Ch20A

Chapter 20

Electric Circuits

This power point presentation is based on the supplement provided by John Wiley & Sons, Inc. for Cutnell & Johnson’s Physics. It is provided for the students in College Physics II class of NC A&T by Dr. Y.-L. Lin and cannot be copied or distributed to any third party.

Ch20A: Sec. 1-6

20.1 Electromotive Force and Current

In an electric circuit, an energy source and an energy consuming deviceare connected by conducting wires through which electric charges move.


Within a battery, a chemical reaction occurs that transfers electrons fromone terminal to another terminal.

The maximum potential difference across the terminals is called the electromotive force (emf).


The electric current is the amount of charge per unit time that passesthrough a surface that is perpendicular to the motion of the charges.

tqIΔΔ

=

One coulomb per second equals one ampere (A).

(20.1)


If the charges move around the circuit in the same direction at all times,the current is said to be direct current (dc).

If the charges move first one way and then the opposite way, the current is said to be alternating current (ac).


Example 1 A Pocket Calculator

The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hourof operation, (a) how much charge flows in the circuit and (b) how much energydoes the battery deliver to the calculator circuit?

(a)

(b)

( ) 3(0.17 10 A)(3600 s) 0.61 Cq I t −Δ = Δ = × =

( )( )0.61 C 3.0 V 1.8 JE = qV = =


Conventional current is the hypothetical flow of positive charges that wouldhave the same effect in the circuit as the movement of negative charges thatactually does occur.

20.2 Ohm’s Law

OHM’S LAW

The ratio V/I is a constant, where V is thevoltage applied across a piece of materialand I is the current through the material:

SI Unit of Resistance: volt/ampere (V/A) = ohm (Ω)

IRVRIV

=== or constant (20.2)

The resistance (R) is defined as the ratio of the voltage V applied across a piece of material to the current I throughthe material.

20.2 Ohm’s Law

A wire or an electrical device offers resistance to electrical flow,it is called a resistor.

For example, the filament in a light bulb is a resistor in the form of a thin piece of wire.

20.2 Ohm’s Law

Example 2 A Flashlight

The filament in a light bulb is a resistor in the formof a thin piece of wire. The wire becomes hot enoughto emit light because of the current in it. The flashlightuses two 1.5-V batteries to provide a current of 0.40 A in the filament. Determine the resistance of the glowing filament.

Ω=== 5.7A 0.40

V 0.3IVR

Solution

(9/9/08)

20.3 Resistance and Resistivity

For a wide range of materials, the resistance of a piece of material is proportional to the length L

ALR ρ=

resistivity (in units of ohm·meter)

(20.3)

and inversely proportional to the cross-sectional area A

R L∝

1/R A∝

Thus, we may write


ALR ρ=


Example 3 Longer Extension Cords

The instructions for an electric lawn mower suggest that a 20-gauge extensioncord can be used for distances up to 35 m, but a thicker 16-gauge cord shouldbe used for longer distances. The cross sectional area of a 20-gauge wire is5.2x10-7 m2, while that of a 16-gauge wire is 13x10-7 m2. Determine the resistance of (a) 35 m of 20-gauge copper wire and (b) 75 m of 16-gauge copper wire.

Ω=×

⋅Ω×==

−

2.1m105.2

m) m)(351072.1(27-

8

ALR ρ(a)

(b) Ω=×

⋅Ω×==

−

99.0m1013

m) m)(751072.1(27-

8

ALR ρ

(9/17/09)


Impedance Plethysmography

calf

2

calf VL

LVL

ALR ρρρ ===

volume

Reading Assignment


( )[ ]oo TT −+= αρρ 1

temperature coefficient of resistivity

The resistivity of a material depends on temperature.

(20.4)

(20.5)

Since R = ρL/A, (20.4) becomes

( )( )

( )

( / )

( / ) 1

( / ) 1

1

o o

o o

o o

R L A

L A T T

L A T T

R T T

ρ

ρ α

ρ α

α

=

⎡ ⎤= + −⎣ ⎦⎡ ⎤= + −⎣ ⎦

⎡ ⎤= + −⎣ ⎦

20.4 Electric Power

Suppose some charge emerges from a battery and the potential differencebetween the battery terminals is V, the electric power can be calculated from V and I.

( ) IVVtq

tVq

tEP =

ΔΔ

=ΔΔ

=Δ

=

energy

power

Time interval

(20.6a)

Eq. (19.3): V=EPE/qo

20.4 Electric Power

IVP =

ELECTRIC POWER

When there is current in a circuit as a result of a voltage, the electricpower delivered to the circuit is:

SI Unit of Power: watt (W = A / voltage)

Alternative ways for calculating the electric power are:

( ) RIIRIP 2==

RVV

RVP

2

=⎟⎠⎞

⎜⎝⎛=

(20.6a)

(20.6b)

(20.6c)

20.4 Electric Power

Example 5 The Power and Energy Used in aFlashlight

In the flashlight, the current is 0.40 A and the voltageis 3.0 V. Find (a) the power delivered to the bulb and(b) the energy dissipated in the bulb in 5.5 minutesof operation.

(a)

(b)

W2.1V) (3.0 A) 40.0( === IVP

J100.4)s330(W)2.1( 2×==Δ= tPE

Solution

Lec. 7 (2/5/09)

20.5 Alternating Current


( )ftVV o π2sin= (20.7)

The voltage (V) produced by ac generator is

Frequency (in cycles/s or Hz) Period = 1/f

Peak voltage

Fig. 20.11

In an AC circuit, the charge flow reverses direction periodically.


In circuits that contain only resistance, the current reverses direction each time the polarity of the generator reverses.

( ) ( )sin 2 sin 2oo

VVI ft I ftR R

π π⎛ ⎞= = =⎜ ⎟⎝ ⎠

peak current

(20.8)

t = 1/ft = 0 t = 2(1/f)


( )ftVIIVP oo π2sin2==

( )ftII o π2sin= ( )ftVV o π2sin=

(20.9)

P

Fig. 20.12


rmsrms222VIVIVIP oooo =⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛==

(20.10)

(20.11)

Root mean square


RIV rmsrms =

rms rmsP I V=

RIP 2rms=

RVP

2rms=

(20.14)

(20.15a)

(20.15b)

(20.15c)

Ohm’s law can be written as:

Average power can be expressed as:


Example 6 Electrical Power Sent to a Loudspeaker

A stereo receiver applies a peak voltage of34 V to a speaker. The speaker behavesapproximately as if it had a resistance of 8.0 Ω.

Determine (a) the rms voltage, (b) the rmscurrent, and (c) the average power for this circuit.

(a) V 242V 34

2rms === oVV

Solution

(b) A 3.0 0.8V 24rms

rms =Ω

==R

VI

(c) ( )( ) W72V 24A 3.0rmsrms === VIP

(9/11/08)


Conceptual Example 7 Extension Cords and a Potential Fire Hazard

During the winter, many people use portable electric space heaters to keepwarm. Sometimes, however, the heater must be located far from a 120-V wall receptacle, so an extension cord must be used. However, manufacturers oftenwarn against using an extension cord. If one must be used, they recommenda certain wire gauge, or smaller. Why the warning, and why are smaller-gauge wires better than larger-gauge wires?

Answer

The extension cord can be heated up when electric current passes through. The resistance of the wire must be kept small.

Wires with smaller gauge (smaller cross section) are better because they offer less resistance than larger-gauge wires.

20.6 Series Wiring

There are many circuits in which more than one device is connected toa voltage source.

Series wiring means that the devices are connected in such a waythat there is the same electric current through each device.

20.6 Series Wiring

( ) SIRRRIIRIRVVV =+=+=+= 212121

+++= 321 RRRRSSeries resistors

Equivalent resistance of the series circuit

20.6 Series Wiring

Example 8 Resistors in a Series Circuit

A 6.00 Ω resistor and a 3.00 Ω resistor are connected in series witha 12.0 V battery. Assuming the battery contributes no resistance to the circuit, find (a) the current, (b) the power dissipated in each resistor,and (c) the total power delivered to the resistors by the battery.

20.6 Series Wiring

(a)

(b)

(c)

Ω=Ω+Ω= 00.9 00.3 00.6SR A 33.1 00.9V 0.12

=Ω

==SR

VI

( ) ( ) W6.10 00.6A 33.1 22 =Ω== RIP

( ) ( ) W31.5 00.3A 33.1 22 =Ω== RIP

W9.15 W31.5 W6.10 =+=P

For 6 Ω resistor:

For 3 Ω resistor:

20.6 Series Wiring

Personal electronic assistants.

Reading Assignment

Ch20A

Documents

Transcript of Ch20A