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Information about Electric current g11

A Presentation used to teach Electric Current to G11 students.

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Current Carrying Conductors Current in same direction. Two parallel current carrying conductors with current flowing in the same direction attract each other by the reduced magnetic flux density between the conductors. SAMPLE ONLY X Magnetic flux cancels out Conductors attract! SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net

Current and Charge The Ampere: The current in two parallel conductors, of infinite length, one meter apart, in a vacuum which will produce a force of 2 x 10 -7 N per meter of length. 1A 2x10-7N 1 meter 1A The Coulomb: Quantity of charge passing any cross section of a conductor in one second when a current of 1A is flowing. SAMPLE ONLY SAMPLE ONLY 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. rheostat – Determine the resistance of a resistor. METHOD: 1. Set up the circuit as shown. A 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. SAMPLE ONLY SAMPLE ONLY V Resistance R SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net

Ohm’s Law - Results Analysis Results V (V) 3.40 3.50 3.70 3.90 4.20 4.40 Voltage (V) I (A) 0.80 0.81 0.85 0.90 0.95 Current vs Voltage 4.20 4.00 3.80 3.60 3.40 3.20 3.00 0.75 0.80 0.85 0.90 Current (A) 0.95 The graph is a straight line showing that the current (I) is directly proportional to the voltage (V). SAMPLE ONLY SAMPLE ONLY 1.00 SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net

EMF - LOST VOLTS 350V Vex EMF = 400V Internal circuit (50) (Vint =50V) lost due to internal resistance Ri in moving across cell Closed Circuit!! A I = IA (400) Lost Volts Vex= 350v Vex 350V (50) Emf = Vex + “lost volts” = IRex+ IRi =I(Rex + ri ) Resistance R Emf SAMPLE ONLY = Vext +V…… ONLY SAMPLE 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 SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net

Parallel Bulb Brightness A 10 W bulb has a LOWER resistance than a 6 W bulb and so draws a higher current and delivers more power UNDER THE SAME VOLTAGE. R10w = V/I = 12/0.833 = 14.4 P6 = VI = (12)(0.500) =6W R6w = V/I = 12/0.5 = 24 P10 = VI = (12)(0.833) = 10 W LOWER resistance gives more power inSAMPLE ONLY parallel! SAMPLE ONLY SAMPLE ONLY Since voltages in parallel are equal the bulb with the highest current (lowest R) For FULL presentation click HERE >> www.warnescience.net will have most power.

Electricity Questions 6V 6.1 The battery has an internal resistance of 1. L1 & L2 each have a resistance of 3 . What will the reading on V be when S is closed? (7) 6.2 An electric motor is rated at 2000W; 250V. Fuses rated at 5A, 10A, & 25A are available. 6.2.1 Which fuse would be most suitable to protect the motor? (5) 6.2.2 Calculate the cost of operating the electric motor for 24 hours if one kilowatt-hour of energy costs 34 cents? (3) SAMPLE ONLY SAMPLE ONLY A V L2 X L1 X S SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net

WarneScience Hi… This is a SAMPLE presentation only. My FULL presentations, which contain loads more slides and other resources, are freely available on my resource sharing website: www.warnescience.net (paste into your browser if link above does not work) Have a look and enjoy! SAMPLE ONLY Keith Warne SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net

A Presentation used to teach Electric Current to G11 students. ... Electric current g11 May 06, 2015 Education keith-warne. System is processing data

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A Presentation used to teach Electric Current to G11 students. A Presentation used to teach Electric Current to G11 students. Docslide.us.

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Current Electricity. Previews available by clicking on item - some formatting and symbols do not show correctly in previews. ... Electric Current G11

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AF-300 G11 ™ User’s Guide GE ... Always connect a ground wire, as electric shock or fire may result. ... • Where a residual-current protective device ...

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IB DP G11 / G12 Dulwich 2008~2010 IB Physics IGCSE Physics ... Electricity How an electric current may exist within a solid material such as a metal wire?

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The PRR GG1 was a class of electric locomotives built for the ... New Haven Railroad by the PRR to compare it to its current standard electric ...

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every time the electric current was switched on, the compass needle moved. This shows that current carrying wire produce an electric current.

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Fuji Electric contribute to the resolution of energy management problems, through pursuit of technological innovation in electric and thermal energy.

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Simple Electric Motor Masco G11 by olivianieves in electronics. ... A solenoid is coil of wire that acts as a magnet while carrying an electric current.

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