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Electric current g11

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A Presentation used to teach Electric Current to G11 students.

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Electric current g11

1. 1. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Electric Current Keith Warne Grade 11 & 12
2. 2. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Prior Knowledge Summary Ix = (ItR//)/Rx Resistors in series potential dividers Parallel resistors Current dividers
3. 3. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Electric Current in a Conductor • Conventional current - positive to negative • Direct current - moves in one direction. • Alternating current - changes direction continuously • Maintaining a current – Conductor – Potential difference – Replacement of charges + - + - <------- electrons “Positive spaces” ------>Conventional current is the movement of from + to - in a conductor. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + e- e- e-e- e- e- e-e- e- e- e-e- e- e- e-e- e-e- e-e-+ - e- e- e- <------- electrons e-
4. 4. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Ohm’s Law - Practical AIM: – Investigate the relationship between the potential difference across a resistor and the current flowing through it. – Determine the resistance of a resistor. V A METHOD: 1. Set up the circuit as shown. 2. Using the rheostat to vary the current in the circuit, obtain a range of readings for the potential difference across R for different currents. Resistance R rheostat
5. 5. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Ohm’s Law - Results The graph is a straight line showing that the current (I) is directly proportional to the voltage (V). I (A) V (V) 0.80 3.40 0.81 3.50 0.85 3.70 0.90 3.90 0.95 4.20 Current vs Voltage 3.00 3.20 3.40 3.60 3.80 4.00 4.20 4.40 0.75 0.80 0.85 0.90 0.95 1.00 Current (A) Voltage(V) Results Analysis
6. 6. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Worked Example A 4 Ω 2 Ω 8 Ω 12 Ω 6 Ω 12v Calculate I = V/R = 12/6 = 2 A
7. 7. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Energy & Power Work = ……… = ………= …….. = Energy Power = the rate at which work is done = ……………….. = ………. = …….. =……… From Joules experiment
8. 8. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Power Investigations Design plan and conduct a simple investigations into the power output of bulbs (or resistors) when connected in series and parallel. The questions you need to answer are: 1) How is the power output of a bulb (or resistor) affected when other similar bulbs are connected to it; a) In series b) In parallel 2) How is the potential difference across the power supply (battery) affected when the resistance in the circuit is changed. CRITERIA You may use circuit boards and croc clips simulations to gather your data. (at least one investigation must be done using a circuit board) You must take measurements of current and voltage and prove your conclusions with calculations and observations of bulb brightness. Circuit diagrams must be drawn and method explained as well as any relevant graphs drawn. Hand in: Plan, Aim, Hyp, Method, (Incl Diagrams), Results (Table), Analysis (Calculations), Conclusions
9. 9. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Series Power • Increasing the resistance in the circuit decreases the current. • Lower current flowing through each bulb results in a lower voltage drop and therefore less power is dissipated. • . (No internal resistance) The voltage drop across the whole circuit remains constant
10. 10. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net A non ohmic resistor (bulb) under varying voltage POWER (W) Volts (V) CURREN T (A) RESISTANCE (W) POWER (W) BULBS R = V/I P=V.I 6 12 0.5 24.0 6.0 6 10 0.453 22.1 4.5 6 8 0.402 19.9 3.2 6 6 0.343 17.5 2.1 6 4 0.274 14.6 1.1 Non Ohmic Conductor/Resistance
11. 11. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net P10 = VI = (3.41)(0.418) = 1.4 W P6 = VI = (8.59)(0.418) = 3.6 W Series Bulb Brightness The higher resistance of the 6 W bulb will limit the current available so the 6 W bulb will deliver more power than the 10 W bulb when in series. Higher resistance gives more power series! Since current equal in series the one with the highest voltage (highest R) has most power. R10w = V/I = 3.41/0.418 = 8.2W R6w = V/I = 8.59/0.418 = 20.6 W
12. 12. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Electricity Questions - SG 4.1 The battery has negligible internal resistance. Calculate: 4.1.1 The total resistance of the circuit. (4) 4.1.2 The current in the 2.4 W resistor. (3) 4.1.3 the potential difference across the 2.4 W resistor. (3) 4.1.4 the potential difference across the parallel combination. (2) A 12V S 2.4W X X X 2W 8W
13. 13. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Vex A Resistance R Vemf Emf 400V I = 0A Vex= 0 v Emf = ………….. • Emf is the ………………… amount of energy that the cell can produce (per unit charge). • Measured when the current in the circuit is ………….. EMF - Electro Motive Force Open Circuit!!
14. 14. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Infinite R No current voltage = EMF Large R Small current voltage < EMF Small R BIG current voltage << EMF Lost volts increase! Lost volts small amount! DECREASING RESISTANCE in the circuit DECREASES VOLTAGE drop in the circuit! Open circuit
15. 15. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Exam Questions QUESTION 1 One word or phrase answers 1.1The law which relates the current in a resistor, maintained at constant temperature, to the potential difference across its ends. (1) 1.2The unit of measure equivalent to one volt per ampere. (1) Reading on V1 Reading on V2 A 12 V 0 V B 12 V 12 V C 0 V 0 V D 0 V 12 V QUESTION 2 Multiple Choice 2.1 A variable resistor, an ammeter, a battery of emf 12 V and voltmeters V1 and V2 are connected as shown in the diagram below. When the switch is open, the readings on voltmeters V1 and V2 respectively are …
16. 16. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Electricity Questions The battery in the circuit diagram below has an EMF of 12 V and an unknown internal resistance r. Voltmeter V1 is connected across the battery and voltmeter V2 is connected across the switch S. The resistance of the connecting wires and the ammeter is negligible. 1 Write down the respective readings on voltmeters V1 and V2 when switch S is open. (2) Switch S is now closed. The reading on voltmeter V1 changes to 9 V. 2. What will the new reading on V2 be?(1) 3. Calculate the total external resistance of the circuit. (4) 4. Calculate the internal resistance, r, of the battery. (5) ANSWERS >>
17. 17. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Hi - This is a SAMPLE presentation only. My FULL presentations, which contain a lot more more slides and other resources, are freely available on my resource sharing website: www.warnescience.net (click on link or logo) Have a look and enjoy! WarneScience
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