8B-Soln06-v1.0.2

8B-Soln06-v1.0.2 - raising the temperature of the resistor...

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d) Now, l goes to 4 l but volume stays the same, so A goes to A / 4 ( r goes to r / 2). R new = 16 R old = 8.56 × 10 -1 Ω . P new = I 2 R new = 16 P old = 123.2 W 4. a) Current moves from the positive terminal of the battery to the negative terminal (but the electron flow, which makes the current, goes in the opposite direction, because electrons are negatively charged), so the current is flowing to the left at Mr. Smiley. b) The positive charges (the protons in the nuclei that make up the material of the wire) are remaining still, which some negative charges (the free electrons in the material) are moving to the right. c) P = IV = I 2 R = V 2 / R = 48 watts. Note that because we used a power expression with an R in it, we are really saying that 48 watts of power are dissipated by the resistor. To keep things in a steady state, however, that means the battery must be continually supplying (or delivering ) 48 watts of power. d) The resistor dissipates energy in the form of heat. Some of that heat will go into
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Unformatted text preview: raising the temperature of the resistor, which is why the resistor ‘heats up’. A resistor will typically increase its resistance as the temperature rises. This is because the higher temperature means the molecules making up the resistor are vibrating more vigorously. The electron’s path becomes more jagged (imagine trying to move through an overcrowded room - if the party is dull and everyone is just standing there, you can move through fairly easily. If the party is exciting and everyone is dancing, it becomes much harder to move through the room - you keep bouncing into everyone and getting deflected from your path to the exit!) e) The battery puts out the same voltage, and we are increasing the resistance, so the current goes down and the battery is putting out less power....
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