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General Physics (PHY 2140)
Lecture 10
 Electrodynamics
Direct current circuits
 parallel and series connections
 Kirchhoff’s rules
Chapter 18
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1
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
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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2

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).
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 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
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


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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

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
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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
6
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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


Two Basic Principles:
Conservation of Charge
Conservation of Energy
a

Resistance Networks
Vab  IReq
Req
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Vab

I
I
Req
b
8
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
Req 
V IR1  IR2

 R1  R2
I
I
Req  R1  R2
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

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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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 _
with
V  V1  V2
V1  IR1 and V2  IR2
v1
Then,
V  I  R1  R2  , so I 
C
Thus,
B
R2
+
v _
+
Ii1
R1
_
V1  V
V
R1  R2
R1
R1  R2
This circuit is known as voltage divider.
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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
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1
1
1
 
Req R1 R2
or
Req 
R1 R2
R1  R2
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

Analogous formula is true for any number of
resistors,
1
1 1
1
    ... (parallel combination)
Req R1 R2 R3
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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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.
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Find the currents I1 and I2 and the voltage Vx in the circuit shown below.
I
Strategy:
7
+
20 V
+
_
Vx
_
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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.
15
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
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•
•
1.
2.
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:
The sum of currents entering any junction must equal the
sum of the currents leaving that junction (current or
junction rule) .
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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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
I a + 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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As a consequence of the Law of the conservation of energy, we have:
•
1.
Assign symbols and directions of currents in the loop
–
2.
The sum of the potential differences across all the
elements around any closed loop must be zero.
If the direction is chosen wrong, the current will come out with a right
magnitude, but a negative sign (it’s ok).
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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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
-
-
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v8
+
+
+
+ v11 -
-
v9 +
20
“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
+
-
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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
21
Example: For the circuit below find I, V1, V2, V3, V4 and the power
supplied by the 10 volt source.
30 V
+
+
_
V1
_
10 V
_
20 
1.
"a"

+
_
_
15 
V3
40 
I
+
V2
+
5
_
_
+
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
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
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Solving this equation gives, I = 0.5 A.
(3)
22
30 V
+
+
_
V1
_
10 V
_
20 

+
_
V3
Using this value of I in Eq. 2 gives:
"a"
_
15 
40 
I
+
V2
V1 = - 10 V
V3 = - 7.5 V
V2 = 20 V
V4 = 2.5 V
+
5
_
+
V4
_
+
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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
Consider the circuit
switch
E
i
t
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
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t RC

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
0.678041728
0.4
0.60.3219583
0.8
0.698805788 0.3011942
0.718230711 0.2817693
0.736402862 0.2635971
t
i
1
1.2
RC is called the time constant
24

Discharge the capacitor
t
switch
E
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
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
0.321958272
0.4
0.60.3219583
0.8
0.301194212 0.3011942
0.281769289 0.2817693
0.263597138 0.2635971
t
i
1
1.2
q  Qet RC
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