Electric Current. Section 13. Intro: Take out a sheet of paper for today’s notes Draw the following diagram and label the parts. Open Switch. Conductor. Closed Switch. +. Battery. Load. . A light is an example of a load. Fuse. Basic Circuit. Resistor. Open Switch.
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Electric Current
Section 13
Intro:Take out a sheet of paper for today’s notesDraw the following diagram and label the parts
Open Switch
Conductor
Closed Switch
+
Battery
Load

A light is an example of a load
Fuse
Basic Circuit
Resistor
Open Switch
Intro:Take out a sheet of paper for today’s notesDraw the following diagram and label the parts
Closed Switch
Conductor
+
Battery
Load

Fuse
Resistor
Write in your notes: A circuit must be closed to have electrons flow (current) and the load on
An open circuit is off
A closed circuit is on
Draw the picture above and label it electron drift
Overall Direction
Electron Drift
Overall Direction
Electron Drift
Direct Current (DC)
Alternating Current (AC)
Load
Battery
Closed Switch
OHM’S LAW
Ohm’s Law
V=IR
P=IV
Draw this table in your notes
Volts or (V)
Voltage
Amperes or (A)
Current
Ohms or (Ω)
Resistance
Watts or (W)
Power
Ohm’s Law
V=IR

+
e
Wire
e
e
e
e
e
e
e
e
e
e
e
electron
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
electron
e
e
e
e
e
e
e
e
Ex: solar cell
Example: turbine
Another factor effecting current
Ohm’s Law
V=IR
e
Wire
e
e
e
e
e
e
e
e
e
e
e
electron
e
Another factor effecting current
Ohm’s Law
V=IR
e
e
e
e
e
e
e
e
e
e
e
Wire
e
electron
e
V=IR
Directly (if V increases, I increases)
Inversely (if R increases , I decreases)
Power (P) is the rate of energy use (change in energy)
Power is also related to current and voltage by this equation.
2. How much current is drawn by a 23Ω lamp when a voltage of 12v is applied?
3. What is the voltage of a battery if it produces a current of 0.75 amps in a 12Ω resistor?
V=IRP=IV
Add to your equation sheet:
P = I2R
P=IV
IR substituted for V
V=IR
V=IRP=IV
Add to your equation sheet:
P = I2R
P=IV
substituted for I
a) what is the power of the appliance
b) What is its resistance
2. What is the power of a light bulb that has a resistance of 190 ohms in a 120 v circuit?
a) what is the power of the appliance
b) What is its resistance
2. What is the power of a light bulb that has a resistance of 190 ohms in a 120 v circuit?
“Power” companies sell us energy. The unit they use for energy is the kilowatt hour.
Physics challenge:
What is the smallest denomination coin you could use to pay for the energy used by a 60w bulb burning for 8 hours? (at a cost 16.0412 cents per kilowatt hour)
Add to your equation sheet
“Power” companies sell us energy. The unit they use for energy is the kilowatt hour.
Physics challenge:
What is the smallest denomination coin you could use to pay for the energy used by a 60w bulb burning for 8 hours? (at a cost 16.0412 cents per kilowatt hour)
Start by converting 60W to kW
60W
1 kW
= 0.060 kW
1000 W
“Power” companies sell us energy. The unit they use for energy is the kilowatt hour.
Physics challenge:
What is the smallest denomination coin you could use to pay for the energy used by a 60w bulb burning for 8 hours? (at a cost 16.0412 cents per kilowatt hour)
kW