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living with the lab
College of Engineering & Science
Ohm’s Law
1.5V
π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ = π‘π‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘‘ βˆ™ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’
1.5V
π‘£π‘œπ‘™π‘‘π‘  = π‘Žπ‘šπ‘π‘  βˆ™ π‘œβ„Žπ‘šπ‘ 
𝑉 =πΌβˆ™π‘…
© 2012 David Hall
living with the lab
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living with the lab
electric current and voltage
electron
shortage
+
electron
supply
-
electric resistance
3
living with the lab
units for electric current
The electric current I is the amount of charge passing a point per unit time.
1 π‘Žπ‘šπ‘ =
𝐢
𝐴=
𝑠
6.24 10
18
π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘  π‘π‘Žπ‘ π‘ π‘–π‘›π‘” π‘Ž π‘π‘œπ‘–π‘›π‘‘
1 πΆπ‘œπ‘’π‘™π‘œπ‘šπ‘
𝐢
=
=
π‘ π‘’π‘π‘œπ‘›π‘‘
π‘ π‘’π‘π‘œπ‘›π‘‘
𝑠
π‘€β„Žπ‘’π‘Ÿπ‘’ 1𝐢 = 6.24 10
18
π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘ 
teams of 2
Class Problem A battery powers a flashlight. If the battery supplies a steady
current of 1.3A over 8 minutes, how many electrons leave the negative terminal of
the battery during this time period?
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living with the lab
1C
units of voltage
1C
the unit for voltage V is the volt
1 π½π‘œπ‘’π‘™π‘’
1 π‘£π‘œπ‘™π‘‘ =
πΆπ‘œπ‘’π‘™π‘œπ‘šπ‘
π‘œπ‘Ÿ
𝐽
𝑉=
𝐢
1.5V
1.5V
π‘€π‘œπ‘Ÿπ‘˜ = 1.5𝑉 βˆ™ 1𝐢 = 1.5𝐽
1.5V
One Coulomb flows through each light bulb.
π‘€π‘œπ‘Ÿπ‘˜ = 1.5𝑉 + 1.5𝑉 βˆ™ 1𝐢 = 3𝐽
The voltage across a bulb is doubled when two batteries are used.
The work done when two batteries are used is twice as much as the work done when one
battery is used, even though the same number of electrons flow through each bulb.
Doubling the voltage doubles the work . . . and gives off much more light.
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living with the lab
voltage / water pressure analogy
depth of water between surface
of lake and surface of river
voltage & pressure are both
measures of a potential difference
Glen Canyon Dam in Arizona, USA
Wikipedia
The work done per gallon of water passing through a turbine will double
(theoretically) when the water depth (pressure) is doubled.
Likewise, the work done per Coulomb of charge passing through a resistor will double
(theoretically) when the voltage is doubled.
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living with the lab
units of resistance
The resistance R is a measure of the frictional resistance encountered by
electrons as they attempt to pass through a material.
R = π‘œβ„Žπ‘šπ‘  π‘œπ‘Ÿ Ω
a pile of electrical resistors
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living with the lab
circuit symbols
We frequently draw diagrams that represent various types of electric circuits.
The most simple diagram is that of a direct current (DC) power source and a resistive element
such as a light bulb or resistor.
+
-
circuit symbol for a DC power source . . . 1.5 volts in this case
1.5V
circuit symbol for a resistor . . . 220 ohms in this case
220 Ω
For resistors, you don’t need to worry about which side is
+ and which side is – (they will work either way).
straight lines depict conductors with an assumed resistance of zero
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living with the lab
circuit diagrams
We can put together a series of circuit symbols to depict an electric circuit.
+
6V
-
12 Ω
The voltages of power sources arranged end-to-end in this way can be
added. Thus, four 1.5V batteries in series has a total voltage of 6V.
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living with the lab
Ohm’s law
Ohm’s law relates voltage, current and resistance as follows . . .
π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ = π‘π‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘‘ βˆ™ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’
π‘£π‘œπ‘™π‘‘π‘  = π‘Žπ‘šπ‘π‘  βˆ™ π‘œβ„Žπ‘šπ‘ 
𝑉 =πΌβˆ™π‘…
+
6V
12 Ω
-
I
conventional
current
teams of 2
Class Problem Use Ohm’s law to determine . . .
(a) the current delivered to the resistor (light bulb) for the circuit above.
(b) the number of electrons that leave the battery pack over 2 minutes.
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