Transcript Series and Parallel Circuits 1_ppt_RevW10

Physics 106 Lesson #11
Series & Parallel Circuits I
Dr. Andrew Tomasch
2405 Randall Lab
[email protected]
Review: Cells, Batteries & Current
• When connected to a circuit,
the potential difference
between the cell or battery
terminals creates an electric
force on the charges in the
conductor causing them to
move and establishing an
electric current
• The conventional current is
from regions of higher
potential to regions of lower
potential, positive to
negative
Conventional
current
Review: Good and Bad Wiring
• Good:
• Bad:
Broken!
• Ugly:
Short!
No device!
=
Review:
Resistors
Resistors are used in appliances
to convert electrical energy into
thermal energy (heat) or light
Toaster
Space Heater
Light Bulb
When an extension cord is used
with a space heater, the cord
must have a resistance that is
sufficiently small to prevent
overheating of the cord
Stove Heating Element
Review:
Series Resistors
• For resistors R1 & R2
connected in series
(sequentially), the
current i passing
through each resistor
must be the same
• The voltages across
R1 & R2 must add up
to V (Loop Rule)
V
+
Review: Parallel Resistors
• Resistors R1 & R2
connected in parallel
have the same potential
difference (voltage) V
across them
• Charge must be
conserved, so the
currents I1 & I2 flowing
through the two
resistors must add up
to the total current I
leaving the battery
(continuity for electric
current!)
V
+
I  I1  I 2
Review: Power
• Power has a precise definition in physics:
Power is the rate at which work is done or
how much work is done per unit of time
W
P
t
Units: J/s  Watt (W)
James Watt
invented the
steam engine
In an electric circuit
power is the product
of current and voltage:
Power  I V
Concept Test #1
The label on a car battery proudly claims
that the battery will provide “250 Ampere hours”. This rating describes the amount of
what quantity that can be drawn from the
battery?
1)
2)
3)
4)
5)
Charge
Current
Voltage
Power
Energy
current =charge/time
 current × time = charge
1 Ampere-hour =
1 Coulomb/second x 3600 s/hr =
3600 Coulombs of charge
Concept Test #2
How much energy is stored in a 2 V battery
with a capacity of 1 Ampere-hour?
1)
2)
3)
4)
5)
3,600 Joules
1,800 Joules
7,200 Joules
10,000 Joules
5,000 Joules
Energy = charge (C) x potential (V)
1 Ampere-hour =
1 Coulomb/second x 3600 s/hr =
3600 Coulombs of charge
3600 Coulombs x 2 V = 7,200 Joules
of energy
Concept Test #3
What is the average power delivered when I
discharge an 8.4 V battery with a capacity of
2 A-hr in 10 minutes?
1) 3,600 Watts
2) 1,800 Watts
3) 7,200 Watts
4) 100 Watts
Power = energy/time = charge (C) x potential (V)/time
2 Ampere-hour = 2 Coulomb/second x 3600 s/hr =
7,200 Coulombs of charge
7,200 Coulombs x 8.4 V = 60,480 Joules of energy
10 min = 600 seconds
60,480 Joules/600 seconds ≈ 100 Watts of power