5.2 Electric Currents
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 source 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.
 Measured in Volts, symbol ε
 Called EMF
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.
 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
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
5.2.3 Apply the equations for resistors in series and in parallel.
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 ….
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 ….
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)
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
5.2.4 Draw circuit diagrams
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.
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.
Example
 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
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
Problem 14:
In the circuit diagram on the right, calculate:
1. The resistance between A and B
2. The resistance between B and C
3. The total circuit resistance
4. The total circuit current
5. The voltage across A and B
6. The current flowing through the 3Ω
7. The power dissipated by the 3Ω resistor
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
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