Download DC Circuits

DC Circuits
October 15, 2008
This is the week that will have been
 Today
 Complete Resistance/Current with some problems
 Friday
 Examination #2: Potential  Resistance & Current
 Next Week: DC Circuits
 Next Topic: Magnetism
 Next Quiz: One week from Friday
The same old closed circuit
Note change in notation: VE
E  IR
I
J
A
L
R
A
2
E
2
PI R
 IE
R
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?
What does the graph tell us??
*The length of the wire is 3 meters.
*The potential difference across the
wire is 12 m volts.
*The wire is uniform.
Let’s get rid of the mm radius and
convert it to area in square meters:
A=pr2 = 3.14159 x 2.452 x 10-6 m2
or
A=1.9 x 10-5 m 2
copper
12 volts
0 volts
Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters
We have all we need….
8
L 1.69 x10 ohm - m  3.0 m
R 
 2.67 m
5
A
1.9 x10
From Ohm' s Law :
6
V
12 10 volts
i 
 4.49 ma
3
R 2.67 10 ohms
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.
It does not obey Ohm's Law.
C. It is a semiconductor.
D. It obeys Ohm's Law.
B.
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?
conductor with lower resistance
B. conductor with higher resistance
C. Equal amount of power delivered to
both conductors.
A.
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 A
What is its resistance?
73.8
This material will NOT be on the exam
DC CIRCUITS
Let’s add resistors …….
SERIES Resistorsi
Series Combinations
i
R1
R2
V1
V2
V
V1  iR1
V2  iR2
and
V  V1  V2  iR  iR1  iR2
R  R1  R2
general :
R ( series )   Ri
i
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??
R1, I1
R2, I2
V
V  iR
V V V
i  i1  i2  

R1 R2 R
so..
1
1
1


R1 R2 R
general
1
1

R
i Ri
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 Source
is NOT an ideal battery
Internal Resistance
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!
A Physical (Real) Battery
Emf
i
rR
Back to which is brighter? (R1=R2)
Back to Potential
Change in potential as one circuits
this complete circuit is ZERO!
Represents a charge in space
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 remember
 In 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.
W  qV
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 EQUATION
 The sum of the currents entering (or leaving) a node in a circuit
is ZERO
TWO resistors again
i
R1
V1
V  iR  iR1  iR2
R2
V2
V
or
R  R1  R2
General for SERIES Resistors
R  Rj
j
A single “real” resistor can be modeled
as follows:
R
a
b
V
position
ADD ENOUGH RESISTORS, MAKING THEM SMALLER
AND 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 = 0
or
E=ir + iR
Circuit Reduction
i=E/Req
Multiple Batteries
Reduction
Computes i
Another Reduction Example
1
1
1
50
1




R 20 30 600 12
R  12
PARALLEL
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
Loop Equation
I need to use E for E
Note RC = (Volts/Amp)(Coul/Volt)
= Coul/(Coul/sec) = (1/sec)
q
 E  iR   0
C
dq
since i 
dt
dq q
R
 E
dt C
or
dq
q
E


dt RC R
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!
General Solution
q  q p  Ke  at
Look at particular solution :
dq
q
E


dt RC R
When the device is fully charged, dq/dt  0 and
q p  CE
When t  0, q  0 and from solution
0  CE  K
K  -CE
dq
q
E

 and q  CE(1 - e -at )
dt RC R
CE (ae  at )  CE(1 - e -at )  E / R
for t  0
CEa  0  E/R
E
1
a

RCE RC
Time Constant
  RC
Result q=CE(1-e-t/RC)
q=CE(1-e-t/RC) and i=(CE/RC) e-t/RC
E t / RC
i e
R
Discharging a Capacitor
qinitial=CE BIG SURPRISE! (Q=CV)
i
iR+q/C=0
dq q
R
 0
dt C
solution
q  q0 e t / RC
q0 t / RC
dq
i

e
dt
RC
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 the
resistor dissipates 0.5 mJ in one second.
Therefore, the POWER =i2R=0.5 mW
P 0.5 mW
i  
 2.38 105 amp 2
R
21 Ω
i  4.88 10 3 amp  4.88 ma
Voltage drop across the reisitor  iR or
2
V  iR  4.88 10-3 amp  21  102mV
If the resistance is doubled what is the
power dissipated by the circuit?
R  42 
V  102 mV
V 102  10 3

 2.43ma
R
42
P  i 2 R  0.248mJ
i