Aim: How can we explain a series circuit?

Download Report

Transcript Aim: How can we explain a series circuit?

Aim: How can we explain
a series circuit?
Do Now:
Find the resistance of a 0.25 m Tungsten wire with a
diameter of 1.5 x 10-3 m.
R = ρL
A
R = (5.6 x 10-8Ω·m) (0.25 m)
1.77 x 10-6 m2
R = 7.9 x 10-3 Ω
Circuits
• A closed path where current
flows (charges flow)
• Has a source of potential
difference (cell or battery)
• Has a resistance (ex: lightbulb)
• Has wires
Lamp
Resistor: hinders
flow of charges
Wire
Provides
complete
path
Switch
Dry cell
Opens and
closes
circuit path
Source of potential
difference – pushes
charges, causing
them to move
Supplies
Voltage
Opens and
closes a circuit
Measures
Voltage
Measures
Current
Supplies
Resistance
Light Bulb Demo
http://www.youtube.com/watch?v=csTlRvqSITI&feature=r
elated
If one goes out,
they all go out
Series Circuits
• Has only 1 path for the current to flow
• Draw a series circuit with a 9 V battery and
3 resistors reading 100Ω, 300Ω, and 50Ω.
Equivalent (Total) Resistance
Req = R1 + R2 + R3
Req = 100 Ω + 300 Ω + 50 Ω
Req = 450 Ω
Draw a new equivalent circuit
Req = 450 Ω
VT = total voltage of circuit
What is the total current?
IT = VT
RT
IT = 9 V
450Ω
IT = 0.02 A
Current is constant for a series circuit
Therefore:
I1 = 0.02 A
I2 = 0.02 A
I3 = 0.02 A
When solving vertically, use the rules in the ref. tables
for series circuits
R(Ω)
When solving
horizontally,
use Ohm’s
Law
1
2
3
Total
Now solve for V1, V2, and V3
I(A)
V(V)
V=IR
100
0.02
2
V=IR
300
50
450
0.02
6
V=IR
0.02
0.02
1
9
Check the voltage: VT = V1 + V2 + V3
Draw a 120 V battery in series
with 3 resistors. R1 = 2 Ω R2 =4Ω,
R3 = 8Ω.
2Ω
4Ω
120 V
8Ω
Find Req, then draw equivalent
circuit with 1 resistor and an
ammeter.
Place directly into
circuit.
120 V
Req = 14 Ω
Current is constant in series so it doesn’t matter where the ammeter is placed
Note: Voltmeters get placed around what they are measuring
Solve for all currents and voltages
R(Ω)
1
I(A)
V1 = I1R1
2
IT = I1 = I2 = I3
8.57
2
IT = I1 = I2 = I3
4
3
Total
V(V)
8
8.57
V1 = (8.57)(2)
17.14
V2 = I2R2
V1 = (8.57)(4)
IT = I1 = I2 = I3
34.28
V3 = I3R3
8.57
V1 = (8.57)(8)
R1+R2+R3
IT =VT / RT
2+4+8
IT =120/14
68.56
120
V1+V2+V3
14
8.57
17.14+34.28+68.56
Remember to check the voltages!