Electric Circuits
IB 12
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.
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?
Schematic
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?
mark potentials at each spot
g) What is ΔVBC?
What is ΔVCD?
What is ΔVDA?
Electric Current
Formula:
I = Δq/Δt
Units:
A (ampere) = C/s
Type:
Scalar
Unofficial definition: rate of flow of electric charge
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.
Closed circuit: complete pathway
for current
Open circuit: incomplete pathway for current –
break in circuit – infinite resistance
Short circuit: circuit with little to no
resistance – extremely high current –
overheating
sketch
sketch
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Resistance
IB 12
Resistance: ratio of potential difference applied
across a piece of material to the current through the
material
Formula:
Units:
For a wire conductor:
Formula:
A short fat cold wire is the best conductor
R = ρL/A
R=V/I
Type:
ohm (Ω) = V/A
scalar
A long hot skinny wire has the most resistance
Unit: W = J/s
Power: energy per unit time
Mechanical Power:
Electrical Power:
P = E/t = (Δq V)/Δt
Alternate Formulas:
Substitute V = IR
P=IV
P = I (IR) = I2 R
P = W/t = F s/t = F v
Type: scalar
P = (V/R)V = V2/ R
Meters in a circuit
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
Schematic diagram
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
IB 12
Series Circuit
Parallel Circuit
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
Number of pathways
for current
Formulas:
VT = V1 + V2 + …
VT = V1 = V2 = . . .
IT = I1 = I2 = …
IT = I1 + I2 + . . .
RT = R1 + R2 + …
1/RT = 1/R1 + 1/R2 + . . .
PT = P1 + P2 + . . .
PT = P1 + P2 + . . .
Series Circuits
Parallel Circuits
Voltage Ratio
Current Ratio
Power Ratio
Power Ratio
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Analyzing Circuits
IB 12
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
IB 12
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.
Describe how the LDR activates the light switch.
Automatic light switch
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
2. Negative Temperature Coefficient (NTC) Thermistor or Temperature Sensor: A sensor
made of semiconducting material whose resistance decreases as its temperature increases.
Describe how the NTC thermistor activates the fire alarm.
Fire alarm
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
3. Strain Gauge or Force Sensor: A long thin metal wire whose resistance increases as it is
stretched since it becomes longer and thinner.
Describe how the strain gauge can measure the strain put on a
section of an airplane body.
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
IB 12
1. Determine the current through and the voltage drop across each resistor.
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.
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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IB 12
4. Determine the current through and the voltage drop across each resistor.
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.
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Ohm’s Law
IB 12
Resistance: ratio of potential difference applied across a piece of material
to the current through the material
Ohm’s Law: for a conductor at constant temperature, the current flowing
through it is proportional to the potential difference across it
Ohmic Device:
Formula:
R=V/I
Relationship:
VαI
a device that obeys Ohm’s law for a wide range of potential differences
Meaning: a device with constant resistance
Example: 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)
2. A resistor is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it.
a) Calculate the resistance of the resistor.
R = V/I
R = 7.5 Ω
b) If the voltage is doubled, what is the new current?
V = IR for resistor
Resistance is constant so double current
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Non-Ohmic Device: a device that does not obey Ohm’s law
Meaning: resistance is not constant
IB 12
Example: filament lamp
1. On the axes at right, sketch the I-V characteristics for a filament lamp.
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.
a) Calculate the resistance of the bulb.
R = V/I
R = 7.5 Ω
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
3. Discuss how the resistance varies with increasing potential difference for devices X, Y, and Z.
X: resistance increases - ratio V/I increases
Y: resistance is constant – ratio V/I is
constant
Z: resistance decreases – ratio V/I
decreases
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Using a Potentiometer to Measure I-V Characteristics
IB 12
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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IB 12
Internal Resistance of Batteries
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
200
20
2
0.2
0.02
0.002
0.0002
0.003
0.03
0.3
3
30
300
3000
30000
0.003
0.03
0.29
2.4
8.8
11.5
12.0
12.0
6.00
5.99
5.85
4.80
1.71
0.23
0.02
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.
Electromotive force (emf): total energy per unit charge supplied by the battery
Symbol: ε or E
Units: V = J/C
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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IB 12
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
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