example 18.3
Two wires P and Q with
circular cross section are
made of the same metal and
have equal length. If the
resistance of wire P is three
times greater than that of wire
Q, determine the ratio of their
diameters.

15. Exercise 18.2
1. A wire 5.0 m long and 3.0 mm in diameter has a
resistance of 100 . A 15 V of potential difference is
applied across the wire. Determine
(a) the current in the wire,
(b) the resistivity of the wire,
(c) the rate at which heat is being produced in the wire.
(College Physics,6th edition, Wilson, Buffa & Lou, Q75,
p.589)
ANS: 0.15 A; 1.414  104  m; 2.25 W

16. 18.3 Variation of Resistance
With Temperature
(a)Explain the effect of temperature on
electrical resistance
(b)Use resistance,
  00 1 TTRR  

17. Effect of Temperature on Electrical
Resistance in Metals
 When the temperature
increases, the number of
free electrons per unit
volume in metal remains
unchanged
 Metal atoms in the crystal
lattice vibrate with greater
amplitude and cause the
number of collisions
between the free electrons
and metal atoms increase
and slowing down the
electron flow
 Hence the resistance in
the metal increases

18. Effect of Temperature on Electrical
Resistance in Metal
 The resistance of a metal
can be represented by the
equation below
where
R = final resistance
Ro= initial resistance
 = the temperature coefficient of
resistivity
Material  (C1)
Silver 4.10  103
Mercury 0.89  103
Iron 6.51  103
Aluminum 4.29  103
Copper 6.80  103
 TαRR  10
 α is defined as the fractional
change in resistance per
Celsius degree
 temperature coefficients of
resistivity for various materials
T
R
R


 0

19. Effect of Temperature on Electrical
Resistance in Metal
19
R
T0
0R
cT
Figure 18.8a : metal Figure 18.8b : semiconductor
R
T0
R
T0
Figure 18.8c : superconductor
R
T0
Figure 18.8d : carbon

20. Example 18.4
A copper wire has a resistance of 25 m at 20 C. When
the wire is carrying a current, heat produced by the current
causes the temperature of the wire to increase by 27 C.
(a) Calculate the change in the wire’s resistance
(b) If its original current was 10.0 mA and the potential
difference across wire remains constant, what is its final
current?
(copper = 6.80  103 C1)

21. Example 18.5
A platinum wire has a
resistance of 0.5 Ω at 0°C.
It is placed in a water bath
where its resistance rises
to a final value of 0.6 Ω.
What is the temperature
of the bath?
(platinum = 3.93  103
C1)

22. Exercise 18.3
1. A wire of unknown composition has a resistance of 35.0 
when immersed in the water at 20.0 C. When the wire is
placed in the boiling water, its resistance rises to 47.6 .
Calculate the temperature on a hot day when the wire has a
resistance of 37.8 . (Physics,7th edition, Cutnell &
Johnson, Q15, p.639)
ANS: 37.78 C
2. A copper wire has a resistance of 25 m at 20 C. When the
wire is carrying a current, heat produced by the current
causes the temperature of the wire to increase by 27 C.
(a) Calculate the change in the wire’s resistance.
(b) If its original current was 10.0 mA and the potential
difference across wire remains constant, what is its final
current? (copper = 6.80  103 C1)
ANS: (a) 4.59×10–3 Ω; (b) 8.45×10–3 A

23. 18.4 Electromotive Force (emf),
Potential Difference & Resistance
(a)Define emf,  of a battery
(b)Explain the relationship between emf of
a battery and potential difference across
the battery terminals
(c)Use terminal voltage,
IrεV 

24. E.m.f,  VS Potential Difference, V
 Terminal voltage, V is
the potential
difference across the
terminals of a battery
when there is a
current flowing
through it
 Electromotive force,
e.m.f (ξ) of a battery is
the maximum potential
difference across its
terminals when it is not
connected to a circuits
Emf in electric
source, p.d
between two
points in a circuit

25. E.m.f,  VS Potential Difference, V
 Consider a circuit
consisting of a battery
(cell) that is connected
by wires to an external
resistor R as shown in
Figure
 In reality, when a battery
is supplying current, its
terminal voltage is less
than its e.m.f, ξ
 This reduces of voltage
is due to energy
dissipation in the battery
 In effect, the battery has
internal resistance (r)

26. E.m.f,  VS Potential Difference, V
 Mathematically
V = IR,
where
 = emf
V = terminal potential difference
I = current
R = external resistance
r = internal resistance
 is the resistance due to
chemicals inside the
battery (cell)
 It will constitutes part of the
total resistance in a circuit
 The emf of a battery is
constant but the internal
resistance of the battery
increases with time as a
result of chemical reaction
 I R r  
V Ir  
Internal resistance, r
rε r ε
OR

27. Example 18.6
A battery has an emf of 9.0
V and an internal resistance
of 6.0 . Determine
(a) the potential difference
across its terminals
when it is supplying a
current of 0.50 A,
(b) the maximum current
which the battery could
supply.

28. Example 18.7
A car battery has an emf of
12.0 V and an internal
resistance of 1.0 . The
external resistor of
resistance 5.0  is
connected in series with the
battery as shown.
Determine the reading of
the ammeter and voltmeter
if both meters are ideal.

29. Exercise 18.4
1. A battery of e.m.f 3.0 V and internal resistance 5.0  is
connected to a switch by a wire of resistance 100 .
The voltage across the battery is measured by a
voltmeter. What is the voltmeter reading when the
switch is
(a) off?
(b) on? ANS: (a) 3V, (b) 2.86 V
2. An idealized voltmeter is connected
across the terminals of a battery while
the current is varied. Figure shows a
graph of the voltmeter reading V as a
function of the current I through the
battery. Find
(a) the emf, ξ and
(b) the internal resistance of the battery
ANS: (a) 9V, 4,5Ω

30. Exercise 18.4
3. A battery of emf 6.0 V is connected across a 10 
resistor. If the potential difference across the resistor is
5.0 V,
(a) Determine
(i) the current in the circuit,
(ii) the internal resistance of the battery.
(b) When a 1.5 V dry cell is short-circuited, a current of 3.0
A flows through the cell. What is the internal resistance
of the cell?
ANS: 0.50 A, 2.0 ; 0.50 

31. Revision
ItQ 
IV 
Laws'Ohm
l
RA
ρ 
ρ
σ
1

 TRR  10
I Battery (cell)
A Brε
R
 I R r  
IRV 

32. Use:
(i) power,
(ii) electrical energy,
IVP 
IVtW 
18.5 Electrical Energy & Power

33. Power, P
 is defined as the energy
liberated per unit time in
the electrical device
 The electrical power P
supplied to the electrical
device is given by
 When the electric current
flows through wire or
passive resistor, hence
the potential difference
across it is
 then the electrical power
can be written as
 It is a scalar quantity and
its unit is watts (W)
t
VIt
t
W
P 
IVP 
IRV 
RIP 2
 OR
R
V
P
2


34. Energy, E
 Consider a circuit
consisting of a battery that
is connected by wires to
an electrical device (such
as a lamp, motor or battery
being charged) where the
potential different across
that electrical device is V
 A current I flows from the terminal A to the
terminal B, if it flows for time t, the charge
Q which it carries from B to A is given by
 Then the work done on this charge Q
from B to A (equal to the electrical
energy supplied) is
 If the electrical device is passive resistor
(device which convert all the electrical
energy supplied into heat), the heat
dissipated H is given by
Electrical device
A B
VI I
QVW 
ItQ 
VItEW 
VItWH  OR RtIH 2


35. Example 18.7
In figure below, a battery
has an emf of 12 V and an
internal resistance of 1.0 .
Determine
(a) the rate of energy
transferred to electrical
energy in the battery,
(b) the rate of heat
dissipated in the battery,
(c) the amount of heat loss
in the 5.0  resistor if the
current flows through it
for 20 minutes.

36. Exercise 18.5
An electric toy of resistance 2.50  is operated by a dry cell
of emf 1.50 V and an internal resistance 0.25 .
(a) What is the current does the toy drawn?
(b) If the cell delivers a steady current for 6.00 hours,
calculate the charge pass through the toy.
(c) Determine the energy was delivered to the toy.
ANS: 0.55 A; 1.19  104 C; 16.3 kJ

37. 18.6 Resistors in Series
& Parallel
Derive and determine effective resistance
of resistors in series and parallel

38. Resistors in
Series
 From the definition of
resistance, thus
Substituting for V1, V2 , V3
and V
where
Reff: effective) resistance
;22 IRV  ;33 IRV ;11 IRV  effIRV 
321eff IRIRIRIR 
321eff RRRR 
 Consider three resistors are
connected in series to the
battery
Characteristics:
 Same current I flows
through each resistor
 Total potential difference,
V (Assumption: the
connecting wires have no
resistance)
321 IIII 
321 VVVV 

39. Example
(a) What is the current in each resistor?
(b) What is the voltage across each resistor?
(c) What is the total resistance?
(d) What is the battery voltage?
ANS: 0.1A, 3V, 4V, 5V; 120; 12V

40. Resistors in
Parallel
 Consider three resistors
are connected in parallel
to the battery
Characteristics:
 Same potential
difference, V across
each resistor
 Total current in the
circuit
 From the definition of
resistance, thus
Substituting for I1, I2 , I3 and I
321 VVVV 
321 IIII 
;
2
2
R
V
I  ;
3
3
R
V
I ;
1
1
R
V
I 
effR
V
I 
321eff R
V
R
V
R
V
R
V

321eff
1111
RRRR


41. Example
(a) What is the total resistance of the circuit (watch
out for the bear trap)?
(b) What is the current through each resistor?
(c) What is the total current?
ANS: 12.8; 0.40A, 0.30A, 0.24A; 0.94A

42. Example 18.8
What is the equivalent
resistance of the resistors
in figure below?
R1= R2= R3= R4= 1 Ω
A
B
R
1
R
2
R
3
R
4
A
B
R1
R2
R34
A
B
R1
R2
R3R4

5
3
ER
A
B
R1R234

43. Example 18.9
Find the current in &
voltage of the 10 Ω
resistor shown below.

44. Series VS Parallel
 When connected to the
same source, two light
bulbs in series draw
less power and glow
less brightly than when
they are in parallel
In the series case the same
current flows through both
bulbs.
If one of the bulbs burns out,
there will be no current at all in
the circuit, and neither bulb will
glow
In the parallel case the potential difference
across either bulb remains equal if one of
the bulbs burns out. The current through
the functional bulb remains equal and the
power delivered to that bulb remains the
same . This is another of the merits of a
parallel arrangement of light bulbs: If one
fails, the other bulbs are unaffected. This
principle is used in household wiring
systems

45. Example 18.10
For the circuit below,
calculate
(a) the effective resistance
of the circuit,
(b) the current passes
through the 12 
resistor,
(c) the potential difference
across 4.0  resistor,
(d) the power delivered by
the battery.
The internal resistance of
the battery may be ignored.

46. Example 18.11
For the circuit above,
calculate the effective
resistance between the
points A and B.

47. Exercise 18.6
1. Determine the equivalent resistances of the resistors
below.
ANS: 0.80 ; 2.7 ; 8.0 
0.2
0.2
0.2
0.2
0.6
01
0.6
0.4
18
16
0.8
0.9
16
0.6
20

48. Exercise 18.6
2. The circuit below includes a
battery with a finite internal
resistance, r = 0.50 .
(a) Determine the current flowing
through the 7.1  and 3.2 
resistors.
(b) How much current flows
through the battery?
(c) What is the potential
difference between the
terminals of the battery?
(Physics,3th edition, James S.
Walker, Q39, p.728)
ANS: 1.1 A, 0.3 A; 1.4 A; 11.3 V

49. Exercise 18.6
3. Four identical resistors are
connected to a battery as shown
below. When the switch is open,
the current through the battery is
I0.
(a) When the switch is closed, will the
current through the battery
increase, decrease or stay the
same? Explain.
(b) Calculate the current that flows
through the battery when the
switch is closed, Give your answer
in terms of I0. (Physics,3th edition,
James S. Walker, Q45, p.728)
ANS: U think

50. Exercise 18.6
4. Figure below
shows the
arrangement of
five equal
resistors in a
circuit. Calculate
(a) the equivalent
resistance
between point x
and y.
(b) the voltage across
point b and c.
(c) the voltage across
point c and y.

51. 18.7 Kirchhoff’s Law
State and use
Kirchhoff’s law

52. 1st Kirchhoff’s Law (KCL)
 the sum of the currents entering any junctions in a
circuit must equal the sum of the currents leaving
that junction
  outin II
1I
2I 3I
321 III  123 III 
3I 2I
1I

53. 2nd Kirchhoff’s Law (KVL)
 in any loop, the sum of emf is equal to the sum of the
products of current and resistance
 For emf, 
 For product of IR
ε
ε
direction of loop
+-
ε
-
ε
+
direction of loop
IR
direction of loop
I
R IR
I
R
direction of loop
IR  

54. Example 18.12
For the circuit
below, determine
the current and its
direction in the
circuit.
1.15
.226
50.8 2,V1.51
4,V5.01
Loop 1
I
I
II

55. Example 18.13
For the circuit below, determine
(a) the currents I1, I2 and I,
(b) the potential difference across the 6.7  resistor,
(c) the power dissipated from the 1.2  resistor.
8.9
9.3
V.09V21
7.6
.21
I1I 2I

56. Example 18.14
Find the value of current, I , resistance R & the e.m.f, ξ
ξ
D
A
F E
B
IR
1
3
AI 11 
AI 22 
C
V12

57. Exercise 18.7
1. For a circuit below, given
1= 8V, R2= 2 , R3= 3
, R1 = 1 and I = 3 A.
Ignore the internal
resistance in each
battery. Calculate
(a) the currents I1 and I2.
(b) the emf, 2.
ANS: 1.0 A, 4.0 A; 17 V

58. Exercise 18.7
2. Determine the current in each resistor in the
circuit below. (College Physics,6th edition,
Wilson, Buffa & Lou, Q57, p.619)
ANS: 3.75 A; 1.25 A; 1.25 A

59. Exercise 18.7
3. Referring to the circuit above, calculate the current I1, I2
and I3
ANS: I1=0.66A, I2 = I3 = 0.33A

60. 18.8 Potential Divider
(a) Explain the principle of a potential divider.
(b) Apply equation of potential divider,
V
RR
R
V 







21
1
1

61. Potential Divider
 A potential divider
produces an output
voltage that is a fraction
of the supply voltage V
 This is done by
connecting two resistors
in series V
1V
1R
I
2V
2R
I 11 IRV  V
RR
R
V 







21
1
1
V
RR
R
V 







21
2
2
 Since the current flowing
through each resistor is
the same,
 Therefore, the potential
difference (voltage)
across R1 is given by
 Similarly,
21eff RRR 
effR
V
I  an
d
21 RR
V
I



62. Potential Divider
 Resistance R1 and R2 can
be replaced by a uniform
homogeneous wire
 The potential difference
(voltage) across the wire
with length l1 is
 Similarly,
V
I
2l1l
I
BA C
2V1V
V
ll
l
V 







21
1
1
V
ll
l
V 







21
2
2

63. Example 18.15
For the circuit below,
(a) calculate the output
voltage
(b) If a voltmeter of
resistance 4000  is
connected across the
output, determine the
reading of the voltmeter.

64. 18.8 Potentiometer &
Wheatstone Bridge
(a) Explain principles of potentiometer and
Wheatstone Bridge and their applications
(b) Use related equations for potentiometer,
and for Wheatstone Bridge,
x
3
2
1
R
R
R
R

2
1
2
1
l
l
ε
ε


65. Potentiometer
 Consider a
potentiometer
circuit below
 The potentiometer
is balanced when
the jockey is at
such a position on
wire AB that there is
no current through
the galvanometer
Thus
Galvanometer
reading = 0
ACx VV 

66. Potentiometer
 When the potentiometer
in balanced, the unknown
voltage (potential
difference being
measured) is equal to
the voltage across AC

67. Application: Compare the emfs of two
cells or find unknown emf
o In this case, a
potentiometer is set up as
illustrated in Figure
below, in which AB is a
wire of uniform resistance
and J is a sliding contact
(jockey) onto the wire.
o An accumulator X
maintains a steady
current I through the wire
AB
X
BA
I
G
I
(2)
(1)
2ε
S
II
1ε
J

68. Application: Compare the emfs of two cells
or find unknown emf
 Initially, switch S is
connected to the terminal (1)
and the jockey moved until
the emf ξ1 exactly balances
the potential difference (p.d.)
from the accumulator
(galvanometer reading is
zero) at point C. Hence
where
and
then
X
BA
I
G
I
(2)
(1)
2ε
S
II
1ε
C
J
1l
1 ACV 
ACAC IRV 
A
ρl
R 1
AC 
1
1
ρl
I
A

 
  
 
(1)

69. Application: Compare the emfs of two cells
or find unknown emf
X
BA
I
G
I
(2)
(1)
2ε
S
II
1ε
C
J
D1l
2l
 After that, the switch S is
connected to the terminal (2)
and the jockey moved until
the emf ξ2 balances the p.d.
from the accumulator at
point D
Hence
where
and
then
2 ADV 
ADAD IRV 
A
ρl
R 2
AD 
2
2
ρl
I
A

 
  
 
(2)

70. Application: Compare the emfs of two cells
or find unknown emf
 By dividing eq. (1) and eq. (2) then
 Since
 Equation above can be written as
1 1
2 2
l
l



1
1
22
ρl
I
A
ρl
I
A


 
 
 
 
 
 
1ρl
R R l
A
  
1 1
2 2
R
R




71. Example 18.16
Consider a potentiometer
with a standard battery with
an e.m.f. of 1.0186 V is
used in the circuit. When the
resistance is 36 Ω, the
galvanometer reads zero. If
the standard battery is
replaced by an unknown
e.m.f. the galvanometer
reads zero when the
resistance is adjusted to 48
Ω. What is the value of the
unknown e.m.f. ?

72. Exercise 18.8(a)
1. In figure below, PQ is a uniform
wire of length 1.0 m and
resistance 10.0 . ξ1 is an
accumulator of emf 2.0 V and
negligible internal resistance.
R1 is a 15  resistor and R2 is a
5.0  resistor when S1 and S2
open, galvanometer G is
balanced when QT is 62.5 cm.
When both S1 and S2 are
closed, the balance length is
10.0 cm. Calculate
(a) the emf of cell ξ2
(b) the internal resistance of cell ξ2.

73. Exercise 18.8(a)
2. Cells A and B and centre-zero
galvanometer G are connected to a
uniform wire OS using jockeys X and Y as
below. The length of the uniform wire OS is
1.00 m and its resistance is 12 . When
OY is 75.0 cm, the galvanometer does not
show any deflection when OX= 50.0 cm. If
Y touches the end S of the wire, OX = 62.5
cm when the galvanometer is balanced.
The emf of the cell B is 1.0 V. Calculate
(a) the potential difference across OY when
OY = 75.0 cm,
(b) the potential difference across OY when Y
touches S and the galvanometer is
balanced,
(c) the internal resistance of the cell A,
(d) the emf of cell A.

74. Wheatstone Bridge
 It is used to measured the
unknown resistance of the
resistor
 Figure below shows the
Wheatstone bridge circuit
consists of a cell of emf , a
galvanometer , know
resistances (R1, R2 and R3)
and unknown resistance Rx.
 The Wheatstone bridge is
said to be balanced when no
current flows through the
galvanometer

75. Wheatstone Bridge
 Hence
then
 Therefore
Since thus
Dividing gives
1CBAC III  2DBAD III and
Potential at C = Potential at D
ADAC VV  BDBC VV and
IRV 
3211 RIRI  and x221 RIRI 
x2
32
21
11
RI
RI
RI
RI

3
1
2
x R
R
R
R 







76. Example 18.17
A wheatstone bridge is used to make a precise
measurement of the resistance of a wire connector. If R = 1
kΩ & the bridge is balanced by adjusting P such that P =
2.5 Q, what is the value of X ?

77. Wheatstone Bridge:
Application (meter bridge)
 The application of the Wheatstone bridge is Metre Bridge
 The metre bridge is balanced when the jockey J is at such a
position on wire AB that there is no current through the
galvanometer. Thus the current I1 flows through the
resistance Rx and R but current I2 flows in the wire AB.
0
Accumulator
Jockey
Thick copper strip
(Unknown resistance)
(resistance box)
Wire of uniform resistance
xR
A
ε
G
B
R
J
2l1lI I
1I
2I
1I

78. Wheatstone Bridge:
Application (meter bridge)
 Let Vx : p.d. across Rx and V : p.d. across R,
 At balance condition,
By applying Ohm’s law, thus
Dividing gives
AJx VV  JBVV and
AJ2x1 RIRI  JB21 RIRI and
A
ρl
R 1
AJ 
JB2
AJ2
1
x1
RI
RI
RI
RI
 where and
A
ρl
R 2
JB 













A
ρl
A
ρl
R
R
2
1
x
R
l
l
R 






2
1
x

79. Exercise 18.8(b)
The circuit shown in figure is
known as a Wheatstone
bridge. Determine the value
of the resistor R such that
the current through the 85.0
 resistor is zero.
(Physics,3th edition, James
S. Walker, Q93, p.731)
ANS: 7.50 

80.  Next
Chapter:
 Chapter 19:
Magnetic
Field