Do now: Paraire, 5 Hui-tanguru 2016. Week 5 + Mon Week 6 Year 13 Physics: Friday 22 nd August 8:45am in the PAC 2 hours: Waves and Mechanics ** Start.

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Presentation transcript:

Do now: Paraire, 5 Hui-tanguru 2016

Week 5 + Mon Week 6 Year 13 Physics: Friday 22 nd August 8:45am in the PAC 2 hours: Waves and Mechanics ** Start revising now ** Plan your time: - review your topic tests - where are your gaps? - use lunchtime tutorials Revision material (old exams): - will be provided around Week 3 Paraire, 5 Hui-tanguru 2016 School Exams

Do now: A 1.5V battery is short-circuited with a piece of copper wire of resistance 1mΩ. 1) Use Ohm’s Law to calculate the theoretical current in the wire 2) In practice, the current in the wire is 3A. Why? Paraire, 5 Hui-tanguru 2016 Internal resistance: r = V/I = 1.5 / 3 = 0.5 Ω I = V/R = 1.5 / (1 x ) = 1,500 A (!!!)

So what does it mean? When you buy a brand new 1.5V cell, the 1.5V has actually been measured when no current is flowing. We call this value the “EMF”. As soon as you start drawing some current, the voltage available to the circuit will be slightly less than this because the battery has some internal resistance.

Internal Resistance A V EMF r I (A) V (V) r EMF voltage when no current is flowing gradient Sketch a graph of V (on the y-axis) versus I (on the x-axis)

The EMF of a cell doesn’t change. So what happens when a battery goes ‘flat’? The internal resistance increases, due to changes in the chemicals of the battery.

Example: An old torch battery is tested with a voltmeter. Voltmeters draw negligible current. The terminal voltage when the voltmeter is connected reads 1.5V. When the battery is connected to a lamp, the battery fails to make it glow. When the battery is connected to the lamp, its terminal voltage is 0.50V and it delivers a current of 0.20A. Calculate the internal resistance of the lamp.

1) Read textbook and do Activity 13B