Transcript Review for Exam 2

DC Circuit
Overview
Examination #2 Review
March 1, 2010
?



Review Today
No problem session after class today
Exam #2 on Wednesday
◦ There will be a class hour after the exam and
we will start some new material.


No quiz on Wednesday
Don’t forget to bring in Lab Notebooks on
Friday

Next topic:
This Week
Things To Remember:
 This chapter (20) does not exist by itself.
 The understanding of POTENTIAL is critical to the
chapter so make sure that you at least look at chapter 19
of you are unsure of the concept.
 The exam structure:
 8 reasonably short conceptual problems, some from
your lab sheets. (Total of 60 points)
 1 Problem on Kirchoff’s Laws (you knew that, right?) (20
points)
 1 Other problem. (20 points)
6 Volts
1.5 Volts
What is a battery and what does it
do??
 The battery has two TERMINALS that maintain a potential difference between
them
 That potential difference is usually referred to as the “voltage of the battery”.
 Any two points in a circuit that are DIRECTLY connected to a battery will have
the battery voltage across it.
 The (+) terminal is defined as having the higher potential and the (–) terminal
is often referred to as the lower potential and sometimes it is referred to as the
ground or reference potential.
 The symbol for the battery is shown above.
 There are MANY types of batteries.
Current
Q
I
 Coul / sec
t
Q  I t
Number of electrons transported
through an area of a wire is
Q I t
Ne 

e
e
Current & Wires
 The electrons are moving in the wire.
 There MUST be an electric field in the wire.
 The field is V/d. d is the length of the wire. V is very small in most
circuits.
 Current goes the opposite way that electrons go. This is rarely a
problem.
 Some wires make it hard to move charge. These are called
Resistors. A simple example is the W wire that you played with.
 For such materials, the relationship between the Voltage across the
element and the current through it is given by Ohm’s Law: V=IR
RESISTANCE
(Ohms –W)
R~
L
A
L
A
  resistivity ohm-meter
R
Typical problems
A wire has a resistance of 21.1 Ω. It is melted down, and
from the same volume of metal a new wire is made that is
four times longer than the original wire. What is the
resistance of the new wire?
The filament in an incandescent light bulb is made from
tungsten. The light bulb is plugged into a 60 V outlet and
draws a current of 1.04 A. If the radius of the tungsten wire
is 0.0024 mm, how long must the wire be?
Two wires are identical, except that one is aluminum and
one is iron. The aluminum wire has a resistance of 0.19 Ω.
What is the resistance of the iron wire?
POWER to the people:
Work qV
P

 I V
time
t
Ohm ' s Law : V=IR
2
E
P=I  IR=I R 
R
2
IV
A cigarette lighter in a car is a resistor that, when activated, is connected
across the 12-V battery. Suppose that a lighter uses 27.1 W of power.
(a) Find the resistance of the lighter.
(b) Find the current that the battery delivers to the lighter.
E2
P  27.1 
R
E2
144
R

6W
27.1 27.1
E
I   12 / 6  2 Amp
R
ADD RESISTORS
SERIES
R  R1  R2  ......
PARALLEL
1
1
1


 .......
R R1 R2
A Hard One?
The circuit in the Figure has been connected for a long time.
What is the voltage across the capacitor?
Determine the power supplied to the 7.0 Ω resistor
in the circuit shown in the drawing. (R1 = 4.0 Ω, R2 =
7.0 Ω and V1 = 15 V.)
Kirchoff’s Law #1
 The sum of all the currents entering a node of a circuit
is equal to the sum of all the currents leaving the node.
 Current Entering= - Current Leaving.
 Sometimes stated: The sum of all the currents entering a
node is equal to zero.
 The sum of the voltage RISES in going around a
current loop is equal to zero.
Kirchoff’s Law #2
H
RISE
DROP
L
Kirchoff’s Law #2
I
L
H
DROP
RISE
Example – The Battery
Emf of Battery
Loop: E –Ir – IR =0
Voltage Supplied = V-Ir