Download presentation

Presentation is loading. Please wait.

Published byVivian Wood Modified about 1 year ago

1
-Electric Power AP Physics C Mrs. Coyle

2
Remember: P= W / t P= dW /d t Power=Work/time W= ΔV q and I = q/t P= I V

3
Electric Power, P= I Δ V Known as Joule’s Law P: is the power consumed by a resistor, R. Unit: Joule/s= Watt

4
kWh kiloWatt hour What does the kWh measure, a) Energy or b) Power ?

5
From P=I Δ V and Ohm’s Law: P=V 2 /R P=I 2 R

6
As a charge moves from a to b, the electric potential energy of the system increases by Q V The chemical energy in the battery must decrease by this same amount The battery “pumps” energy to the +charges

7
As the current flows through the resistor (c to d), the system loses electric potential energy Energy is transformed into heat energy in the resistor

8
The power is the rate at which the energy is delivered to the resistor

9
Resistors Expend Thermal Energy Wasted heat energy is called “Joule Heating” or “I 2 R” loss.

10
Why is long distance power transmitted at high voltages? Hint: P = I V Answer: For a given P, keep the current, I, low to minimize “I 2 R” loss in the transmitting wires, so increase V.

11
Electric heaters(Coil Heaters) P= V 2 /R The lower the R the greater the heat given off by the resistor for a given voltage.

12
Brightness of a Light bulb and Power The greater the power actually used by a light bulb, the greater the brightness. Note: the power rating of a light bulb is indicated for a given voltage and the bulb may be in a circuit that does not have that voltage.

13
Wattage and Thickness of Filament For a given V, (P = IV) the higher the wattage of a light bulb, the larger the current and therefore the smaller the resistance of the filament (V=I R). Thus, the higher wattage bulb will have a filament of lower resistance and therefore a larger cross-sectional area (R=ρ L / A).

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google