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Electric Circuits In the electric circuit shown below, energy is transferred from the battery to the light bulb by charges

that move through a conducting wire because of a potential difference set up in the wire by the battery.

Schematic

mark potentials at each spot

The circuit shown contains a typical 9-volt battery.

a) What is the emf of the circuit?

b) How much energy does one coulomb of charge carry around the circuit?

c) How much energy do two coulombs of charge carry around the circuit?

d) How much energy does each coulomb of charge have at point B?

e) How much energy does each coulomb of charge have at point C?

f) What is VB? What is VC?

g) What is ∆VBC? What is ∆VCD? What is ∆VDA?

Official Definition of One Ampere (1 A) of current – a fundamental unit

One ampere is the amount of current flowing in each of two infinitely-long parallel wires of negligible cross-sectional area

separated by a distance of one meter in a vacuum that results in a force of exactly 2 x 10-7

N per meter of length of each wire.

Short form – Current is defined in terms of the force per unit length between parallel current-carrying conductors.

Formula: Units: Type:

I = ∆q/∆t A (ampere) = C/s Scalar

Unofficial definition: rate of flow of electric charge

Electric Current

Closed circuit: complete pathway

for current

Open circuit: incomplete pathway for current –

break in circuit – infinite resistance

sketch

Short circuit: circuit with little to no

resistance – extremely high current –

overheating

sketch

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Resistance

Resistance: ratio of potential difference applied

across a piece of material to the current through the

material

Formula: Units: Type:

R = V / I ohm (Ω) = V/A scalar

For a wire conductor:

A short fat cold wire is the best conductor

A long hot skinny wire has the most resistance

Formula:

R = ρL/A

Power: energy per unit time Unit: W = J/s Type: scalar

Mechanical Power:

P = W/t = F s/t = F v

Electrical Power:

P = E/t = (∆q V)/∆t

P = I V

Alternate Formulas:

Substitute V = IR

P = I (IR) = I2 R

P = (V/R)V = V2/ R

Meters in a circuit

Schematic diagram

Ammeter: measures current

Placement: Must be placed in series to allow

current to flow through it

Circuit must be broken to insert ammeter

Ideal ammeter: has zero resistance so it will not

affect current flowing through it

Voltmeter: measures potential difference

Placement: Must be placed in parallel to measure

potential difference between two points

circuit does not to be broken

Ideal voltmeter: has infinite resistance so it will not

allow any current to flow through it and disrupt circuit

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Series and Parallel Circuits

Characteristic Series Circuit Parallel Circuit

Number of pathways

for current one More than one

Current Same everywhere – same for all

devices Current splits – shared among devices

Potential Difference

(Voltage)

Voltage shared among devices –

voltage splits Same for all devices

Overall resistance high low

Power low high

VT = V1 + V2 + …

IT = I1 = I2 = …

RT = R1 + R2 + …

PT = P1 + P2 + . . .

VT = V1 = V2 = . . .

IT = I1 + I2 + . . .

1/RT = 1/R1 + 1/R2 + . . .

PT = P1 + P2 + . . .

Formulas:

Voltage Ratio Current Ratio

Power Ratio Power Ratio

Series Circuits Parallel Circuits

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Analyzing Circuits

Determine the current through and the voltage drop across each resistor in each circuit below.

1. 2.

3. 4.

Potential Divider: Resistors in series act as a “potential (voltage) divider.” They split the potential of the source between them.

5. A 20Ω device requires 40 V to operate properly but no 40 V source is available. In each case below, determine

the value of added resistor R1 that will reduce the voltage of the source to the necessary 40V for device R2.

(A) (B) (C) (D)

6. A mini light bulb is rated for 0.60 W at 200 mA and is placed in series with a variable resistor. Only a 9.0 volt battery is

available to power it. To what value should the variable resistor be set to power the bulb correctly?

Bulb needs only 3 V

Bulb has resistance of 15 Ω at rated

power

Added resistance should be 30

ohms

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The Use of Sensors in Circuits

1. Light-Dependent Resistor (LDR) or Light Sensor: A photo-conductive cell made of

semiconducting material whose resistance decreases as the intensity of the incident light increases.

2. Negative Temperature Coefficient (NTC) Thermistor or Temperature Sensor: A sensor

made of semiconducting material whose resistance decreases as its temperature increases.

3. Strain Gauge or Force Sensor: A long thin metal wire whose resistance increases as it is

stretched since it becomes longer and thinner.

Automatic light switch Describe how the LDR activates the light switch.

Describe how the NTC thermistor activates the fire alarm. Fire alarm

Describe how the strain gauge can measure the strain put on a

section of an airplane body.

As ambient light decreases, resistance of LDR increases

Potential difference across LDR increases

Switch needs minimum PD to turn on

When light intensity drops to desired level, PD is high

enough to turn on switch

As external temperature increases, resistance of NTC

decreases

Potential difference across R2 increases

Switch needs minimum PD to turn on

When temperature increases to desired level, PD is high

enough to turn on switch

As strain increases, resistance of strain gauge increases

Potential difference across R2 decreases

Voltmeter can read change in voltage which can be used

to determine amount of strain on part

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Combination Series-Parallel Circuits

2. The battery has an emf of 12 V and negligible internal resistance and the voltmeter has an internal resistance of 20 kΩ.

Determine the reading on the voltmeter.

1. Determine the current through and the voltage drop across each resistor.

3. A cell with negligible internal resistance is connected to three resistors as shown. Compare the currents in each

part of the circuit.

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5. A battery with emf E and negligible internal resistance is connected in a circuit with three identical light bulbs.

a) Determine the reading on the voltmeter when the switch is open and when it is closed.

b) State what effect closing the switch has on the current through each bulb and the brightness of each bulb.

4. Determine the current through and the voltage drop across each resistor.

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Ohm’s Law

Ohm’s Law: for a conductor at constant temperature, the current flowing

through it is proportional to the potential difference across it

Resistance: ratio of potential difference applied across a piece of material

to the current through the material

Formula:

R = V / I

Relationship:

V α I

Example: resistor

Ohmic Device: a device that obeys Ohm’s law for a wide range of potential differences

2. A resistor is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it.

b) If the voltage is doubled, what is the new current?

V = IR for resistor

Resistance is constant so double current

R = V/I

R = 7.5 Ω

a) Calculate the resistance of the resistor.

1. On the axes at right, sketch the I-V characteristics for a resistor.

Resistance:

a) R = V/I at any point

b) related to slope of graph

(Reciprocal = resistance)

Meaning: a device with constant resistance

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Non-Ohmic Device: a device that does not obey Ohm’s law

Example: filament lamp

1. On the axes at right, sketch the I-V characteristics for a filament lamp.

Meaning: resistance is not constant

Resistance:

a) R = V/I at any point

b) as current increases, wire filament heats up and

resistance increases

c) Resistance is NOT related to the slope

d) except in initial linear region

2. A flashlight bulb is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it.

b) If the voltage is doubled, what is the new current?

V = IR for bulb but resistance is not 7.5 ohms any

more – R increases with T so less than double current

R = V/I

R = 7.5 Ω

a) Calculate the resistance of the bulb.

X: resistance increases - ratio V/I increases

Y: resistance is constant – ratio V/I is

constant

Z: resistance decreases – ratio V/I

decreases

3. Discuss how the resistance varies with increasing potential difference for devices X, Y, and Z.

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Using a Potentiometer to Measure I-V Characteristics

Potentiometer: a type of variable resistor with three contact points

Common use: as a potential divider to measure the I-V characteristics of a device

The schematic shows how a potentiometer can be used as a potential divider to measure the I-V characteristics of a

filament lamp. It is placed in parallel with the lamp and the slider (center contact point) effectively splits the

potentiometer into two separate resistors AB and BC. By moving the slider, the ratio of the voltage drops across the

resistors AB and BC is varied.

Redraw the schematic with an ammeter and a voltmeter correctly

placed to measure the I-V characteristics of the filament lamp.

Comment on the circuit characteristics as the slider is moved from A to B to C.

Slider at A:

Slider at B:

Slider at C:

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Internal Resistance of Batteries

Electromotive force (emf): total energy per unit charge supplied by the battery Symbol: ε or E

Units: V = J/C

A 6 volt battery is connected to a variable resistor and the current in the circuit and potential difference

across the terminals of the battery are measured over a wide range of values of the resistor. The results

are shown in the table.

Resistance (Ω) Predicted

Current (A)

Actual

Current (A)

Voltage across

battery (V)

2000 0.003 0.003 6.00

200 0.03 0.03 5.99

20 0.3 0.29 5.85

2 3 2.4 4.80

0.2 30 8.8 1.71

0.02 300 11.5 0.23

0.002 3000 12.0 0.02

0.0002 30000 12.0 0.00

Why does the current seem to be limited to a maximum of 12.0 amperes and why does the voltage

across the battery not remain constant at 6.0 volts?

The battery has some internal resistance. As the

external resistance decreases, more and more of the

energy supplied by the battery is used up inside the

battery.

Terminal Voltage (Vterm): potential difference across the terminals of the battery

Ideal Behavior: Vterm always equals emf since no internal resistance

Real Behavior:

1) Think of battery as internal E and tiny internal resistor r

2) Vterm only equals the emf when no current is flowing

3) E is split between R and r

4) When R>>r, Vterm ≈ emf

5) As R decreases, Vr increases and VR decreases

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Relationship between emf and terminal voltage

Treat internal resistance as a series resistor

ε = I RT

ε = I (R + r)

ε = IR + Ir

Note that in the absence of internal resistance, ε = Vterm

1. A resistor is connected to a 12 V source and a switch. With the switch open, a voltmeter reads the potential difference

across the battery as 12 V yet with the switch closed, the voltmeter reads only 9.6 V and an ammeter reads 0.40 A for the

current through the resistor. Sketch an appropriate circuit diagram and calculate the internal resistance of the source.

2. Discuss the expected I-V characteristics for this battery and how they can be

experimentally determined.

R can be adjusted from 0 to its max value

A graph of Vterm vs. I can be drawn

Specific equation of graph can be compared to math model to derive

internal resistance

Emf = Vterm + Ir

Vterm = -Ir + emf so slope = -r and y-intercept = emf