General Physics (PHY 2140)
Lecture 10
¾ Electrodynamics
9Direct current circuits
9 parallel and series connections
9 Kirchhoff’s rules
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 18
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Department of Physics and Astronomy
announces the Fall 2003 opening of
The Physics Resource Center
on Monday, September 22 in
Room 172 of Physics Research Building.
Hours of operation:
Mondays, Tuesdays, Wednesdays
Thursdays and Fridays
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Undergraduate students taking PHY2130-2140 will be able to get assistance in this
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The Center will be open: Monday, September 22 to Wednesday, December 10, 2003.
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2
Lightning Review
Last lecture:
1. Current and resistance
∆Q
I=
∆t
9 Temperature dependence of resistance R = Ro 1 + α (T − To ) 


9 Power in electric circuits
( ∆V )
P = I ∆V = I 2 R =
2
R
Review Problem: Consider a moose
standing under the tree during the lightning
storm. Is he ever in danger? What could
happen if lightning hits the tree under which he
is standing?
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Introduction: elements of electrical circuits
A branch: A branch is a single electrical element or device (resistor, etc.).
b
b
b
b
b
A circuit with 5 branches.
A junction: A junction (or node) is a connection point between two or more
branches.
b
•
•
b
•b
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).
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18.1 Sources of 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
+ -
V = IR = E
E
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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
R
E
A
D
Note: E is the same as the
terminal voltage when the
current is zero (open circuit)
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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
E
I=
R+r
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B
C
r
R
E
D
A
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
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Direct Current Circuits
Two Basic Principles:
¾
Conservation of Charge
¾
Conservation of Energy
a
I
Resistance Networks
V ab = IR eq
R eq
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V ab
≡
I
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
=
= R1 + R2
Req ≡
I
I
Req = R1 + R2
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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
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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.
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18.3 Resistors in parallel
I
A
I2
+
V
R2
_
1. Since both R1 and R2 are
connected to the same battery,
potential differences across R1 and
R2 are the same,
I
I1
R1
+
V
Req
_
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
V
Req
By definition,
I=
Thus, Req would be
V1 V2 V V
V
=
+
=
+
I=
Req R1 R2 R1 R2
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1
1
1
=
+
Req R1 R2
or
Req =
R1 R2
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
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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
+
I = I1 + I 2
I2
I1
with
V
_
R2
R1
Then,
Thus,
V
V
, and I 2 =
R1
R2
IReq
RR
, with Req = 1 2
I1 =
R1
R1 + R2
I1 =
I1 = I
R2
R1 + R2
This circuit is known as current divider.
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Direct current circuits: example
Find the currents I1 and I2 and the voltage Vx in the circuit shown below.
I
Strategy:
7Ω
+
20 V
+
_
Vx
_
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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.
16
Direct current circuits: example
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
_
Then find current I by,
First find the equivalent resistance seen
by the 20 V source:
4Ω(12Ω)
Req = 7Ω +
= 10 Ω
12Ω + 4Ω
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
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).
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a. 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.
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b. 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 a right
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
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b. Loop rule: 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
-
v1
+
-
v3
- v5 +
v4
-
Path 1
+
Path 2
v6
+
+ v7 -
+
Path 3
v12
v10
-
-
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v8
+
+
+
+ v11 -
-
v9 +
21
b. Loop rule: illustration
“b”
•
-
Using sum of the drops = 0
+ v 2
-
v1
+
-
v3
v4
v10
-
+ v11 -
- v7 + v10 – v9 + v8 = 0
• “a”
v8
+
+
v12
Blue path, starting at “a”
+
+ v7 -
+
-
v6
+
+
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- 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
22
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
+
+
_
V1
20 Ω
_
10 V
_
"a"
•
+
_
V3
1.
_
15 Ω
40 Ω
I
+
V2
+
5Ω
+
_
V4
_
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
We note that: V1 = - 20I, V2 = 40I, V3 = - 15I, V4 = 5I
(2)
We substitute the above into Eq. 1 to obtain Eq. 3 below.
10 + 20I – 30 + 15I + 5I – 20 + 40I = 0
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Solving this equation gives, I = 0.5 A.
(3)
23
Kirchhoff’s Rules: Single-loop circuits (cont.)
30 V
+
+
_
V1
20 Ω
_
10 V
_
•
+
_
V3
Using this value of I in Eq. 2 gives:
"a"
_
15 Ω
40 Ω
I
+
V2
+
5Ω
+
_
V4
_
V1 = - 10 V
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.
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18.5 RC circuits
Consider the circuit
switch
t
(
i
q
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
0.22
0.24
0.26
0.28
0.3
0.32
0.340
0.36
0.38
0.4
q
C
+q -q
vc = q/C
R
vR = iR
q = Q (1 − e −t RC )
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0.2
i
0
1
0.064493015
0.935507
0.124826681 0.8751733
0.181269247 0.8187308
0.234071662 0.7659283
0.283468689 0.7165313
0.329679954
0.67032
0.372910915 0.6270891
0.41335378 0.5866462
0.451188364 0.5488116
0.486582881 0.5134171
0.519694699 0.4803053
0.550671036
0.449329
0.579649615 0.4203504
0.606759279 0.3932407
0.632120559 0.3678794
0.655846213 0.3441538
0.4
0.60.3219583
0.8
0.678041728
0.698805788 0.3011942
0.718230711 0.2817693
0.736402862 0.2635971
t
1
1.2
RC is called the time constant
25
RC circuits
Discharge the capacitor
t
switch
(
i
C
+q -q
vc = q/C
q
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
0.22
0.24
0.26
0.28
0.3
0.32
0.340
0.36
0.38
0.4
q
R
vR = iR
0.2
i
1
1
0.935506985
0.935507
0.875173319 0.8751733
0.818730753 0.8187308
0.765928338 0.7659283
0.716531311 0.7165313
0.670320046
0.67032
0.627089085 0.6270891
0.58664622 0.5866462
0.548811636 0.5488116
0.513417119 0.5134171
0.480305301 0.4803053
0.449328964
0.449329
0.420350385 0.4203504
0.393240721 0.3932407
0.367879441 0.3678794
0.344153787 0.3441538
0.4
0.60.3219583
0.8
0.321958272
0.301194212 0.3011942
0.281769289 0.2817693
0.263597138 0.2635971
t
1
1.2
q = Qe −t RC
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