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October 15, 2008
DC Circuits
This is the week that will have beenToday
Complete Resistance/Current with some problemsFriday
Examination #2: Potential Resistance & CurrentNext Week: DC CircuitsNext Topic: MagnetismNext Quiz: One week from Friday
The same old closed circuit
Note change in notation: VE
IER
ERIP
A
LR
A
IJ
IRE
22
The figure below gives the electrical potential V(x) along a copper wire carrying a uniform current, from a point at higher potential (x=0m) to a point at a lower potential (x=3m). The wire
has a radius of 2.45 mm. What is the current in the wire?
copper
12 volts 0 volts
What does the graph tell us??
*The length of the wire is 3 meters.*The potential difference across the
wire is 12 volts.*The wire is uniform.
Let’s get rid of the mm radius and convert it to area in square meters:A=r2 = 3.14159 x 2.452 x 10-6 m2
orA=1.9 x 10-5 m 2
Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters
We have all we need….
ma 49.41067.2
1012
R
Vi
:Law sOhm' From
67.2 109.1
0.3m-ohm 1069.1
3
6
5
8
ohms
volts
mx
mx
A
LR
When the potential difference across a certain conductor is doubled, the current is observed to increase by a factor of three. What can you conclude about the conductor?
A. It is a perfect conductor.B. It does not obey Ohm's
Law.C. It is a semiconductor.D. It obeys Ohm's Law.
Two conductors of the same length and radius are connected across the same potential difference. One conductor has twice the resistance of the other. To which conductor is more power delivered?
A. conductor with lower resistanceB. conductor with higher resistanceC. Equal amount of power delivered
to both conductors.
An electric utility company supplies a customer's house from the main power lines (120 V) with two copper wires, each of which is 50.0 m long and has a resistance of 0.108 Ω per 300 m.
(a) Find the potential difference at the customer's house for a load current of 110 A. 116 V
(b) For this load current, find the power delivered to the customer. 12.8 kW
(c) Find the rate at which internal energy is produced in the copper wires.436 W
A toaster is rated at 780 W when connected to a 240-V source.
What current does the toaster carry?3.25 AWhat is its resistance?73.8
This material will NOT be on the exam
DC CIRCUITSLet’s add resistors …….
Series Combinations
iiRseriesR
general
RRR
iRiRiRVVV
and
iRV
iRV
)(
:21
2121
22
11
R1 R2
i i
V1 V2V
SERIES Resistors
The rod in the figure is made of two materials. The figure is not drawn to scale. Each conductor has a square cross section 3.00 mm on a side. The first material has a resistivity of 4.00 × 10–3 Ω · m and is 25.0 cm long, while the second material has a resistivity of 6.00 × 10–3 Ω · m and is 40.0 cm long. What is the resistance between the ends of the rod?
Parallel Combination??
i iRR
general
RRR
so
R
V
R
V
R
Viii
iRV
11
111
..
21
2121
R1, I1
R2, I2
V
What’s This???In this Figure, find the equivalent resistance between points (a) F and H and [2.5] (b) F and G. [3.13]
(a) Find the equivalent resistance between points a and b in the Figure.
(b) A potential difference of 34.0 V is applied between points a and b. Calculate the current in each resistor.
Power Source in a Circuit
The ideal battery does work on charges moving them (inside) from a lower potential to one that is V higher.
A REAL Power Sourceis NOT an ideal battery
V
E or Emf is an idealized device that does an amount of work E to move a unit charge from one side to another.
By the way …. this is called a circuit!
Internal Resistance
A Physical (Real) Battery
Rr
Emfi
Internal Resistance
Back to which is brighter? (R1=R2)
Back to Potential
Represents a charge in space
Change in potential as one circuitsthis complete circuit is ZERO!
Consider a “circuit”.
This trip around the circuit is the same as a path through space.
THE CHANGE IN POTENTIAL FROM “a” AROUND THE CIRCUIT AND BACK TO “a” is ZERO!!
To rememberIn a real circuit, we can neglect the
resistance of the wires compared to the resistors.We can therefore consider a wire in a circuit to be
an equipotential – the change in potential over its length is slight compared to that in a resistor
A resistor allows current to flow from a high potential to a lower potential.
The energy needed to do this is supplied by the battery.
VqW
NEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE.LOOP EQUATION
The sum of the voltage drops (or rises) as one completely travels through a circuit loop is zero.
Sometimes known as Kirchoff’s loop equation.
NODE EQUATIONThe sum of the currents entering (or leaving)
a node in a circuit is ZERO
TWO resistors again
jj
21
21
RR
Resistors SERIESfor General
RRR
or
iRiRiRV
i
R1 R2
V1 V2
V
A single “real” resistor can be modeledas follows:
R
a b
V
position
ADD ENOUGH RESISTORS, MAKING THEM SMALLERAND YOU MODEL A CONTINUOUS VOLTAGE DROP.
We start at a point in the circuit and travel around until we get back to where we started.
If the potential rises … well it is a rise.If it falls it is a fall OR a negative rise.We can traverse the circuit adding each rise
or drop in potential.The sum of all the rises around the loop is
zero. A drop is a negative rise.The sum of all the drops around a circuit is
zero. A rise is a negative drop.Your choice … rises or drops. But you must
remain consistent.
Take a trip around this circuit.
Consider voltage DROPS:
-E +ir +iR = 0or
E=ir + iRrise
Circuit Reduction
i=E/Req
Multiple Batteries
Reduction
Computes i
Another Reduction Example
PARALLEL
1212
1
600
50
30
1
20
11
RR
START by assuming a DIRECTION for each Current
Let’s write the equations.
In the figure, all the resistors have a resistance of 4.0 and all the (ideal) batteries have an emf of 4.0 V. What is the current through resistor R?
The Unthinkable ….
RC Circuit Initially, no current
through the circuit Close switch at (a) and
current begins to flow until the capacitor is fully charged.
If capacitor is charged and switch is switched to (b) discharge will follow.
Close the Switch
I need to use E for E
Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)
Really Close the Switch
I need to use E for E
R
E
RC
q
dt
dq
or
EC
q
dt
dqR
C
qiRE
dt
dqi since
0
Equation Loop
Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)
This is a differential equation.
To solve we need what is called a particular solution as well as a general solution.
We often do this by creative “guessing” and then matching the guess to reality.
You may or may not have studied this topic … but you WILL!
RC
REaeCE
R
E
RC
q
dt
dq
CEq
R
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RC
q
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E/R0CEa
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-CEK
KCE0
solution from and 0q 0,When t
and 0dq/dt charged,fully is device When the
:solution particularat Look
Solution General
at-
at-
Time Constant
RC
Result q=CE(1-e-t/RC)
q=CE(1-e-t/RC) and i=(CE/RC) e-t/RC
RCteR
Ei /
Discharging a Capacitor
qinitial=CE BIG SURPRISE! (Q=CV)i
iR+q/C=0
RCt
RCt
eRC
q
dt
dqi
eqq
solutionC
q
dt
dqR
/0
/0
0
In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase of thermal energy Eth in the resistor as a function of time t.
(a)What is the electric potential across the battery? (60)(b) If the resistance is doubled, what is the POWER dissipated by the circuit? (39)(c) Did you put your name on your paper? (1)
Looking at the graph, we see that theresistor dissipates 0.5 mJ in one second.
Therefore, the POWER =i2R=0.5 mW
ma 88.41088.4
1038.2Ω 21
mW 0.5
3
252
ampi
ampR
Pi
mVamp 10221104.88iRV
or iR reisitor theacross drop Voltage3-
If the resistance is doubled what is the power dissipated by the circuit?
mJRiP
ma
R
248.0
43.242
10102
R
Vi
mV 102V 42
2
3