Transcript Document

General Physics (PHY 2140)
Lecture 6
 Electrodynamics
Direct current circuits
 parallel and series connections
 Kirchhoff’s rules
 RC circuits
http://www.physics.wayne.edu/~alan/2140Website/Main.htm
Chapter 18
7/17/2015
1
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.
Hours of operation:
Monday and Tuesday
Wednesday and Thursday
Friday, Saturday and Sunday
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:
7/17/2015
Monday, May 21 to Thursday, August 2, 2007.
2
Lightning Review
Last lecture:
1. Current and resistance


Q
I
t
Temperature dependence of resistance
Power in electric circuits
R  Ro 1   T  To  
 V 
P  I V  I 2 R 
2
R
2.
DC Circuits

EMF

Resistors in series
7/17/2015
V  E  Ir
Req  R1  R2  R3  ...
3
Introduction: elements of electrical circuits
A branch: A branch is a single electrical element or device (resistor, etc.).





A circuit with 5 branches.
A junction: A junction (or node) is a connection point between two or more
branches.





A circuit with 3 nodes.
If we start at any point in a circuit (node), proceed through connected
electric devices back to the point (node) from which we started, without
crossing a node more than one time, we form a closed-path (or loop).
7/17/2015
4
18.1 Sources of EMF
 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
 Electromotive Force (EMF)
• provides increase in potential E
• converts some external form of energy into electrical energy
 Single emf and a single resistor: emf can be thought of as a
“charge pump” V = IR
I
+ -
V = IR = E
E
7/17/2015
5
EMF
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 V is
V  E  Ir
B
C
r
E
A
R
D
Note: E is the same as the
terminal voltage when the
current is zero (open circuit)
7/17/2015
6
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
V  E  Ir  IR, or
E  Ir  IR
Thus, the current in the circuit is
I
7/17/2015
E
Rr
B
C
r
R
E
A
D
Power output:
I E  I 2r  I 2 R
Note: we’ll assume r negligible unless otherwise is stated
7
Measurements in electrical circuits
Voltmeters measure Potential Difference (or voltage) across
a device by being placed in parallel with the device.
V
Ammeters measure current through a device by being
placed in series with the device.
A
7/17/2015
8
Direct Current Circuits
Two Basic Principles:

Conservation of Charge

Conservation of Energy
a
I
Resistance Networks
Ohm’s Law:
Vab  IReq
Vab
Req 
I
7/17/2015
Req
b
9
18.2 Resistors in series
A
+
v2 _
B
R2
+
v _
+
Ii1
R1
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,
I1  I 2  I
v1
_
C
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,
V  IR1  IR2
By definition,
V  IReq
Thus, Req would be
V IR1  IR2
Req 

 R1  R2
I
I
Req  R1  R2
7/17/2015
10
Resistors in series: notes
Analogous formula is true for any number of resistors,
Req  R1  R2  R3  ...
(series combination)
It follows that the equivalent resistance of a series
combination of resistors is greater than any of the
individual resistors
7/17/2015
11
Resistors in series: example
In the electrical circuit below, find voltage across the resistor R1 in terms of
the resistances R1, R2 and potential difference between the battery’s
terminals V.
Energy conservation implies:
A
+
v2 _
B
with
R2
+
v _
+
Ii1
R1
v1
Then,
C
Thus,
_
V  V1  V2
V1  IR1 and V2  IR2
V
V  I  R1  R2  , so I 
R1  R2
V1  V
R1
R1  R2
This circuit is known as voltage divider.
7/17/2015
12
18.3 Resistors in parallel
I
A
I2
+
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,
R1
+
V
V1  V2  V
Req
2. Because of the charge conservation,
current, entering the junction A, must
equal the current leaving this junction,
_
I  I1  I 2
V
Req
By definition,
I
Thus, Req would be
V1 V2 V V
V
I




Req R1 R2 R1 R2
7/17/2015
1
1
1


Req R1 R2
or
Req 
R1R2
R1  R2
13
Resistors in parallel: notes
Analogous formula is true for any number of resistors,
1
1 1
1
    ...
Req R1 R2 R3
(parallel combination)
It follows that the equivalent resistance of a parallel
combination of resistors is always less than any of the
individual resistors
7/17/2015
14
Resistors in parallel: example
In the electrical circuit below, find current through the resistor R1 in terms of
the resistances R1, R2 and total current I induced by the battery.
Charge conservation implies:
I  I1  I 2
I
+
I2
I1
with
V
_
R2
R1
Then,
Thus,
V
V
, and I 2 
R1
R2
IReq
RR
I1 
, with Req  1 2
R1
R1  R2
I1 
I1  I
R2
R1  R2
This circuit is known as current divider.
7/17/2015
15
Direct current circuits: example
Find the currents I1 and I2 and the voltage Vx in the circuit shown below.
I
Strategy:
7
+
20 V
+
_
Vx
_
7/17/2015
I2
I1
4
12 
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.
16
Direct current circuits: example
Find the currents I1 and I2 and the voltage Vx in the circuit shown below.
I
7
+
20 V
+
_
Vx
I2
I1
4
12 
_
Then find current I by,
First find the equivalent resistance seen
by the 20 V source:
4(12)
Req  7 
10 
12  4
20V 20V
I 

 2A
Req
10
We now find I1 and I2 directly from the current division rule:
I1 
2 A(4)
 0.5 A, and I 2  I  I1  1.5 A
12  4
Finally, voltage Vx is Vx  I 2  4  1.5 A  4  6V
7/17/2015
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
E=18V
+
R1=4
R2=3
7/17/2015
R3=6
18
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=4
R1=4
+
E=18V
R2=3
E=18V
R3=6
+
Req=6
R23=2
So, Ieq= E/Req=3A
P=IeqE = 108W
7/17/2015
19
18.4 Kirchhoff’s rules and DC currents
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’s rules:
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 closed-circuit loop must be zero
(voltage or loop rule).
7/17/2015
20
1. Junction rule
As a consequence of the Law of the conservation of charge, we have:
•
The sum of the currents entering a node (junction point)
equal to the sum of the currents leaving.
Ia
Id
Ic
Ib
Ia + Ib = Ic + Id
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.
7/17/2015
21
2. Loop rule
As a consequence of the Law of the conservation of energy, we have:
•
The sum of the potential differences across all the
elements around any closed loop must be zero.
1. Assign symbols and directions of currents in the loop

If the direction is chosen wrong, the current will come out with the
correct magnitude, but a negative sign (it’s ok).
2. Choose a direction (cw or ccw) for going around the loop.
Record drops and rises of voltage according to this:

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

If EMF is traversed “from + to – ”: - E


7/17/2015
22
Simplest Loop Rule Example:
Single loop, start at point A
Battery traversed from – to +

So use + E
B
So use – IR (a voltage drop)
For the loop we have:
+
E
-
0  E  IR
A
7/17/2015
C
(a voltage gain)
Resistor traversed from + to –

I
+
R
-
D
23
Loop rule: more complex illustration
Loops can be chosen arbitrarily. For example, the circuit below contains a
number of closed paths. Three have been selected for discussion.
Suppose that for each element, respective current flows from + to - signs.
-
+ v 2
- v5 +
-
v1
v4
+
v6
+
-
v3
Path 1
+
Path 2
+ v7 -
+
Path 3
v12
v10
-
-
7/17/2015
v8
+
+
+
+ v11 -
-
v9 +
24
Loop rule: illustration
“b”
•
-
Using sum of the drops = 0
+ v 2
-
v1
v4
+
v3
v12
v10
-
+ v11 -
+ v7 - v10 + v9 - v8 = 0
• “a”
v8
+
+
+
Blue path, starting at “a”
+
+ v7 -
+
v6
+
-
7/17/2015
- v5 +
-
v9 +
Red path, starting at “b”
-v2 + v5 + v6 + v8 - v9 + v11
+ v12 - v1 = 0
Yellow path, starting at “b”
- v2 + v5 + v6 + v7 - v10 + v11
+ v12 - v1 = 0
25
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.
30 V
+
V3
15 
V1
_
_
20 
V4
1.
"a"

+
40 
I
5
10 V
_
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.
-10 - 20I + 30 - 15I - 5I + 20 - 40I = 0
7/17/2015
Solving this equation gives, I = 0.5 A
(3)
26
Kirchhoff’s Rules: Single-loop circuits (cont.)
30 V
+
-
_
V1
20 
10 V
+
_

+
+
V3
Using this value of I in Eq. 2 gives:
"a"
V1 = 10 V
_
15 
40 
I
-
V2
+
5
_
+
V4
_
V3 = 7.5 V
V2 = 20 V
V4 = 2.5 V
+
20 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.
7/17/2015
27
18.5 RC circuits
Charge across capacitor
When switch is closed, current flows because
capacitor is charging
Q
0.63 Q
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
RC is called the time constant
7/17/2015
28

Discharging the capacitor in RC circuit
Charge across capacitor
If a capacitor is charged and the switch is closed, then
current flows and the voltage on the capacitor gradually
decreases.
Q
0.37Q
This leads to decreasing charge
7/17/2015
q  Qe
t RC
29
Example : charging an unknown capacitor
A series combination of a 12 k 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 capacitance of
the capacitor.
7/17/2015
30
A series combination of a 12 k 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 capacitance of the
capacitor.
E
I
Given:
R =12 k
E = 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
7/17/2015
t
 V
 log 1  
RC
 E
t
1s

 46.5 F
 V
 10 V 
R log 1  
12, 000   log 1 

31
E


12
V


Chapter 19: Magnetism
Magnetic field pattern for a horse-shoe magnet
7/17/2015
32