University Physics: Waves and Electricity Ch26. Ohm’s Law Lecture 10 Dr.-Ing. Erwin Sitompul ...

University Physics: Waves and Electricity

Ch26. Ohm’s LawLecture 10

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

2013

10/2Erwin Sitompul University Physics: Wave and Electricity

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?


Solution of Homework 8

(1.2 cm)(1.2 cm)A21.44 cm

4 21.44 10 m

15 cmL 0.15 m

8iron 10 10 m

iron

LR

A

84

(0.15)(10 10 )

(1.44 10 )

41.042 10

104.2

3 15 10 ( C)

iron iron,0 iron 01 ( )T T 8 3(10 10 ) 1 (5 10 )(35 20) 8(10 10 )(1.075)

810.75 10 m

iron

LR

A

84

(0.15)(10.75 10 )

(1.44 10 )

41.120 10

112

(a) (b)

15

1.21.2


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.


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.


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.


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 Vi1watt 1W 1V A 1volt ampere

• Rate of electrical energy transfer

W Pt

1 joule 1 J 1W s 1watt s ..... kWh

• Electrical energy transfer or electrical work


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.

2P i R2V

PR

• Resistive dissipation (energy lost)


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(d) R is doubled with i unchanged

(a) and (b) tie, (d), (c)

22

0

VP i R

R

2

a

(2 )VP

R

2

c 2

VP

R

2b (2 )P i R

2d (2 )P i R

2

4V

R 04P

24i R 04P2

0.5V

R 00.5P

22i R 02P


Example

You are given a length of uniform heating wire made of a nickel-chromium-iron alloy 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.

2

1

VP

R

2(120)

72 200 W

2 2

2 1 1

2 2

V VP

R R

2(120)4

72 800 W

2

4V

R

• The power dissipated by the wire cut in half is four times the power dissipated by the full wire.

• Advantage: The heating time reduced to one-fourth.

• Disadvantage: The current is doubled, may destroy the wire


Electricity Rates

Daily electricity consumption of a household (tariff group R-2)• Air conditioner, 850 W, 8 hour 6.80 kWh• Lights, 8 x 25 W, 12 hours 2.40 kWh• Television, 180 W, 18 hours 3.24 kWh

12.44 kWh Rp11,000

University Physics: Waves and Electricity

Ch27. Circuit TheoryLecture 10

Dr.-Ing. Erwin Sitompulhttp://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.


– +

b

a

E iR

a b a

Va

Vb

i

Work, Energy, and Emf

• The battery operates as a “pump” that moves positive charges from lower (–) to higher (+) electric potential.

Water Analogy

ai

b


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.


Single-Loop Circuit

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.

Circuit Loop Hiking Loop


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 a0iR E

• Counterclockwise move, starting from a0iR E

Loop Rule: The algebraic sum of the changes in potential encountered in a complete path of any loop of a circuit must be zero.


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

At points a, b, and c, rank (greatest first): (b) The magnitude of the current(c) The electric potential

E

All the same b, then a and c tie


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:

0ir iR E iR r

E • For ideal battery, r = 0

• In ideal battery, there is no potential drop across the battery


1eq

1 1n

i iR R

Resistance in Series and in Parallel

Resistance in Series

Resistance in Parallel

eq 1 2 3R R R R

eq1

n

ii

R R

Þ

Þ eq 1 2 3

1 1 1 1

R R R R


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?

iA,closed > iA,open Light bulb A becomes brighter when S is closed.

eq A B DR R R R

• The lamps have identical resistance of R

• P = Vi = i2R (brightness)

• S open3R

openeq

iR

E

3RE

• S closed

eq A B C DR R R R R 1

22 R

closedeq

iR

E

1

22 R

E

0.4R

E

0.333R

E


Trivia: Switching Theory

There is a room with one locked door and closed windows. Inside the room is a light bulb. Outside the room there are three switches.

You are allowed to flip any switches as many times as you wish. But, you may only open the door and enter the room once.

How do you find out which switch turns on the light?


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.

10 5 9 24

eq

iR

E 12

24 0.5 A

eq 1 2 3R R R R


Potential Difference Between Two Points

12a bV V • a ® b, cw

12a bV V 12 VabV

• a ® b, ccw

3 2 1a bV iR iR iR V 3 2 1a bV V iR iR iR

0.5(9 5 10)a bV V 12 VabV

• Clockwise ® cw• Counterclockwise ® ccw

0.5 A

1 2b dV iR iR V • b ® d, cw

1 2( )b dV V i R R 0.5(10 5)b dV V 7.5 VbdV

3b dV iR V E• b ® d, ccw

3b dV V iR E12 (0.5)(9)b dV V 7.5 VbdV


Potential Difference Across a Battery

The value given to E indicates the difference between the positive terminal and negative term of a battery.

–+

ba

1 6 VE

6 Va bV V

– +

a b

2 8 VE

8 Vb aV V

–+

ba

3 2 VE

2 Va bV V

2 Vb aV V

≡

≡


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.


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.


Homework 9A

1. Find the equivalent resistance between nodes a and b for the circuit shown below.R1 = R9 = R10 = 2 Ω

R3 = R5 = R7 = R11 = 4 Ω

R2 = R4 = R6 = R8 = 10 Ω

2. A circuit containing five resistors connected to a battery with a 12 V emf is shown below. What is the potential difference across the 5 Ω resistor? What is the current flowing through the 12 Ω resistor?


Homework 9B

1. If the emf of the battery is 6.60 V and the current flowing through the 4.00-Ω resistor at the bottom is 0.78 A, determine the value of unknown resistors R.

University Physics: Waves and Electricity Ch26. Ohm’s Law Lecture 10 Dr.-Ing. Erwin Sitompul ...

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