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Electric Field – the space around a charge where electric forces can be detected. Field lines show the direction a positive charge would move in. The closer.

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Presentation on theme: "Electric Field – the space around a charge where electric forces can be detected. Field lines show the direction a positive charge would move in. The closer."— Presentation transcript:

1 Electric Field – the space around a charge where electric forces can be detected. Field lines show the direction a positive charge would move in. The closer the lines the stronger the force. link

2 Volts per division = 2 V Peak Voltage = 1.5 x 2 = 3 V Volts per division = 5 mV Volts per division = 20 V Peak voltage = 3 x 5 = 15 mV ( 0.015 V) Peak voltage = 1.5 x 20 = 30 V

3 Q = ItV = IR Series I 1 = I 2 = I 3 V T = V 1 + V 2 + V 3 R T = R 1 + R 2 + R 3 Parallel I = I + I 2 V 1 = V 2 = V 3 1 / R T = 1 / R 1 + 1 / R 2 Short cut if resistors are identical, divide by the number of resistors Electricity E = Pt P = IVP = I 2 R

4 1.What would happen if the LED was the other way around? 2.Why is there a resistor in the circuit? 3.What size of resistor would you need if the LED is to work correctly? LED maximum current 30 mA and forward voltage 3 V. 12 V

5 1 Energy = 2kJTime = 5 sPower = ? 2 Current = 2 AVoltage = 2VResistance = ? 3 Current = 15 ATime = 120 sCharge = ? 4 Voltage = 12 VCurrent = 0.5 AResistance 5 Charge = 13 CTime = 2 minutesCurrent = ? 6 Current = 25 mAPower = 1000WVoltage = ? 7 Power = 1000 WTime = 150 sEnergy = ? 10  8. Find the resistance (a) 10  (b) (c) X 4  Y 12  Y 10 Ω 20 Ω (d)

6 1.P = E ÷ t = 2000 ÷ 5 = 400 W 2. R = V ÷ I = 2 ÷ 2 = 1 Ω 3. Q = It = 15 x 120 = 1800 C 4. R = V ÷ I = 12 ÷ 0.5 = 24  5. I = Q ÷ t = 13 ÷ 120 = 0.11 A 6. V = P ÷ I = 1000 ÷ 0.025 = 40 000 V 7. E = Pt = 1000 x 150 = 150 000J 8. (a) R t = R 1 + R 2 = 10 + 10 = 20  (b) 10 ÷ 2 = 5  (c) 1/R T = 1/R 1 +1/R 2 = ¼ + 1/12 = 0.25 + 0.08333= 0.333 R T = 1/0.333 = 3  (d) 1/R T = 1/R 1 +1/R 2 = 1/20 + 1/10 = 0.05 + 0.1= 0.15 R T = 1/0.15 = 6.7  add to series part 6.7 + 20 = 26.7 Ω

7 12 V 3 1.5 3 5 1. What is this meter measuring? 2. What is the current going through the bulb? 3. If both resistors are identical, what current is going through each? 4. How would you measure the voltage across a bulb? 5. What would be the voltage across each resistor? 7

8 V  ~ A M

9 Activity 24 Light Controlled Switch to switch on street lights when it gets dark Block diagram Light sensorTransistor switchLED Block diagram Circuit diagram +5 V 0 V The variable resistor is used to set the brightness at which the light is switched on.

10 Explanation When light level, resistance LDR, voltage across LDR. The transistor conducts. The LED lights. +5 V 0 V


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