Download Review Exam 4

Review Exam 4
Chapter 18
Electric Currents
Resistance and Resistors
The ratio of voltage to current is called the
resistance:
(18-2a)
(18-2b)
18.4 Resistivity
The constant ρ, the resistivity, is
characteristic of the material.
18.4 Resistivity
18.4 Resistivity
For any given material, the resistivity changes
with temperature:
Since
R T  R 0 1   T - T0 
R0 = Resistance at some reference temperature (T0)
α = temperature coefficient and R = resistance
at some new temperature T
If  positive, if T , R 
If  negative if T , R 
18.5 Electric Power
18.5 Electric Power
The unit of power is the watt, W.
1Watt = 1 Joule/second
Chapter 19
DC Circuits
Conservation Laws
• Conservation of Charge (I = Q/t) and I
• Conservation of Energy (PE = QV) and
V
• Conservative Forces (pages 148-149)
• ∑ of Work around a closed path = 0
• Independent of Path
• For a closed loop ∑ of PE gained and
lost or ∑ of V gained and lost = 0
I
ε
+
-
ε-V = 0
ε –IR =0
ε = IR
Simple Circuit
R
V = IR
Resistors in Series
From this we get the equivalent resistance (that
single resistance that gives the same current in
the circuit).
Resistors in Parallel
This gives the reciprocal of the equivalent
resistance:
(19-4)
Repeat this Mantra
• The current is the same for
elements connected in series
• The voltage is the same for
elements connected in parallel
Multiple Loop Game Rules
• Draw picture
• Define number of loops
• Pick ARBITRARY directions for the
currents
• Indentify branch points (where currents
divide)
• Currents into branch points = currents out
of branch points
• Circle each loop and ∑ V’s = 0
How to circle a loop
• Start at some ARBITRARY point
• Circle clockwise or counterclockwise (your
choice)
• End at starting point
How to Add V’s
When you come to
a Battery as you circle
the loop
ε
-ε
Direction you take
+ε
When you come to
a Resistor as you circle
the loop you will travel
with or against the
current
R
I
- IR
+ IR
19.5 Circuits Containing Capacitors in
Series and in Parallel
In this case, the total capacitance is the sum:
(19-5)
19.5 Circuits Containing Capacitors in
Series and in Parallel
In this case, the reciprocals of the
capacitances add to give the reciprocal of the
equivalent capacitance:
(19-6)
Ammeters – Very small resistance
so voltage across it is very small
In series
Voltmeters - Very large resistance, takes away
little current.
In Parallel
What would happen if I put an
Ammeter in Parallel?
A
ConcepTest 19.8
Kirchhoff’s Rules
The lightbulbs in the
1) both bulbs go out
circuit are identical. When
2) intensity of both bulbs increases
the switch is closed, what
3) intensity of both bulbs decreases
happens?
4) A gets brighter and B gets dimmer
5) nothing changes
ConcepTest 19.8
Kirchhoff’s Rules
The lightbulbs in the
1) both bulbs go out
circuit are identical. When
2) intensity of both bulbs increases
the switch is closed, what
3) intensity of both bulbs decreases
happens?
4) A gets brighter and B gets dimmer
5) nothing changes
When the switch is open, the point
between the bulbs is at 12 V. But so is
the point between the batteries. If
there is no potential difference, then
no current will flow once the switch is
closed!! Thus, nothing changes.
Follow-up: What happens if the bottom
battery is replaced by a 24 V battery?
24 V
ConcepTest 19.10
More Kirchhoff’s Rules
1) 2 – I1 – 2I2 = 0
Which of the equations is valid
2) 2 – 2I1 – 2I2 – 4I3 = 0
for the circuit below?
3) 2 – I1 – 4 – 2I2 = 0
4) I3 – 4 – 2I2 + 6 = 0
5) 2 – I1 – 3I3 – 6 = 0
1
I2
2
6V
22 VV
4V
I1
1
I3
3
ConcepTest 19.10
More Kirchhoff’s Rules
1) 2 – I1 – 2I2 = 0
Which of the equations is valid
2) 2 – 2I1 – 2I2 – 4I3 = 0
for the circuit below?
3) 2 – I1 – 4 – 2I2 = 0
4) I3 – 4 – 2I2 + 6 = 0
5) 2 – I1 – 3I3 – 6 = 0
Eqn. 3 is valid for the left loop:
The left battery gives +2V, then
there is a drop through a 1
resistor with current I1 flowing.
Then we go through the middle
battery (but from + to – !), which
gives –4V. Finally, there is a
drop through a 2 resistor with
current I2.
1
I2
2
6V
22 VV
4V
I1
1
I3
3