Download Lecture Notes: Y F Chapter 26

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Valve RF amplifier wikipedia , lookup

Test probe wikipedia , lookup

Integrating ADC wikipedia , lookup

Wien bridge oscillator wikipedia , lookup

Integrated circuit wikipedia , lookup

Mathematics of radio engineering wikipedia , lookup

Surge protector wikipedia , lookup

Flexible electronics wikipedia , lookup

Multimeter wikipedia , lookup

Electric charge wikipedia , lookup

Power MOSFET wikipedia , lookup

Switched-mode power supply wikipedia , lookup

Nanofluidic circuitry wikipedia , lookup

Resistive opto-isolator wikipedia , lookup

Opto-isolator wikipedia , lookup

Two-port network wikipedia , lookup

Operational amplifier wikipedia , lookup

Current source wikipedia , lookup

RLC circuit wikipedia , lookup

TRIAC wikipedia , lookup

Ohm's law wikipedia , lookup

Rectiverter wikipedia , lookup

Current mirror wikipedia , lookup

Network analysis (electrical circuits) wikipedia , lookup

Transcript
Resistors in Series
1) There must be the same current in both resistors
I1 = I 2 = I
What is the Current?
2)
V1 ≠ V2
so
V1 = IR1
V2 = IR2
But we know the total potential
ε = V1 + V2
ε = IR1 + IR2
ε = I ( R1 + R2 )
3) The current is the same as if the circuit had
one resistor with
R = R1 + R2
We say R is the electrically “equivalent resistor.
For
Req = R1 + R2 + R3 + R4 + R5 + ...
Resistors in Parallel
1) The ends of the two resistors are connected by conductors
so they must be at the same potential
V1 = I1R1 = V
V2 = I 2 R2 = V
What is the Current?
V
I1 =
R1
V
I2 =
R2
2) Since all of the current must flow through the resisitors
⎛1
V V
1 ⎞
I1 =
+
= V ⎜⎜ + ⎟⎟
R1 R2
⎝ R1 R2 ⎠
3) The two resistors act like one resistor with
1
1
1
=
+
R eq R1 R2
=
1
1
1
1
= +
+
R eq R1 R2 R3
Reminder: Series/parallel rules for resistors
are just the “opposite” of that for capacitors
Analysis of a More Complex Circuit
Question: What are the Currents
example 26.1
Resistors in Series and Parallel
R = R1 + R2 + R3
1 1
1
1
=
+
+
R R1 R2 R3
Sample Exercises: 26.4, 26.8, 26.12, 26.13, 26.18
Many complex circuits can be resolved using repeated application
of the series and parallel rules…BUT NOT ALL!
Examples of circuits that CANNOT be solved that way:
Kirchoff’s Rules
∑I =0
∑V = 0
Junction Rule – Valid at any Junction
Loop Rule – Valid for any closed loop
example
Many complex circuits can be resolved using repeated application
of the series and parallel rules…BUT NOT ALL!
Examples of circuits that CANNOT be solved that way:
Kirchhoff’s Rules
∑I =0
∑V = 0
Junction Rule – Valid at any Junction
Loop Rule – Valid for any closed loop
example
Sampe Problem:
What is the current in each resistor?
What is the equivalent resistance of the 5 resistors?
[See Problem Solving Strategy using Kirchhoff’s Rules (Y&F p987)]
Step 1- Lay out a clear diagram of circuit.
Step 2- Label ALL Currents – Use Kirchhoff’s Junction Rule to simplify
The directions of currents you chose is NOT IMPORTANT
If you guess wrong – Mr. Kirchhoff will give you a minus
sign for the current.
Step 3- Identify Positive and Negative sense for each component
+
-
+ +
+
+
The sense of + and – depends on your choice of current
Step 4- Draw Loops around circuit –
every resistor must lie on at least one loop
+
+ +
+
+
The directions of the loops is NOT IMPORTANT
It does NOT have to be in the direction of the currents!
Step 5 – Go around the loops to get the equationsNOTE – If we go from + to – potential change is NEGATIVE
If we go from - to + potential change is POSITIVE
+
+
+ +
13V −
13V − I 1 (1Ω ) − (I 1 − I 3 )(1Ω ) = 0
− I 2 (1Ω ) − (I 2 + I 3 )(2Ω ) + 13V = 0
− I 1 (1Ω ) − I 3 (1Ω ) + I 2 (1Ω ) = 0
+
Loop 1
Loop 2
Loop 3
Step 3- Solve the Simultaneous Equations
13V − I 1 (1Ω ) − (I 1 − I 3 )(1Ω ) = 0
Loop 1
− I 2 (1Ω ) − (I 2 + I 3 )(2Ω ) + 13V = 0
Loop 2
− I 1 (1Ω ) − I 3 (1Ω ) + I 2 (1Ω ) = 0
Loop 3
Solve Equation 3 for I2:
I 2 = I1 + I 3
Plug into Equations 1 and 2 (this gives 2 equations and 2 unknowns):
13V = I 1 (2Ω ) − I 3 (1Ω )
13V = I 1 (3Ω ) + I 3 (5Ω )
Multiply top equation by 5 and add two equations:
65V = I 1 (10Ω ) − I 3 (5Ω )
13V = I 1 (3Ω ) + I 3 (5Ω )
78V = I 1 (13Ω )
I1 = 6 A
I 3 = −1 A
Negative sign here means we
guessed wrong about direction of I3
I2 = 5 A
Time Dependent Circuits – The RC Circuit
Example 1 – Charging a Capacitor-
Capacitor has no net charge at t=0
Voltage and Current are functions of Time –
However – Kirchhoff’s Rules are still
valid at instant in Time
vab = iR
q
vbc =
C
Ohms Law
Definition of Capacitance
q
ε − iR − = 0
C
+
ε
q
i= −
R RC
Close Switch at t=0
Kirchhoff;s Loop Rule
At t=0, there is no charge on
the capacitor so at the “instant”
the switch is closed, we expect
a current of I0=ε/R.
As the capacitor charges, we
the current to fall.
ε
q
i= −
R RC
dq ε
q
= −
dt R RC
This is a “Differential” Equation –
Here are the steps to solve a Differential Equation –
1)
2)
3)
4)
5)
Guess a solution and check to see if it works.
Look it up in Math Book
Use “Mathematica” or some other numerical computer tool
Ask a Mathematician
If steps 1-4 fail…
Give up…Find another problem!
dq ε
q
= −
dt R RC
(
q = Cε 1 − e − t / RC
see page 999 in Y&F for solution
)
(
q = Cε 1 − e − t / RC
)
dq d
ε −t / RC
−t / RC
I=
= Cε 1 − e
= e
dt dt
R
(
I = I 0 e − t / RC
)
d Bt
e = Be Bt
dt
Time Dependent Circuits – The RC Circuit
Example 1 – Charging a Capacitor-
Capacitor has no net charge at t=0
Voltage and Current are functions of Time –
However – Kirchhoff’s Rules are still
valid at instant in Time
vab = iR
q
vbc =
C
Ohms Law
Definition of Capacitance
q
ε − iR − = 0
C
+
ε
q
i= −
R RC
Close Switch at t=0
Kirchhoff;s Loop Rule
At t=0, there is no charge on
the capacitor so at the “instant”
the switch is closed, we expect
a current of I0=ε/R.
As the capacitor charges, we
the current to fall.
ε
q
i= −
R RC
dq ε
q
= −
dt R RC
This is a “Differential” Equation –
Here are the steps to solve a Differential Equation –
1)
2)
3)
4)
5)
Guess a solution and check to see if it works.
Look it up in Math Book
Use “Mathematica” or some other numerical computer tool
Ask a Mathematician
If steps 1-4 fail…
Give up…Find another problem!
dq ε
q
= −
dt R RC
(
q = Cε 1 − e
− t / RC
see page 999 in Y&F for solution
)
τ = RC
Time Constant
(
q = Cε 1 − e − t / RC
)
ε −t / RC
dq d
−t / RC
I=
= Cε 1 − e
= e
dt dt
R
(
I = I 0 e − t / RC
)
d Bt
e = Be Bt
dt
Example 2 – Discharging a CapacitorMr. Kirchhoff says:
+
q = Q0 e − t / RC
q
iR + = 0
C
I = I 0 e − t / RC
Exponential Decay – Which Curve is right?
Strategy for understanding which curve is correct:
Try to figure out what the current (or charge) would be just when the
switch is closed (or opened)
t=0
Then try to figure out what the current (or charge) would be after
a very long time
t=∞
At t=0 there is NO charge on the capacitor –
and therefore no voltage across capacitor
I0
I0=ε/R
At t=∞ the capacitor is fully charged and there
is no current:
I0
I∞=0
At t=0 there is NO charge on the capacitor –
therefore
Q0=0
At t=∞ the capacitor is fully charged and therefor
Q∞=CV
QI∞
0
End of Chapter 26
You are responsible for the material covered in T&F Sections 26.1-26.4
You are expected to:
•
Understand the following terms:
Equivalent Resistance, Junction, Kirchhoff junction rule, Kirchhoff’s loop rule, RC circuit,
RC time constant
•
Be able to calculate equivalent resistances in simple geometries
•
Be able to apply Kirchoff’s rules to determine currents and voltages in more complex
geometries
•
Determine the current and voltage as a function of time in R-C circuits.
Recommended F&Y Exercises chapter 26:
•
4, 6, 11, 12, 22,24, 44,46, 48