Download Lecture 09 - Electric Circuits

Lecture 07
Current & Circuits
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Quiz on Friday (This weeks
material)
Examination #2 one week from
Friday: March 11th.
Last time
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We discussed current, resistance,
power and the elements of the
microscopic theory of resistivity.
Today we start electric circuits after
a short diversion.
The Spectrum of Conductors
ENGINEERED MATERIALS!!!
Semiconductors
Materials
VERY IMPORTANT MATERIAL
SILICON
Silicon Crystal Lattice
Electron Freed!
Add an impurity
Special Impurities
Effect of Impurities on r
Importance
P
N
P
N
P
N
P
Diode
current
voltage
breakdown
Who Cares??
Silicon
Transistor
CHIPS
History
Thinking Chips?
Back to the Mundane
Fun and Frolic With Electric Circuits
Power Source in a Circuit
V
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 Battery
Emf
i
rR
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
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In a real circuit, we can neglect the
resistance of the wires compared to the
resistors.
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We can therefore consider a wire in a circuit to
be an equipotential.
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.
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LOOP EQUATION
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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 resistor can be modeled
as follows:
R
a
b
V
position
ADD ENOUGH RESISTORS, MAKING THEM SMALLER
AND YOU MODEL A CONTINUOUS VOLTAGE DROP.
Take a trip around this circuit.
Consider voltage DROPS:
-E +ir +iR = 0
or
E=ir + iR
Might as well repeat the parallel thing!
i  i1  i2
V
R1 i1
R2 i2
V V V
 
R R1 R2
1 1 1
 
R R1 R2
General :
1
1

R
j Rj
Circuit Reduction
i=E/Req
Multiple Batteries
Reduction
Computes i
Another Example
1
1
1
50
1




R 20 30 600 12
R  12
PARALLEL
NOTICE ASSUMED DIRECTION OF
TRAVEL
Voltage Drops:
-E1 –i1R1 + i2R2 + E2 +i1R1 = 0
From “a”
-i3R1 + E2 – E2 –i2R2 =0
NODE
I3 +i2 = i1
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
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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
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.
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To solve we need what is called a
particular solution as well as a
general solution.
We often do this by creative
“guessing” and thenb matching the
guess to reality.
You may oir 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 amp2
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
3
V 102  10
i 
 2.43ma
R
42
2
P  i R  0.248mJ