Download Electric Current

Electric Current
Current
 Rate at which charge passes a given point in a
circuit in one second
 Flow of coulombs
 Ampere (A) is unit of current (I)
 I = Δq
t
 Ammeter measures current
 Electric Circuit is a closed path along which
charged particles move
Current Calculation
A total of 20.0C of charge pass a given point in a
conductor in 4.0 seconds. Determine the current in
the conductor.
I = Δq = 20.0C = 5A
t
4.0s
Conditions Necessary For An Electric Circuit
 A potential difference between two points. A
voltage source for energy supplied by a battery
(cell), power company, or transformer.
A battery supplies the force or push on electrons
to move
Voltmeter measures potential energy.
 A closed path
Conductivity In Solids
 Conductivity – availability of charges that are
relatively free to move under the influence of an
electric field
 Conductors – lots of free electrons, metals
 Insulators – tightly held electrons, nonmetals
Ohms Law
 Opposition that a device or conductor offers to
the flow of electric current
 R=V
R = resistance
I
V = volts
I = current
 Ohm(Ω) is the unit of Resistance
 1Ω = volt
amp
Example:
A current of 0.10 ampere flowing through a lamp
connected to 12.0 volt source. What is the resistance
of the lamp?
V = 12.0V
R = V = 12.0V = 120Ω
I = 0.10A
I
0.10A
Factors That Affect The Resistance Of A
Conductor
Resistivity
 Temperature affects
o solids, temperature increases R increases
o Liquids, temperature increases R decreases
 Length and cross-section of wire
R = рL
A
р = resistivity of material
A = cross-sectional area
R = resistance
Example:
A piece of wire has a resistance of 8 ohms. What is
the resistance of a second piece of wire of the same
composition, same diameter and at the same
temperature, but with one half the length of the first
wire?
4 Ohms
Example:
A 5.00M long tin wire has a cross-sectional area of
2.00 x 10^-6m² and a 0.35 ohm resistance. Determine
the resistivity of tin.
p = AR = (2.00 x 10^-6m²)(0.35Ω) = 1.4 x 10^-7Ω·m
L
5.00m
 Resistor – limits current flow or provides a
potential (voltage) drop
Electric Circuits
Electric Circuit Symbols
Switch
A
Ammeter
Cell
Battery
V
Voltmeter
Resistor
Variable Resistor
Series Circuit
 Only one current path
 Current is the same everywhere in the circuit
IT = I1 = I 2 = I3
 The sum of all the potential drops (individual
voltage readings) equals the total voltage in the
circuit
V T = V 1 + V 2 + V3
 The sum of individual resistor readings equals
the total resistance of the circuit
RT = R1 + R2 + R3
RT = equivalent(eq) R
 Conservation of Charge
VT out = VT in
IT out = IT in
8v
2A
R1
4Ω
+
36V
12v
16v
R2
6Ω
R3
8Ω
2A
Voltmeters are placed across components of a circuit
or “in parallel”
Ammeters are placed in the circuit or “in series”
2A
V1
V2
R1
4Ω
R2
6Ω
+
V3
R3
8Ω
2A
36V
Find the total resistance (equivalent resistance):
RT = R1 + R2 + R3 = 4Ω + 6Ω + 8Ω = 18Ω
Find the total current:
IT = VT = 36V = 2A
RT 18Ω
Find the current in each circuit:
IT = I1 = I2 = I3 2A!
Find the voltage drop across each resistor:
V1 = I1R1 = (2A)(4Ω) = 8V
V2 = I2R2 = (2A)(6Ω) = 12V
V3 = I3R3 = (2A)(8Ω) = 16V
Find the total power used by the circuit:
PT = VTIT = (36V)(2A) = 72W
Find the power used by the 6Ω resistor:
P2 = V2I2 = (6Ω)(2A) = 12W
Find the amount of heat given off by the 6Ω resistor
in 30 seconds:
W = Pt = (12W)(30s) = 360J
Same Circuit, Different Drawing
12V
R2
6Ω
R14Ω
8V
16V
8Ω R3
36V
2A
2A
Parallel Circuit
 More than one current path
 The total current is the sum of the branch
currents
IT = I1 + I 2 + I3
 The total voltage is the same as the voltage
reading across each resistor
V T = V 1 = V 2 = V3
 The sum of the reciprocals of the branch resistors
is equal to the total resistance (equivalent
resistance)
1 = 1 + 1+ 1
RT R1 R2 R3
12V
V
3A
R1 4Ω
12V
2A
R2 6Ω
12V
1A
R3 12Ω
6A
6A
+
12V
-
12V
3A
R1 4Ω
12V
2A
R2 6Ω
12V
1A
R3 12Ω
6A
6A
+
_
12V
Find the total resistance (equivalent resistance)
1 = 1 + 1 + 1 = 1 + 1 + 1
RT R1 R2 R3
4Ω 6Ω 12Ω
1 = .25Ω + .17Ω + .08Ω
RT
1 = .5Ω
RT
RT = 2Ω
RT is smaller than any one resistor in the circuit!
Find the voltage drop across each resistor
VT = V1 = V2 = V3 VT = 12V
Find the total current in the circuit
IT = VT = 12V = 6A
RT 6Ω
Find the current in each of the resistors
I1 = V1 = 12V = 3A
R1
4Ω
I2 = V2 = 12V = 2A
R2
6Ω
I3 = V3 = 12V = 1A
R3 12Ω
IT = 3A + 2A + 1A = 6A
Find the total power used by the circuit
PT = VTIT = (12V)(6A) = 72W
Find the power used by R2
P2 = V2I2 = (12V)(2A) = 24W
Adding more branches with more resistors increases
the total current in the circuit