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Electric Currents
Topic 5
These notes were typed in association with Physics for use
with the IB Diploma Programme by Michael Dickinson
Electric Circuits
5.2.1 Define electromotive force (emf)
IB Definition: electromotive Force(emf) – electromotive force is
defined as the potential difference across the terminals of a
source of electricity when there is no current flowing(open circuit
p.d.)
 Example: Connect a voltmeter across a sourse of electricity(a AA
battery). It will show the available potential difference(voltage).
For the AA battery it will read 1.5V.
 Now add a the battery to a circuit, maybe a light bulb and the
voltmeter reads slightly lover, possible 1.3V
 The term “force” is a misnomer.
 Measuered in Volts, symbol ε
 Called EMF
Electric Circuits
5.2.2 Describe the concept of internal resistance.
 So what happened to these “lost volts”?
 If the circuit is allowed to run for some time, the batter will heat
up.
 This loss of energy in the battery is due to the “INTERNAL
RESISTANCE” within the battery itself.
***Draw Internal resistance circuit***
 Battery has internal resistance, r, and bulb with resistance, R,
 emf = terminal voltage + “lost volts”
 ε = Vterm + Vlost
 ε = IR + Ir
 1.5V = 1.3V + 0.2V
 All electrical sources, power supplies, generators etc. have internal
resistance
Electric Circuits
5.2.2 Describe the concept of internal resistance.
IB Formula: ε = I(R + r)
Practice 9
 The open circuit terminal voltage of a battery is measured to be
9.0V. When connected to a light bulb, the voltage is seen to drop to
8.6V. If the current flowing in the circuit is measured at 0.02A,
calculate the internal resistance of the battery.
 Answer: 20Ω
Electric Circuits
5.2.3 Apply the equations for resistors in series and in parallel.
***See Diagram on board***
Series Circuits
 Current flowing is the same at all points
 Potential difference across the terminals of the source of emf is = to
the sum of the individual components in the external circuit.
 Vtotal = V1 + V2 + V3 ….
since V = IR
 IRtotal = IR1 + IR2 + IR3 ….
 IRtotal = I(R1 + R2 + R3….)
 Rtotal = R1 + R2 + R3 ….
Electric Circuits
5.2.3 Apply the equations for resistors in series and in parallel.
***See Diagram on board***
Parallel Circuits
 The potential difference(voltage) across each branch of the circuit
is equal
 The total current entering a junction is equal to the total current
leaving the junction.
 Itotal = I1 + I2 + I3 ….
since I = V/R
 V/Rtotal = V/R1 + V/R2 + V/R3 ….
 V/Rtotal = V(1/R1 + 1/R2 + 1/R3….)
 1/Rtotal = 1/R1 + 1/R2 + 1/R3 ….
Electric Circuits
5.2.3 Apply the equations for resistors in series and in parallel.
IB Physics Formula:
Resistors in series
 Rtotal = R1 + R2 + R3 ….
Resistors in parallel
 1/Rtotal = 1/R1 + 1/R2 + 1/R3 ….
 Non-IB formula worth remembering
Product over sum - parallel
 Rtotal = (R1 x R2)/(R1 + R2)
Electric Circuits
Problem 10
 Three resistors are connected in series. Their values are 2Ω, 4Ω and
6Ω. Calculate their combined resistance.
 Answer: 12Ω
Problem 11
 Three resistors are connected in parallel. There values are 20Ω,
30Ω and 40Ω. Calculate their combined resistance.
 Answer: 9.23Ω
Problem 12
 Two resistors are combined in parallel. There values are 6Ω and
3Ω. Calculate their combined resistance.
 Answer: 2Ω
Electric Circuits
5.2.4 Draw circuit diagrams
Electric Circuits
5.2.5 Describe the use of ideal ammeters and ideal voltmeters.
Ammeter
 Measures amount of current flowing in a circuit.
 Connected IN SERIES with the components in the circuit.
 Should have low resistance.
Voltmeter
 Measures amount of potential difference across an electrical
component.
 Connected IN PARALLEL with the components in the circuit
 Should have high resistance.
Electric Circuits
5.2.6 Describe a potential divider
 Resistors in series split the voltage amongst the components in the
circuit.
 Example: Two identical light bulbs with resistance, R, are
connected to a 9.0V battery. The available voltage would be
divided so each filament received 4.5V. If there were three then
each bulb would receive 3V.
 If they have different resistance then the available voltage is
divided up proportionally among the components.
Electric Circuits
5.2.6 Describe a potential divider
Electric Circuits
5.2.6 Describe a potential divider
 Total available potential difference = 9.0V
 Total resistance = R + 2R + 3R = 6R
 The resistance of the left hand resistor is 1/6 fo the total resistance,
so it receives 1/6 of the available potential difference.(9÷6 = 1.5V)
 If you sum the individual voltages it equals the total available
voltage
Electric Circuits
5.2.8 Solve problems involving electric circuits
Problem 13
 What is the potential difference across the filament bulb in the
below drawing.
 Answer: 3.0V
Electric Circuits
5.2.8 Solve problems involving electric circuits
Solution
 The 20Ω lamp is in parallel with a 20Ω resistor. This gives a
combined resistance of 10Ω. Together with the other 20Ω
resistor(in series), the total circuit resistance is 30Ω.
 The voltage is split with 10/30th going to the lamp and resistor on
the left and 20/30th going to the resistor on the right.
 This means that the lamp will receive 1/3 of the available voltage =
3.0V
Electric Circuits
5.2.8 Solve problems involving electric circuits
Problem 14:
Answers
1. Rtotal = 2Ω
2. Rtotal 2Ω
3. Rtotal = 8Ω
4. I = 1.5A
5. V = 3V
6. I = 1.0A
7. P = 3.0W
Electric Circuits
5.2.8 Solve problems involving electric circuits
Problem 14:
Formulas
1. Rtotal = (R1 x R2)/(R1 + R2)
2. 1/Rtotal = 1/R1 + 1/R2 + 1/R3
3. Rtotal = R1 + R2 + R3
4. I = V/R
5. V = IR
6. I = V/R
7. P = I2 R