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KS4 Electricity – Resistance, Power and Energy © Boardworks Ltd 2003 Use a textbook or other resource to fill in the table below: Component Circuit symbol Diode Variable Resistor Thermistor Light Dependent Resistor © Boardworks Ltd 2003 How are current and voltage related for a resistor? Set up the circuit as shown below: Slowly move the variable resistor and for each setting record the current in amps and the voltage in volts. Plot a graph of your results. A V P.D. (V) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 I (A) 0 0.6 1.1 1.8 2.5 3.0 3.5 4.2 © Boardworks Ltd 2003 Current and voltage x x x x x x x x Potential difference / V Draw adoes line of best fit for Plot your points on the graph. If you get a your straight line it means What graph look like? your graph. that the two quantities current and voltage are proportional. What does proportional mean? If you double the voltage then the current doubles. This fact was put into a law: Ohm’s Law: The current flowing through a wire is proportional to the potential difference (voltage) across it provided the temperature remains constant. Georg Ohm 1826 © Boardworks Ltd 2003 Electron flow and Resistance Electricity in wires is a flow of electrons along the wire. As the electrons move along the wire they collide with the metal atoms. These collision make the atoms vibrate more…which makes the metal hotter. Resistance is a measure of how much a material tries to stop electricity passing through it. © Boardworks Ltd 2003 V=IR We can express Ohm’s Law mathematically using the equation: Voltage = Current x Resistance V=IR Voltage measured in Volts (V) Current measured in Amps (A) Resistance measured in Ohms () © Boardworks Ltd 2003 Formula triangles Formula triangles help you to rearrange formula, the triangle for the Ohm’s Law is shown below: Whatever quantity you are trying to find cover it up and it will leave you with the calculation required. So if you were trying to find current, I….. …you would cover I up… …and you are left with the sum… V I R I = V R x © Boardworks Ltd 2003 Formula triangles © Boardworks Ltd 2003 Resistance for a bulb If you have a filament bulb and it has a current of 20A running through it, with a potential difference of 100V across it, what is the resistance of the bulb? V = IR R = V/I R = 100V/20A R =5 © Boardworks Ltd 2003 Current for a diode A diode has a current of 5A running through it, and a resistance of 5. What is the potential difference across the diode? V = IR V = 5A x 5 V = 25V © Boardworks Ltd 2003 Resistors in series Total resistance = R1 + R2 4 2 What is the total resistance for the circuit shown? Total resistance = R1 + R2 Total resistance = 4 + 2 Total resistance = 6 © Boardworks Ltd 2003 Resistors in series What is the total resistance for the circuit shown? 6 34 Total resistance = R1 + R2 Total resistance = 6 + 34 Total resistance = 40 © Boardworks Ltd 2003 Resistors in parallel Total resistance = R1xR2 R1+ R2 4 2 What is the total resistance for the circuit shown? Total resistance = 4 x 2 4 + 2 = 1.33 © Boardworks Ltd 2003 Resistors in parallel What is the total resistance for the circuit shown? 8 6 Total resistance = 8 x 6 8 + 6 = 3.4 © Boardworks Ltd 2003 Voltage/current graphs I 1 I 2 V I 3 V V 4 V 1. A wire or resistor. 2. A filament lamp. 3. Wires of different materials. 4. A diode. ..partly x I Which of the components obeys Ohms Law? © Boardworks Ltd 2003 P=IV We can express the relationship between current, voltage and power mathematically using the equation: Power = Current x Voltage P=IV Voltage measured in Volts (V) Current measured in Amps (A) Power measured in Watts (W) © Boardworks Ltd 2003 Formula triangles Formula triangles help you to rearrange formula, the triangle for the Power Law is shown below: Whatever quantity you are trying to find cover it up and it will leave you with the calculation required. So if you were trying to find current, I….. …you would cover I up… …and you are left with the sum… P I V I = P V x © Boardworks Ltd 2003 Power calculations If you have a filament bulb and it has a potential difference of 200V across it and a current of 0.2A running through it. At what power is the bulb operating at? P = IV P = 0.2A x 200V P = 40W © Boardworks Ltd 2003 Power calculations If you have a filament bulb and it operates at a power of 60W and it has a potential difference of 240V across it, what is the current running through the bulb? P = IV I = P/V I = 60W / 240V I = 0.25A © Boardworks Ltd 2003 kV, kJ, kW 1 kV = 1000 V 1 kJ = 1000 J 1 kW = 1000 W How many Volts in 6kV? 6 000 V _________ How many Joules in 12.3kJ? 12 300 J _________ How many Watts in 0.6kW? 600 _________ W © Boardworks Ltd 2003 kV, kJ, kW 1 kV = 1000 V 1 kJ = 1000 J 1 kW = 1000 W How many kiloVolts in 9 000V? 9.0 _________ kV 23.5 How many kiloJoules in 23 500J? _________ kJ How many kiloWatts in 325W? 0.325 kW _________ © Boardworks Ltd 2003 What are the units of resistance? A. Amps B. Ohms C. Volts D. Watts © Boardworks Ltd 2003 What does the circuit symbol shown represent? A. Voltmeter B. Variable resistor C. Light dependent resistor D. Thermistor © Boardworks Ltd 2003 What causes resistance? A. Inertia B. Friction C. Buoyancy D. Viscosity © Boardworks Ltd 2003 If two resistors of 6 and 4 are placed in parallel, what is their total resistance in the circuit? A. 10 B. 2 C. 2.4 D. 24 © Boardworks Ltd 2003 If a resistor that obeys Ohm’s Law has a potential difference of 10V across it and a current of 5A running through it. What is its resistance? A. 50 B. 2 C. 0.5 D. 15 © Boardworks Ltd 2003