DC Circuits
•
•
•
•
Resistance Review
Following the potential around a circuit
Multiloop Circuits
RC Circuits
Homework for today:
Read Chapters 26, 27
Chapter 26 Questions 1, 3, 10
Chapter 26 Problems 1, 17, 35, 77
Homework for tomorrow:
Chapter 27 Questions 1, 3, 5
Chapter 27 Problems 7, 19, 49
WileyPlus assignment: Chapters 26, 27
Review: Series and Parallel Resistors
Series:
R = R1 + R2 + R3
Why?
Parallel:
1 / R = 1 / R1 + 1 / R2 + 1 / R3
Why?
Following the Potential
Study Fig. 27-4 in the text to see how the potential
changes from point to point in a circuit.
Note the net change around the loop is zero.
Following the Potential
Note the net change around the loop is zero.
Q.27-1
With the current i flowing as
shown, which is at the higher
potential, point b or point c?
1)
2)
3)
4)
B is higher
C is higher
They are the same
Not enough
information
Solution
With the current i flowing as shown, which is
at the higher potential, point b or point c?
Solution: Current flows from high to low
potential just like water flows down hill.
(1) b is higher
(3) they’re the same
(2) c is higher
(4) not enough info
Example: Problem 27-30
ε = 6.0V
R1 = 100 Ω
R2 = R3 = 50 Ω
R4 = 75 Ω
(a) Find the equivalent resistance of the network.
(b) Find the current in each resistor.
i1
Problem 27-30 (part a)
ε = 6.0V
R234
R1 = 100 Ω
R2 = R3 = 50 Ω
R4 = 75 Ω
(a) Find the equivalent resistance of the network.
1 / R234 = 1 / R2 + 1 / R3 + 1 / R4 = 16 / 300
So R234 = 300 / 16 = 19 Ω
Now R1 and R234 are in series so
Req = R1 + R234 = 100 Ω + 19 Ω = 119 Ω
(b) Now current
i1 = ε / Req = 50 mA
Problem 27-30 (part b)
(b) Find the current in each resistor.
First note that i2R2 = i3R3 = i4R4.
V = iR234
V
So
i2 = V / R2 = .95 / 50 = 19 mA
i3 = V / R3 = .95 / 50 = 19 mA
i4 = V / R4 = .95 / 75 = 12 mA
= .050 A × 19 Ω
= 0.95 V
Check: These
three add up
to i1 = 50 mA.
Q.27-2
?
R2 = 2 Ω
R3 = 3 Ω
i2 = 6 A
i3 = ?
Calculate i3, find the closest single-digit number (0-9).
Q.27-2
?
R2 = 2 Ω
R3 = 3 Ω
i2 = 6 A
1)
2)
3)
4)
5)
6)
7)
8)
9)
1A
2A
3A
4A
5A
6A
7A
8A
9A
i3 = ?
Q.27-2
?
R2 = 2 Ω
R3 = 3 Ω
i2 = 6 A
Vb − Vc = i2 R2 = i3 R3
i2 R2 2
∴ i3 =
= i2 = 4 A
R3
3
4
More Complicated Circuits
How do we solve a problem with more
than one emf and several loops? We
can’t do it just by series and parallel
resistor combinations.
?
Rules for Multiloop Circuits
• The net voltage change around any loop is zero.
“Energy conservation”
• The net current into any junction is zero.
“Charge conservation”
Using these two rules we can always get
enough equations to solve for the currents if
we are given the emfs and resistances.
Example
i3
i2
ε
ε
1
= 24V
2
= 12V
R1 = 5 Ω
i1
R2 = R3 = 30 Ω
Find all currents!
First define unknowns: i1 , i2 , i3
Example (continued)
i3
i2
Left-hand loop:
ε
i1
1
− i3 R3 − i1 R1 = 0
Right-hand loop:
ε
2
− i2 R2 − i1 R1 = 0
Junction:
i1 = i2 + i3
Algebra: solve 3 equations for 3 unknowns i1 , i2 , i3
Loop and junction equations:
ε
ε
− i3 R3 − i1 R1 = 0
2 − i2 R2 − i1 R1 = 0
i1 = i2 + i3
1
Put in the given numbers and also replace i1 by i2+i3:
5i1 + 30i3 = 5i2 + 35i3 = 24
5i1 + 30i2 = 35i2 + 5i3 = 12
Solve two equations in two unknowns to get:
i2 = 250 mA
Add to get
i3 = 650 mA
i1 = i2 + i3 = 900 mA
Check by using outer loop:
i3
i2
ε
1
− i3 R3 + i2 R2 − ε 2
=0?
i1
24 − 30(.65 − .25) − 12
= 12 − 30 × .40
= 12 − 12
=0
i3
Repeat with a different R1
i2
i1
ε
ε
1
= 24 V
2
= 12V
R1 = 40 Ω
R2 = R3 = 30 Ω
Exercise for the student: Same equations give
negative i2 in this case! This means current
going downward through right-hand battery.
Back to Basics
• Examples that don’t involve so much algebra,
but focus on the ideas of current and voltage.
• Even though you have a multiloop circuit so
you need to write down the equations from the
loop rule and the junction rule, you may not
have to actually solve simultaneous equations.
Simpler Examples
Textbook homework
problem 27-19
Both these problems can be solved for one
unknown at a time, without messy algebra.
9 + 18 × I 2 = 0
I 2 = −0.5 A
12
I1 =
= 2.4 A
9 − 5 × I1 + 3 = 0
5
I 3 = I1 − I 2 = 2.9 A
Check:
Pin = 9 I 3 + 3 I1 = 33.3 W
2
2
Pout = 5 I1 + 18 I 2 = 33.3W
Discharging a Capacitor
Capacitor has charge Q0.
At time t=0, close switch.
What is charge q(t) for t>0?
Obviously q(t) is a function which decreases gradually,
approaching zero as t approaches infinity.
What function would do this?
q( t ) = Q0e
− t /τ
But what is the time constant ?
Analyze circuit equation: find
τ = RC
Charging a
Capacitor
V
q(t )
For small t, q=0 and i=V/R.
For large t, q=CV and i=0.
[
q( t ) = CV 1 − e
τ = RC
− t /τ
]
DC Circuits II
• Circuits Review
• RC Circuits
• Exponential growth and decay
Circuits review so far
• Resistance and resistivity
R= ρl/A
• Ohm’s Law and voltage drops
∆V = − iR
• Power and Joule heating
P = iV
• Resistors in series and parallel
• Loop and junction rules
Review: Series and Parallel Resistors
Series:
R = R1 + R2 + R3
Why?
Parallel:
1 / R = 1 / R1 + 1 / R2 + 1 / R3
Why?
Review: Rules for Multiloop Circuits
• The net voltage change around any loop is zero.
“Energy conservation”
• The net current into any junction is zero.
“Charge conservation”
Using these two rules we can always get
enough equations to solve for the currents if
we are given the emfs and resistances.
Review example
i3
i2
Left-hand loop:
ε
i1
1
− i3 R3 − i1 R1 = 0
Right-hand loop:
ε
2
− i2 R2 − i1 R1 = 0
Junction:
i1 = i2 + i3
Algebra: solve 3 equations for 3 unknowns i1 , i2 , i3
If any i < 0, current flows the opposite direction.
Simpler Examples
Textbook homework
problem 27-19
Both these problems can be solved for one
unknown at a time, without messy algebra.
Q.27-3
Resistors R1 and R2 are connected in series.
If R2 > R1, what can you say about the
resistance R of the combination?
1.
2.
3.
4.
R > R2
R2 > R > R1
R1 > R
None of the above
Q.27-3
• Resistors R1 and R2 are connected in series.
• R2 > R1.
• What can you say about the resistance R of
this combination?
Solution:
R = R1 + R2
so
R > R2
(1) R > R2
( 2) R2 > R > R1
( 3) R1 > R
( 4) None of the above
Q.27-4
Resistors R1 and R2 are connected in parallel.
If R2 > R1, what can you say about the
resistance R of the combination?
1.
2.
3.
4.
R > R2
R2 > R > R1
R1 > R
None of the above
Q.27-4
• Resistors R1 and R2 are connected in parallel.
• R2 > R1.
• What can you say about the resistance R of
this combination?
Solution:
1
1
1
=
+
R R1 R2
so
1
1
>
R R1
(1) R > R2
( 2) R2 > R > R1
( 3) R1 > R
( 4) None of the above
Discharging a Capacitor
Capacitor has charge Q0.
At time t=0, close switch.
What is charge q(t) for t>0?
Obviously q(t) is a function which decreases gradually,
approaching zero as t approaches infinity.
What function would do this?
q( t ) = Q0e
− t /τ
But what is the time constant ?
Analyze circuit equation: find
τ = RC
Discharging a Capacitor
+
i
−
But
dQ
i=−
dt
Sum voltage changes
around loop:
Q
Q / C − iR = 0 , i =
RC
dQ
Q
=−
dt
RC
Get differential
equation for Q(t):
Solution:
Q ( t ) = Q0e
− t /τ
Where is the time constant
τ = RC
Charging a
Capacitor
ε
q(t )
For small t, q=0 and i = ε / R
For large t, i=0 and q = Cε
But what are q( t ), i ( t ) ?
Charging a
Capacitor
i(t) →
Sum voltage changes:
ε − iR − Q / C = 0
dQ
i=
dt
Solution
ε
Get diff. eq.:
+Q(t)
-Q(t)
dQ ε Q
= −
dt R RC
t
/
τ
−


Q ( t ) = Cε  1 − e



τ = RC
Charging a Capacitor
See solution gives desired behavior:
−
/
t
τ


Q ( t ) = Cε  1 − e


→ 0 as t → 0
→ Cε as t → ∞
Exponential Growth and Decay
This simple differential equation occurs in
many situations:
dQ
= (Const .) Q
dt
If dQ/dt = +KQ, we have the “snowball” equation:
growth rate proportional to size. Population growth.
dQ
+ Kt
Q = Q0 e
= + KQ
dt
If dQ/dt = -KQ, we have rate of decrease proportional
to size. For example radioactive decay.
dQ
− Kt
Q = Q0 e
= − KQ
dt
Try Exponential Solution
Exponential Growth
200
150
Q
We know we want a
result which increases
faster and faster. One
function which does this
is the exponential
function. So try that:
100
50
0
0
20
40
t
To solve
dQ
t /τ
= KQ try Q ( t ) = Q0e
dt
Questions:
1. Is this a solution?
2. If so, what is the “time constant” τ ?
60
To solve
dQ
t /τ
= KQ try Q ( t ) = Q0e
dt
Q ( t ) = Q0 e
t /τ
dQ
d t /τ Q0 t /τ Q
= Q0 e =
e =
dt
dt
τ
τ
dQ
But we want
= KQ
dt
So we DO have a solution IF
τ = 1/ K
Doubling Time
t /τ
If Q ( t ) = Q0 e
how long does it take for Q to double?
( t + ∆ t ) /τ
Q ( t + ∆t ) e
=
t /τ
Q(t )
e
∆t /τ
And e
= 2 if
So ∆t ≅ 0.7τ
=e
∆t /τ
∆t / τ = ln( 2) = 0.693
Radioactive Decay
For an unstable isotope, a certain fraction of the
atoms will disintegrate per unit time.
For
dQ
− t /τ
= − KQ use Q ( t ) = Q0 e
dt
Now τ is called the mean life, and the half-life
is T1/2 = τ ln(2) = time for half the remaining
atoms to disintegrate, and
T1 / 2 ≅ 0.7τ
Discharge of a Capacitor
Back to electricity. From the loop rule we got
dQ
Q
=−
= − KQ
dt
RC
So the solution is Q ( t ) = Q e − t /τ
0
But what are Q0 and τ?
Initial condition: Q0 = Q ( 0 )
Time constant:
τ = 1 / K = RC
Example
A 40 pF capacitor with a charge of 20 nC is
discharged through a 50 MΩ resistor.
( a ) What is the time constant?
τ = RC = 40 × 10
−12
× 50 × 10 = 2.0 × 10
6
−3
( b ) At what time will ½ the charge remain?
T1 / 2 = 0.7τ = 0.7 × 2.0 × 10
−3
s = 1.4 ms
( c ) How much charge will remain after 5 ms?
Q ( t ) = Q0e
− t /τ
= 20 × e
−2.5
= 1.64 nC
s
Circuits Summary
Things to remember about DC circuits:
•
•
•
•
•
•
•
Resistance and resistivity
R= ρl/A
Ohm’s Law and voltage drops
∆V = − iR
Power and Joule heating
P = iV
Resistors in series and parallel
Loop and junction rules
RC circuits: charging and discharging a capacitor
RC time constant
− t /τ
Q ( t ) = Q0e
Quiz tomorrow on Chapters 26,27.
τ = RC