Department of Physics and Astronomy
announces the Spring-Summer 2007 opening of
The Physics Resource Center
on Monday, May 21 in
Room 172 of Physics Research Building.
General Physics (PHY 2140)
Lecture 6
Hours of operation:
operation:
¾ Electrodynamics
Monday and Tuesday
Wednesday and Thursday
Friday, Saturday and Sunday
9Direct current circuits
9 parallel and series connections
9 Kirchhoff’s rules
9 RC circuits
10:00 AM to 5:00 PM
10:00 AM to 4:00 PM
Closed
Undergraduate students taking PHY1020, PHY2130, PHY2140, PHY2170/2175 and
PHY2180/2185 will be able to get assistance in this Center with their homework,
labwork and other issues related to their physics course.
The Center will be open:
Monday, May 21 to Thursday, August 2, 2007.
http://www.physics.wayne.edu/~alan/2140Website/Main.htm
Chapter 18
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1
Lightning Review
2
Introduction: elements of electrical circuits
A branch:
branch: A branch is a single electrical element or device (resistor, etc.).
etc.).
Last lecture:
1. Current and resistance
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I=
ΔQ
Δt
b
b
b
9 Temperature dependence of resistance R = Ro ⎡1 + α (T − To ) ⎤
⎣
⎦
9 Power in electric circuits
2.
DC Circuits
9 Resistors in series
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( ΔV )
R=
b
A circuit with 5 branches.
2
A junction:
junction: A junction (or node) is a connection point between two or more
branches.
b
R
•
ΔV = E − Ir
b
•
9 EMF
P = I ΔV = I
2
b
b
•
Req = R1 + R2 + R3 + ...
A circuit with 3 nodes.
If we start at any point in a circuit (node), proceed through connected
connected
electric devices back to the point (node) from which we started,
started, without
crossing a node more than one time,
time, we form a closedclosed-path (or loop).
loop).
3
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4
1
18.1 Sources of EMF
EMF
¾ Steady current (constant in magnitude and direction)
• requires a complete circuit
• path cannot be only resistance
cannot be only potential drops in direction of current flow
¾ Electromotive Force (EMF)
• provides increase in potential E
• converts some external form of energy into electrical energy
¾ Single emf and a single resistor: emf can be thought of as a
“charge pump” V = IR
+ -
I
Each real battery has some
internal resistance
AB: potential increases by E
on the source of EMF, then
decreases by Ir (because of
the internal resistance)
Thus, terminal voltage on the
battery ΔV is
B
C
r
E
A
ΔV = E − Ir
R
D
Note: E is the same as the
terminal voltage when the
current is zero (open circuit)
V = IR = E
E
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5
ΔV = E − Ir = IR, or
E = Ir + IR
Thus, the current in the circuit is
E
I=
R+r
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6
Measurements in electrical circuits
EMF (continued)
Now add a load resistance R
Since it is connected by a
conducting wire to the battery →
terminal voltage is the same as
the potential difference across the
load resistance
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Voltmeters measure Potential Difference (or voltage) across
a device by being placed in parallel with the device.
B
C
V
r
R
E
A
Ammeters measure current through a device by being
placed in series with the device.
D
Power output:
I E = I 2r + I 2 R
Note: we’ll assume r negligible unless otherwise is stated
A
7
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8
2
18.2 Resistors in series
Direct Current Circuits
¾
¾
A
Two Basic Principles:
Conservation of Charge
Conservation of Energy
+
v2 _
R2
+
+
v _
Ii1
R1
Resistance Networks
V ab = IR eq
R eq
V
≡ ab
I
I1 = I 2 = I
v1
_
a
Ohm’s Law:
1. Because of the charge conservation, all
charges going through the resistor R2 will
also go through resistor R1. Thus, currents
in R1 and R2 are the same,
B
C
I
Req
2. Because of the energy conservation, total
potential drop (between A and C) equals to
the sum of potential drops between A and B
and B and C,
ΔV = IR1 + IR2
b
By definition,
ΔV = IReq
Thus, Req would be
Req ≡
ΔV IR1 + IR2
=
= R1 + R2
I
I
Req = R1 + R2
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9
10
Resistors in series: example
Resistors in series: notes
In the electrical circuit below, find voltage across the resistor
resistor R1 in terms of
the resistances R1, R2 and potential difference between the battery’
battery’s
terminals V.
Analogous formula is true for any number of resistors,
Req = R1 + R2 + R3 + ...
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Energy conservation implies:
(series combination)
A
It follows that the equivalent resistance of a series
combination of resistors is greater than any of the
individual resistors
+
v _
v2 _
+
R2
with
V = V1 + V2
V1 = IR1 and V2 = IR2
v1
Then,
V = I ( R1 + R2 ) , so I =
C
Thus,
B
+
Ii1
R1
_
V1 = V
V
R1 + R2
R1
R1 + R2
This circuit is known as voltage divider.
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11
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12
3
18.3 Resistors in parallel
A
I
+
I1
+
V
R2
_
1. Since both R1 and R2 are
connected to the same battery,
potential differences across R1 and
R2 are the same,
I
I2
V
R1
Resistors in parallel: notes
Req
_
1
1 1
1
= + + + ...
Req R1 R2 R3
V1 = V2 = V
2. Because of the charge conservation,
current, entering the junction A, must
equal the current leaving this junction,
I = I1 + I 2
By definition,
I=
V
Req
Thus, Req would be
I=
V V
V
V V
= 1+ 2 =
+
Req R1 R2 R1 R2
1
1
1
=
+
Req R1 R2
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Analogous formula is true for any number of resistors,
Req =
or
It follows that the equivalent resistance of a parallel
combination of resistors is always less than any of the
individual resistors
R1 R2
R1 + R2
13
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In the electrical circuit below, find current through the resistor
resistor R1 in terms of
the resistances R1, R2 and total current I induced by the battery.
Find the currents I1 and I2 and the voltage Vx in the circuit shown below.
I
Charge conservation implies:
+
I = I1 + I 2
V
_
I2
R2
I1
R1
I1 = I
Strategy:
7Ω
+
V
V
, and I 2 =
with I1 =
R1
R2
IReq
RR
, with Req = 1 2
I1 =
Then,
R1
R1 + R2
Thus,
14
Direct current circuits: example
Resistors in parallel: example
I
(parallel combination)
20 V
+
_
Vx
_
I2
I1
4Ω
12 Ω
1.
Find current I by finding the
equivalent resistance of the circuit
2.
Use current divider rule to find the
currents I1 and I2
3.
Knowing I2, find Vx.
R2
R1 + R2
This circuit is known as current divider.
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15
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16
4
Direct current circuits: example
Example:
Determine the equivalent resistance of the circuit as shown.
Determine the voltage across and current through each resistor.
Determine the power dissipated in each resistor
Determine the power delivered by the battery
Find the currents I1 and I2 and the voltage Vx in the circuit shown below.
I
7Ω
+
+
_
20 V
Vx
I2
I1
4Ω
12 Ω
_
First find the equivalent resistance seen
by the 20 V source:
Req = 7Ω +
4Ω(12Ω)
=10 Ω
12Ω + 4Ω
E=18V
+
R1=4Ω
R2=3Ω
Then find current I by,
R3=6Ω
20V 20V
I =
=
= 2A
10Ω
Req
We now find I1 and I2 directly from the current division rule:
I1 =
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2 A(4Ω)
= 0.5 A, and I 2 = I − I1 = 1.5 A
12Ω + 4Ω
Finally, voltage Vx is Vx = I 2 ( 4Ω ) = 1.5 A ( 4Ω ) = 6V
17
Example:
Determine the equivalent resistance of the circuit as shown.
Determine the voltage across and current through each resistor.
Determine the power dissipated in each resistor
Determine the power delivered by the battery
+
R1=4Ω
E=18V
R2=3Ω
E=18V
+
Req=6Ω
18
18.4 Kirchhoff’s rules and DC currents
+
R3=6Ω
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The procedure for analyzing complex circuits is based on
the principles of conservation of charge and energy
They are formulated in terms of two Kirchhoff’
Kirchhoff’s rules:
R1=4Ω
R23=2Ω
So, Ieq= E/Req=3A
1. The sum of currents entering any junction must equal the
sum of the currents leaving that junction (current or
junction rule) .
2. The sum of the potential differences across all the
elements around any closedclosed-circuit loop must be zero
(voltage or loop rule).
P=IeqE = 108W
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19
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20
5
1. Junction rule
2. Loop rule
As a consequence of the Law of the conservation of charge, we have:
have:
As a consequence of the Law of the conservation of energy, we have:
have:
The sum of the potential differences across all the
The sum of the currents entering a node (junction point)
• elements around any closed loop must be zero.
• equal to the sum of the currents leaving.
Ia
1. Assign symbols and directions of currents in the loop
Ic
Id
„
Ia + Ib = Ic + Id
Ib
2. Choose a direction (cw
(cw or ccw)
ccw) for going around the loop.
Record drops and rises of voltage according to this:
„
Similar to the water flow in a pipe.
„
„
I a, I b, I c , and I d can each be either a positive
or negative number.
„
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21
„
For the loop we have:
I
B
So use + E (a voltage gain)
So use – IR (a voltage drop)
If EMF is traversed “from + to – ”: - E
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22
Loops can be chosen arbitrarily. For example, the circuit below contains a
number of closed paths. Three have been selected for discussion.
Single loop, start at point A
Battery traversed from – to +
Resistor traversed from + to –
If a resistor is traversed in the direction of the current: V = -IR
If a resistor is traversed in the direction opposite to the current: V=+IR
If EMF is traversed “from – to + ”: + E
Loop rule: more complex illustration
Simplest Loop Rule Example:
„
If the direction is chosen wrong, the current will come out with the
correct magnitude, but a negative sign (it’
(it’s ok).
+
E
-
0 = +E − IR
Suppose that for each element, respective current flows from + to - signs.
C
+
-
R
+ v 2
v4
v1
-
+
A
+
-
D
v3
v6
Path 1
+
Path 2
+ v7 -
+
Path 3
v10
-
-
23
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v8
+
+
+
v12
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- v5 +
-
+ v11 -
- v9 +
24
6
Loop rule: illustration
“b”
Kirchhoff’s Rules: Single-loop circuits
Example: For the circuit below find I, V1, V2, V3, V4 and the power
supplied by the 10 volt source.
Using sum of the drops = 0
•
+ v 2
v1
- v5 +
v4
+
+
-
v3
+ v7 -
+ v11 -
+
+ v7 - v10 + v9 - v8 = 0
V3
15 Ω
• “a”
20 Ω
- v2 + v5 + v6 + v7 - v10 + v11
+ v12 - v1 = 0
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"a"
•
_ +
40 Ω
V2
For convenience, we start at
point “a” and sum voltage
drops =0 in the direction of
the current I.
-10 - V1 + 30 - V3 - V4 + 20 - V2 = 0 (1)
_ +
20 V
2. We note that: V1 = 20I, V2 = 40I, V3 = 15I, V4 = 5I
(2)
3. We substitute the above into Eq. 1 to obtain Eq. 3 below.
Yellow path, starting at “b”
- v9 +
1.
10 V
I
V4
-v2 + v5 + v6 + v8 - v9 + v11
+ v12 - v1 = 0
V1
_
5Ω
Red path, starting at “b”
v8
+
+
v10
-
-
Blue path, starting at “a”
+
+
+
v12
30 V
v6
-10 - 20I + 30 - 15I - 5I + 20 - 40I = 0
25
Kirchhoff’s Rules: Single-loop circuits (cont.)
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(3)
Solving this equation gives, I = 0.5 A
26
18.5 RC circuits
When switch is closed, current flows because
capacitor is charging
+
-
_
V1
20 Ω
10 V
+
_ +
•
+
V3
Using this value of I in Eq. 2 gives:
"a"
V1 = 10 V
_
15 Ω
40 Ω
I
-
_
+
V4
V3 = 7.5 V
V2 = 20 V
V4 = 2.5 V
_ +
20 V
Q
0.63 Q
V2
+
5Ω
Charge across capacitor
30 V
P10(supplied) = -10I = - 5 W
(We use the minus sign in –10I because the current is entering the + terminal)
In this case, power is being absorbed by the 10 volt supply.
As capacitor becomes charged, the current
slows because the voltage across the resistor is
E - Vc and Vc gradually approaches E.
q = Q (1 − e −t RC )
Once capacitor is charged the current is zero
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27
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RC is called the time constant
28
7
Discharging the capacitor in RC circuit
Example : charging an unknown capacitor
Charge across capacitor
If a capacitor is charged and the switch is closed, then
current flows and the voltage on the capacitor gradually
decreases.
A series combination of a 12 kΩ
kΩ resistor and an unknown capacitor is
connected to a 12 V battery. One second after the circuit is
completed, the voltage across the capacitor is 10 V. Determine the
the
capacitance of the capacitor.
Q
0.37Q
This leads to decreasing charge
q = Qe −t RC
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29
5/30/2007
A series combination of a 12 kΩ
kΩ resistor and an unknown capacitor is
connected to a 12 V battery. One second after the circuit is completed,
completed, the
voltage across the capacitor is 10 V. Determine the capacitance of the
capacitor.
30
Chapter 19: Magnetism
E
I
Given:
R =12 kΩ
Ε = 12 V
V =10 V
t = 1 sec
C
Recall that the charge is
building up according to
R
q = Q (1 − e −t RC )
Q = CV
Thus the voltage across the capacitor changes as
Find:
V=
C=?
q Q
= (1 − e −t RC ) = E (1 − e −t RC )
C C
This is also true for voltage at t = 1s after the switch is closed,
V
V
= 1 − e −t RC ⇒ e −t RC = 1 − ⇒
E
E
C=−
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−t
⎛ V⎞
= log ⎜1 − ⎟
RC
⎝ E⎠
t
1s
=−
= 46.5μ F
⎛ V⎞
⎛ 10 V ⎞
R log ⎜ 1 − ⎟
12,
000
log
1
Ω
−
(
) ⎜
⎟
31
⎝ E⎠
⎝ 12 V ⎠
Magnetic field pattern for a horse-shoe magnet
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32
8