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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.

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Presentation on theme: "Electric Currents Topic 5 These notes were typed in association with Physics for use with the IB Diploma Programme by Michael Dickinson."— Presentation transcript:

1 Electric Currents Topic 5 These notes were typed in association with Physics for use with the IB Diploma Programme by Michael Dickinson

2 Electric Circuits 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

3 Electric Circuits 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 = V term + V lost  ε = IR + Ir  1.5V = 1.3V + 0.2V  All electrical sources, power supplies, generators etc. have internal resistance

4 Electric Circuits 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

5 Electric Circuits 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.  V total = V 1 + V 2 + V 3 …. since V = IR  IR total = IR 1 + IR 2 + IR 3 ….  IR total = I(R 1 + R 2 + R 3 ….)  R total = R 1 + R 2 + R 3 ….

6 Electric Circuits 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.  I total = I 1 + I 2 + I 3 …. since I = V/R  V/R total = V/R 1 + V/R 2 + V/R 3 ….  V/R total = V(1/R 1 + 1/R 2 + 1/R 3 ….)  1/R total = 1/R 1 + 1/R 2 + 1/R 3 ….

7 Electric Circuits Apply the equations for resistors in series and in parallel. IB Physics Formula: Resistors in series  R total = R 1 + R 2 + R 3 …. Resistors in parallel  1/R total = 1/R 1 + 1/R 2 + 1/R 3 ….  Non-IB formula worth remembering Product over sum - parallel  R total = (R 1 x R 2 )/(R 1 + R 2 )

8 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

9 Electric Circuits Draw circuit diagrams

10 Electric Circuits 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.

11 Electric Circuits 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.

12 Electric Circuits Describe a potential divider

13 Electric Circuits 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

14 Electric Circuits Solve problems involving electric circuits Problem 13  What is the potential difference across the filament bulb in the below drawing.  Answer: 3.0V

15 Electric Circuits 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 10/30 th going to the lamp and resistor on the left and 20/30 th going to the resistor on the right.  This means that the lamp will receive 1/3 of the available voltage = 3.0V

16 Electric Circuits Solve problems involving electric circuits Problem 14: Answers 1.R total = 2Ω 2.R total 2Ω 3.R total = 8Ω 4.I = 1.5A 5.V = 3V 6.I = 1.0A 7.P = 3.0W

17 Electric Circuits Solve problems involving electric circuits Problem 14: Formulas 1.R total = (R 1 x R 2 )/(R 1 + R 2 ) 2.1/R total = 1/R 1 + 1/R 2 + 1/R 3 3.R total = R 1 + R 2 + R 3 4.I = V/R 5.V = IR 6.I = V/R 7.P = I 2 R


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