Ks4 Electricty Resistance, Power And Energy

KS4 Electricity – Resistance

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Use a textbook or other resource to fill in the table below: Variable Resistor Thermistor Light Dependent Resistor Diode Circuit symbol Component

How are current and voltage related for a resistor? V A 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. 0 0.6 1.1 1.8 2.5 3.0 3.5 4.2 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0 I (A) P.D. (V)

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 Current, I / amps Potential difference / V What does proportional mean? Current and voltage If you double the voltage then the current doubles. x x x x x x x x Plot your points on the graph. Draw a line of best fit for your graph. What does your graph look like? If you get a straight line it means that the two quantities current and voltage are proportional. This fact was put into a law:

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.

V=IR We can express Ohm’s Law mathematically using the equation: Voltage measured in Volts (V) Current measured in Amps (A) Resistance measured in Ohms (  ) Voltage = Current x Resistance V=IR

Formula triangles V I  R   x 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… I = V R

Formula triangles

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 

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

Resistors in series 4  2  Total resistance = R 1 + R 2 What is the total resistance for the circuit shown? Total resistance = R 1 + R 2 Total resistance = 4  + 2  Total resistance = 6 

6  34  What is the total resistance for the circuit shown? Total resistance = R 1 + R 2 Total resistance = 6  + 34  Total resistance = 40  Resistors in series

Resistors in parallel 2  4  Total resistance = R1xR2 R1+ R2 What is the total resistance for the circuit shown? Total resistance = 4  x 2  4  + 2  = 1.33 

Resistors in parallel 6  8  What is the total resistance for the circuit shown? Total resistance = 8  x 6  8  + 6  = 3.4 

Voltage/current graphs 1. A wire or resistor. 2. A filament lamp. 3. Wires of different materials. 4. A diode. Which of the components obeys Ohms Law? I V I V I V I V 1 2 3 4    ..partly x

P=IV We can express the relationship between current, voltage and power mathematically using the equation: Voltage measured in Volts (V) Current measured in Amps (A) Power measured in Watts (W) Power = Current x Voltage P=IV

Formula triangles P I  V   x 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… I = P V

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 200 V P = 40W

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

kV, kJ, kW 1 kV = 1000 V 1 kJ = 1000 J 1 kW = 1000 W How many Volts in 6kV? _________ V How many Joules in 12.3kJ? _________ J How many Watts in 0.6kW? _________ W 6 000 12 300 600

kV, kJ, kW 1 kV = 1000 V 1 kJ = 1000 J 1 kW = 1000 W How many kiloVolts in 9 000V? _________ kV How many kiloJoules in 23 500J? _________ kJ How many kiloWatts in 325W? _________ kW 9.0 23.5 0.325