Transcript (a) and (b) - Erwin Sitompul

Lecture 10
Ch26. Ohm’s Law
University Physics: Waves and Electricity
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Homework 8
A rectangular block of iron has dimensions 1.2 cm  1.2 cm 
15 cm. The temperature of the surrounding air is 20°C. A
potential difference is to be applied to the block between
parallel sides.
(a) What is the resistance of the block if the two parallel sides
are the square ends (with dimensions 1.2 cm  1.2 cm)?
(b) The temperature of the iron block increases up to 35°C due
to the flowing current. What is the resistance of the block
now?
Erwin Sitompul
University Physics: Wave and Electricity
10/2
Solution of Homework 8
15
1.2
cm)
(a) A  (1.2 cm)(1.2
2
 1.44 cm
 1.44 104 m2
L  15 cm
 0.15 m
iron  10 108 m
L
A
 (10 108 )
R  iron
(0.15)
(1.44 104 )
 1.042 10 4 
 104.2 
Erwin Sitompul
(b)
  5 103 (C)1
iron  iron,0 1 iron (T  T0 )
 (10 108 ) 1  (5 103 )(35  20) 
 (10 108 )(1.075)
 10.75 108 m
L
A
 (10.75 108 )
R  iron
(0.15)
(1.44 104 )
 1.120 10 4 
 112 
University Physics: Wave and Electricity
10/3
Ohm’s Law
 As we just discussed, a resistor is a conductor with a
specified resistance. It has that same resistance no matter
what the magnitude and direction (polarity) of the applied
potential difference are.
 Other conducting devices, however, might have resistance
that change with the applied potential difference.
 First, we must define how to assign polarity to a terminal and
how to describe current direction.
• The terminal with higher potential is
given a positive sign, while the
terminal with lower potential is given
a negative sign.
• The current will flow from higher
potential to lower potential. This is
taken as direction of positive current.
Erwin Sitompul
University Physics: Wave and Electricity
10/4
Ohm’s Law
 The i-V plot of a 1000 Ω resistor
is shown next.
 The slope of the line (i/V) is the
same for all V.
 This means that the resistance
of the device is independent of
the magnitude and polarity of V.
 The i-V plot of a pn junction
diode is shown next.
 The relation between i and V is
not linear. The slope of the line
(i/V) varies throughout V.
 This means that the resistance
of the device depends on the
magnitude and the polarity of V.
Erwin Sitompul
University Physics: Wave and Electricity
10/5
Ohm’s Law
 Ohm’s law:
A conducting device obeys Ohm’s law when the resistance of
the device is independent of the magnitude and polarity of the
applied potential difference. Otherwise, it does not obey
Ohm’s law.
• Resistors obey the Ohm’s law.
• Diodes do not obey the Ohm’s law.
Erwin Sitompul
University Physics: Wave and Electricity
10/6
Power in Electric Circuits
 The figure below shows a circuit consisting of
a battery, connected by wires to an
unspecified conducting device.
 The wires are assumed to have negligible
resistance.
 The unspecified device might be a resistor, a
rechargeable battery, a motor, or some other
electrical device.
 The rate at which energy is transferred from the battery to the
unspecified device is given by:
P  Vi
• Rate of electrical energy transfer
1 watt  1 W  1 V  A  1 volt ampere
Erwin Sitompul
University Physics: Wave and Electricity
10/7
Power in Electric Circuits
 The principle of conservation of energy tells
us that the decrease in electric potential
energy from a to b is accompanied by a
transfer of energy to some other form.
 If the unspecified device is a motor, the
energy is transferred as work done on the
load.
 If the device is a rechargeable battery that is being charged,
the energy is transferred to stored chemical energy in the
storage battery.
 If the device is a resistor, the energy is transferred to internal
thermal energy, tending to increase the resistor’s
temperature.
P  i2 R
Erwin Sitompul
V2
P
R
• Resistive dissipation (energy lost)
University Physics: Wave and Electricity
10/8
Checkpoint
A potential difference V is connected across a device with
resistance R, causing current i through the device.
Rank the following variations according to the change in the
rate at which electrical energy is converted to thermal energy
due to the resistance, greatest change first:
(a) V is doubled with R unchanged
(b) i is doubled with R unchanged
(c) R is doubled with V unchanged (a) and (b) tie, (d), (c)
(d) R is doubled with i unchanged
V2 2
P0 
i R
R
(2V )2
V2
4
Pa 
 4P0
R
R
Pb  (2i)2 R  4i 2 R  4P0
V2
V2
 0.5
Pc 
 0.5P0
R
2R
Pd  i 2 (2R)  2i 2 R  2P0
Erwin Sitompul
University Physics: Wave and Electricity
10/9
Example
You are given a length of uniform heating wire made of a
nickel-chromium-iron allow called Nichrome. It has a
resistance R of 72 Ω. At what rate is energy dissipated in each
of the following situations?
(1) A potential difference of 120 V is applied across the full
length of the wire.
(2) The wire is cut in half, and a potential difference of 120 V is
applied across the length of each half.
V 2 (120) 2
 200 W
P1 

R
72
• The power dissipated by the
wire cut in half is four times
the power dissipated by the
2
2
2
2
full wire.
V
(120)
V
V
 800 W • Advantage: The heating time
4
P2  1  1  4
R
72
R 2R
reduced to one-fourth.
2
• Disadvantage: The current is
doubled, may destroy the
wire
Erwin Sitompul
University Physics: Wave and Electricity 10/10
Lecture 10
Ch27. Circuit Theory
University Physics: Waves and Electricity
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Work, Energy, and Emf
 Charge carriers will only flow through a conductor if we
establish a potential difference between its two ends.
 To produce a steady flow of charge, we need a “charge
pump”, a device that maintains a potential difference between
a pair of terminals by doing work and the charge carriers.
 Such a device is called an electromotive force device (emf
device).
 Emf devices come in various kinds. All transform one source
of energy into electrical energy.
 A common emf device is the battery, electric generator, solar
cells, and fuel cells.
Erwin Sitompul
University Physics: Wave and Electricity 10/12
Work, Energy, and Emf
b
• The battery operates as a “pump” that
moves positive charges from lower (–)
to higher (+) electric potential.
i
– +
a
b
a
i
Vb
E
iR
Va
Water Analogy
Erwin Sitompul
a
b
a
University Physics: Wave and Electricity 10/13
Energy Transfers in the Circuit
• A and B are two ideal rechargeable batteries, R is a
resistance, and M is an electric motor that can lift an
object.
• EB > EA, so battery B determines the direction of current.
• Battery B is charging battery A. It also provides energy
to motor M and energy that is being dissipated by
resistance R.
Erwin Sitompul
University Physics: Wave and Electricity 10/14
Single-Loop Circuit
Circuit Loop
Hiking Loop
 Suppose we start at any point in the circuit above and
proceed in either direction.
 As we move, we add algebraically the potential differences
that we encounter.
 When we return to our starting point, we must also have
returned to our starting potential.
 Loop Rule
The algebraic sum of the changes in potential encountered in
a complete path of any loop of a circuit must be zero.
Erwin Sitompul
University Physics: Wave and Electricity 10/15
Single-Loop Circuit
 Resistance Rule
For a move through a resistance in
the direction of the current, the
change in potential is –iR (downhill);
in the opposite direction +iR (uphill)
 Emf Rule
For a move through an ideal emf
device in the direction of the emf
arrow, the change in potential is +E;
in the opposite direction –E.
• Clockwise move, starting from a
E  iR  0
• Counterclockwise move, starting from a
iR  E  0
Erwin Sitompul
University Physics: Wave and Electricity 10/16
Checkpoint
The figure shows the current i in a single-loop circuit with a
battery B and a resistance R (and wires of negligible
resistance).
(a) Should the emf arrow at B be drawn pointing leftward or
rightward?
Rightward, the same as
the direction of current
E
At points a, b, and c, rank (greatest first):
(b) The magnitude of the current All the same
(c) The electric potential b, then a and c tie
Erwin Sitompul
University Physics: Wave and Electricity 10/17
Ideal and Real Battery
 The figure below left shows a real battery, with internal
resistance r, wired to an external resistor of resistance R.
 The internal resistance r of the battery is the electrical
resistance of the materials that build the battery and thus
unremovable.
 If we apply the loop rule clockwise beginning at point a, the
changes in potential give us:
E
• For ideal battery, r = 0
E  ir  iR  0  i 
• In ideal battery, there is no
Rr
potential drop across the battery
Erwin Sitompul
University Physics: Wave and Electricity 10/18
Resistance in Series and in Parallel
Req  R1  R2  R3

n
Req   Ri
i 1
Resistance in Series
1
1 1
1
  
Req R1 R2 R3

n
1
1

Req i 1 Ri
Resistance in Parallel
Erwin Sitompul
University Physics: Wave and Electricity 10/19
Checkpoint
Consider a circuit with an ideal battery and four identical light
bulbs connected a shown in the figure. Initially, the switch S is
open. Then, the switch is closed.
What happens to light bulb A?
• S open
Req  RA  RB  RD  3R
E
E
E
 0.333

iopen 
R
Req 3R
• S closed
• The lamps have identical
resistance of R
• P = Vi = i2R (brightness)
Req  RA  RB RC  RD  2 12 R
E
E
E
iclosed 
 1  0.4
R
Req 2 2 R
iA,closed > iA,open
Light bulb A becomes brighter when S is closed.
Erwin Sitompul
University Physics: Wave and Electricity 10/20
Potential Difference Between Two Points
 To find the potential between any two points in a circuit, we
start at one point and go through the circuit to the other point,
following any path.
 Along the way, the changes in potential we encounter must
be added algebraically.
 The voltage difference is independent of the path chosen.
Req  R1  R2  R3
 10  5  9
 24 
12
E

 0.5 A
i
Req 24
Erwin Sitompul
University Physics: Wave and Electricity 10/21
Potential Difference Between Two Points
 0.5 A
• Clockwise  cw
• Counterclockwise  ccw
• a  b, cw
Va  12  Vb
Va  Vb  12
Vab  12 V
• a  b, ccw
• b  d, cw
Vb  iR1  iR2  Vd
Vb  Vd  i( R1  R2 )
Vb  Vd  0.5(10  5)
Vbd  7.5 V
Erwin Sitompul
• b  d, ccw
Va  iR3  iR2  iR1  Vb
Va  Vb  iR3  iR2  iR1
Va  Vb  0.5(9  5  10)
Vab  12 V
Vb  E  iR3  Vd
Vb  Vd  E  iR3
Vb  Vd  12  (0.5)(9)
Vbd  7.5 V
University Physics: Wave and Electricity 10/22
Potential Difference Across a Battery
 The value given to E indicates the difference between the
positive terminal and negative term of a battery.
E1  6 V
+
–
b
a
Va  Vb  6 V
≡
Vb  Va  8 V
≡
E2  8 V
– +
a
b
E3  2 V
+
–
a
Erwin Sitompul
Va  Vb  2 V
b
Vb  Va  2 V
University Physics: Wave and Electricity 10/23
Ammeter and Voltmeter
 An instrument used to measure current is called an ammeter.
It should be connected serially, means the current to be
measured must pass through the meter.
 An instrument used to measure potential difference is called
a voltmeter. To find the potential difference between any two
points, the voltmeter should be connected in parallel, means
the voltmeter terminals are connected between those points.
Erwin Sitompul
University Physics: Wave and Electricity 10/24
Homework 9
(a) Find the equivalent resistance between points a and b in
the circuit diagram below.
(b) Calculate the current in each resistor if a potential
difference of 34 V is applied between points a and b.
Erwin Sitompul
University Physics: Wave and Electricity 10/25
Homework 9
New
A circuit containing five resistors connected to a battery with a
12 V emf is shown below.
(a) What is the potential difference across the 5 Ω resistor?
(b) What is the current flowing through the 12 Ω resistor?
Erwin Sitompul
University Physics: Wave and Electricity 10/26