Electric Circuits. Ohm’s Law The relationship between the potential difference in a circuit and...

Electric Circuits

Ohm’s Law

The relationship between the potential difference in a circuit and the resulting current.

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20.1 Electromotive Force and Current

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

http://phet.colorado.edu/en/simulation/signal-circuit

http://phet.colorado.edu/en/simulation/signal-circuit


The electric current is the amount of charge per unit time that passesthrough a surface that is perpendicular to the motion of the charges. By convention, we use the flow of positive charge or “electron holes”.

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One coulomb per second equals one ampere (A).

20.2 Ohm’s Law

To the extent that a wire or an electrical device offers resistance to electrical flow,it is called a resistor.

Kirchoff’s Junction Law

The sum of the currents entering a junction are equal to the sum of the currents leaving.

What current flows through the 4th wire and which way does it go?

A.1 A into the junction

B.1 A out of the junction

C.9 A into the junction

D.9 A out of the junction

In which direction does the potential decrease?

A.Clockwise

B.Counterclockwise

C.Potential is the same everywhere in the circuit.

Ohm’s Law in the Cell

The resistance between the walls of a biological cell is 1.00 x 1010 Ω.

(a)What is the current when the potential difference between the walls is 95 mV?

(b)If the current is composed of Na+ ions (q = +e), how much charge moved between the cell walls in 0.2 s?

Ohm’s Law in the Cell

Suppose that the resistance between the walls of a biological cell is 1.00 x 1010 Ω.

(a)What is the current when the potential difference between the walls is 95 mV? 9.5 x 10-12 A

(b)If the current is composed of Na+ ions (q = +e), how much charge moved between the cell walls in 0.2 s? 1.9 x 10-12 C

Power in a circuit

Through which resistor is the most power dissipated?

A. a B. b C. b,d D. d

Power rating of a household applicance

• Commercial and residential electrical systems are set up so that each individual appliance operates at a potential difference of 120 V.

• Power Rating or Wattage is the power that the appliance will dissipate at a potential difference of 120 V (e.g. 100 W bulb, 1000 W space heater).

• Power consumption will differ if operated at any other voltage.

• Energy is often expressed as kilowatt-hours, for metering purposes.

kW-h

How much energy in a kilowatt-hour?

A)1000 Joules

B)0.28 Joules

C)3.6 million Joules

D)60,000 Joules

Resistors in Series• Series wiring means that the resistors are connected so that there is

the same current through each resistor.• The potential difference through each resistor is :

|∆VR |= IR• The equivalent resistor (Rs ) is made by adding up all resistors in

series in the circuit.∆Vbat = |∆Vcircuit |= IRs

20.6 Series Wiring

•By understanding that the potential difference across any circuit resistor will be a voltage DROP, while the potential difference back through the battery (charge escalator) will be a GAIN, one can replace the ΔV with “V” , and dispense with the absolute value signs.

V1 (voltage drop across the resistor1) = IR1

V2 (voltage drop across the resistor2) = IR2

Vbat = Vcir = IReq

20.6 Series Wiring

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)

00.9 00.3 00.6SR A 33.1 00.9

V 0.12

SR

VI

W6.10 00.6A 33.1 22 RIP

W31.5 00.3A 33.1 22 RIP

W9.15 W31.5 W6.10 P

Series ResistorsWhat is the value of R (3rd resistor in circuit)?

a. 50 Ω b. 25Ω c. 10 Ω d. 0

20.7 Parallel Wiring

Parallel wiring means that the devices areconnected in such a way that the same voltage is applied across each device.

When two resistors are connected in parallel, each receives current from the battery as if the other was not present.

Therefore the two resistors connected inparallel draw more current than does eitherresistor alone.


The two parallel pipe sections are equivalent to a single pipe of thesame length and same total cross sectional area.


parallel resistors

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1111

RRRRP

note that for 2 parallel resistors this equation is equal to :

Rp = (R1R2 ) / (R1 + R2)

Note that the potential difference of the battery is the same as the potential difference across each resistor in parallel.


Example 10 Main and Remote Stereo Speakers

Most receivers allow the user to connect to “remote” speakers in additionto the main speakers. At the instant represented in the picture, the voltageacross the speakers is 6.00 V. Determine (a) the equivalent resistanceof the two speakers, (b) the total current supplied by the receiver, (c) thecurrent in each speaker, and (d) the power dissipated in each speaker.


(a)

00.8

3

00.4

1

00.8

11

PR 67.2PR

(b) A 25.2 67.2

V 00.6rmsrms

PR

VI


A 750.0 00.8

V 00.6rmsrms

R

VI(c) A 50.1

00.4

V 00.6rmsrms

R

VI

(d) W50.4V 00.6A 750.0rmsrms VIP

W00.9V 00.6A 50.1rmsrms VIP

Circuits Wired Partially in Series and Partially in ParallelExample 12

Find a) the total current supplied by the the battery and b) the voltage dropbetween points A and B.

Circuits Wired Partially in Series and Partially in ParallelExample 12

a) Itot = V/Rp = 24V/240Ω = .10Ab) now go back to the 1st circuit in part c to

calculate the voltage drop across A-B:VAB = IRAB = (.10A)(130 Ω) = 13V

Your Understanding

• What is the ratio of the power supplied by the battery in parallel circuit A to the power supplied by the battery in series circuit B?

a.¼ c. 2 e. 1

b. 4 d. 2

A.

B.

Analyze the identical bulbsWhich bulbs will light

a.All bulbs

b.A, B

c.A only

d.none

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What happens to A?With switch closed, A

is______________ it was when switch was open.

a.Brighter than

b.Less bright than

c.Equal to what http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=4678&itemId=0470223553&resourceId=15300

closed




Complex Circuit

The current through the 8 Ω resistor is .590 A. Determine the following:

a.Current in the 20 Ω resistor

b.Current in the 9 Ω resistor

c.Battery voltage

Complex Circuit

The current through the 8 Ω resistor is .590 Determine the following:

a.Current in the 20 Ω resistor: .885 A

b.Current in the 9 Ω resistor: 2.49 A

c.Battery voltage: 22.4 V

20.11 The Measurement of Current and Voltage

An ammeter must be inserted intoa circuit so that the current passesdirectly through it.The resistance of the ammeter changes the current through the circuit.The ideal ammeter has a very low resistance

20.11 The Measurement of Current and Voltage

To measure the voltage between two points, in a circuit, a voltmeter is connected in parallel, between the points.A voltmeter takes some current away from the circuit it measures.The ideal voltmeter has a very large resistance so it diverts a negligible current.

• A battery consists of chemicals, called electrolytes, sandwiched in between 2 electrodes, or terminals, made of different metals.

• Chemical reactions do positive work and separate charge. The electric field does the same amount of negative work, which translates as a potential difference between positive and negative terminals

• A physical separator keeps the charge from going back through the battery.

The Battery

ε and ∆V• For an ideal battery potential

difference between the positive and negative terminals, ∆V, equals the chemical work done per unit charge.

∆V = Wch /q = ε• ε is the emf of the battery• Due to the internal resistance of

a real battery, ∆V is often slightly less than the emf.

• A capacitor can store charge, but has no way to do the work to separate the charge. It has a potential difference, but no ε.

How do we know there is a current?


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).

EOC #3A fax machine uses 0.110 A of current in its normal mode

of operation, but only 0.067 A in the standby mode. The machine uses a potential difference of 120 V.

(a)In one minute (60 s!), how much more charge passes through the machine in the normal mode than in the standby mode?

(b)How much more energy is used?

EOC #3A fax machine uses 0.110 A of current in its normal mode

of operation, but only 0.067 A in the standby mode. The machine uses a potential difference of 120 V.

(a)In one minute (60 s!), how much more charge passes through the machine in the normal mode than in the standby mode? .043 C/s x 60 s = 2.6 C

(b)How much more energy is used? 2.58 C x 120 J/C = 310 J


Conventional current is the hypothetical flow of positive charges (electron holes) that would have the same effect in the circuit as the movement of negative charges that actually does occur.

20.2 Ohm’s Law

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

To the extent that a wire or an electrical device offers resistance to electrical flow,it is called a resistor.

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 of0.40 A in the filament. Determine the resistance ofthe glowing filament.

5.7A 0.40

V 0.3

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Electric Circuits. Ohm’s Law The relationship between the potential difference in a circuit and...

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