1. Electric current and Ohm’s Law
AP Physics
Section 18-1 to 18-5
Electric current and Ohm’s Law

2. When a conductor connects a region of high potential to a region of low potential an electric current will flow through the conductor.
The flow of positive charge from high to low potential (from + to –) is called conventional current.
Current is the movement of any charged particles. These may be electrons in a metal, or ions in an electrolyte solution. ∆q
I =
The symbol for electric current is I. ∆t
The unit of current is one of the base SI units, the ampere (A). 1A = 1 coulomb/second (1 C/s).

3. Production of electric current
conventional current I
Conventional current is the flow of positive charges from positive to negative (high to low potential). In metals, direct current is actually electron flow from negative to positive. + + – – + + – – + + + – – – + +
flow of electrons
in metal conductor – – + –
Production of electric current
In the diagram above, the current flow would eventually stop when the spheres reach equilibrium. However, a device may be used to “pump” the current back to high potential again, just like a water pump raising water to a higher elevation.

4. Sources of electromotive force
Electromotive force (EMF, symbol: ℰ) is the potential difference created by any source of electrical energy including electrochemical cells and batteries, dynamos (DC generators), photovoltaic cells, thermoelectric generators, etc.
photovoltaic
cell
DC power
supply
cell
battery
hand dynamo

5. Resistance I ∆V = IR ∆V R =
Resistance is a property of materials that measures how much current flows compared to the potential difference across the material.
Symbol: R
Units: ohm (Ω)
∆V = IR ∆V
often written,
Ohm’s Law
R = ∆V I
AP sheet:
I = R
Properties of conductor that affect the resistance
length (ℓ) — higher R for longer length.
cross-sectional area (A) — higher R for smaller area.
temperature — higher R at higher temperatures.
material resistivity (ρ) — the material used affects R

6. Resistors
resistivity, ρ
materials from most to least conductive
In electronics, special components with known resistances are used to control the flow of current. These are called resistors. The values in ohms are printed on the resistor, or a color code is used.
silver
copper
gold
aluminum
tungsten
iron
platinum
lead
nichrome
brown = 1
carbon
black = 0
orange = 000
germanium
gold = 5%
silicon
For given T,
R = ρ ℓ A
10kΩ
resistor symbol

7. Various types of resistors with different values and tolerances (possible degree of error).
Resistor construction.

8. Measuring devices Ω V A Ammeter Voltmeter Ohmmeter Multimeter
Various measuring devices are used to measure potential difference (voltage), current, and resistance.
Ammeter
Voltmeter V A Ω
Ohmmeter
Multimeter

9. Other common circuit symbols
switch
(single throw)
variable resistor
(trimmer)
ground (earth)
capacitor
bulb
potentiometer
diode

10. Simple circuit problems
Calculate the current in the circuits in milliamperes. 1) 2)
15kΩ
330Ω I I 9V 3V
I = 0.6 mA
I = 9.1 mA
Calculate the resistance of the resistors.
RV is very high. Why?
12V 6V V V 3) 4) R1 R2 I I
R1 = 62Ω
R2 = 470Ω
12V 6V A A
0.194 A
RA is very low. Why?
12.8 mA

11. Electric Power I = ∆V2 UE = PEe UE ∆UE = W = q ∆V ∆V = q = It P = I∆V
We learned that: UE
∆UE = W = q ∆V
∆V =
so, q0 ∆q
q = It
I =
also,
so, ∆t
Substitute q:
∆UE = W = It∆V
Since power is energy or work per time in seconds:
P = I∆V
Units: watt (W)
∆V2
P = I2R
P =
and
Also, since ∆V = IR: R

12. André-Marie Ampère
Alessandro Volta
Georg Simon Ohm
James Watt

13. Calculating Power 1) 2) 3) 4)
Calculate the power of all four previous circuits, 1–4. 1) 2)
P = W
P = W 3)
P = 2.3 W 4)
P = W
If all four circuits run for 5.0 minutes, how much energy is used by each circuit. 1)
energy = 1.6 J 2)
energy = 8.1 J 3)
energy = 690 J 4)
energy = 23 J

14. I = ∆V ∆V2 ∆V2 P = energy/time energy = P•t = 100 W (3600 s) energy =
A car has two 50 watt fog lights running, for a total of 100 W of power. The electrical system runs off a 12V lead acid battery.
1. How much energy do the fog lights use in one hour?
P = energy/time
energy = P•t
= 100 W (3600 s)
energy = J
= 360 kJ
2. What is the current in the circuit? ∆V P
12 V
100 W
P = I∆V
I = = =
8.3 A
3. What is the total resistance of the two lights?
P = R
∆V2 P
∆V2
100 W
(12 V)2
R = = =
1.4 Ω