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3.1
I.
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21
ing
d Te ch n ol o
Voltage and Current Laws
ee
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12
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Un
VOLTAGE AND CURRENT LAWS
an
3
u
i
Ohm’s Law
• R = ρ Al
ρ=Resistivity in Ω.m
ρ for Silver=1.64e-8, Copper=1.72e-8, Carbon (semiconductor)=4e-5,
ρ for Silicon=6.4e2, Paper (insulator)=1e+10, Glass=1e+12
• Ohm’s Law: v ∝ i
v = Ri, R in Ω. 1Ω = 1V /A
• Short Circuit= resistance approaching zero
• Open Circuit= resistance approaching ∞
• Nonlinear resistors do not follow Ohm’s law.
• Conductance G =
1S = 1A/V .
1
R
measured in mho while in SI units it is Siemens.
• i = Gv
• Power dissipated in a resistor= p = vi =
v2
R
=
i2
G
• Power is a nonlinear function in i or v.
• Power dissipated in a resistor is a positive quantity ( R, v 2 are positive quantities).
• A resistor absorbs power. It is a passive quantity and can not generate energy
but it absorbs.
3.2
Nodes, Branches, and Loops
• Branch represents a single circuit element (resistor, voltage source, ...)
• Node is the point of connection between two or more branches.
• Loop is any closed path in a circuit.
3
VOLTAGE AND CURRENT LAWS
13
• Series elements: two elements are said to be in series if they
Exclusively share a single node and hence carry the same current.
• Parallel elements: two elements are said to be in parallel if they are connected
to the same two nodes and hence have the same voltage across them.
3.3
Kirchhoff’s Laws
• KCL: the algebraic sum of currents entering a node (or a closed boundary) is
zero.
PN
n=1 in = 0
Convention: Entering current is +ve, leaving current is -ve.
Examples:
1. Ex.3.1/p31
2. PP3.1/p32
• KVL: the algebraic sum of voltages around a closed path (or loop) is zero.
PN
m=1 vm = 0
Sum of voltage drops=sum of voltage rises
Examples:
1. Ex.3.2/p33
2. Ex3.3/p34
3. Ex3.4/p35
4. PP3.3/p36
3.4
Series Resistors and Voltage Division
Equivalent resistance of any number of resistors connected in series is the sum of
the individual resistances
3.4.1
Parallel Resistors and Current Division
The equivalent resistance of two parallel resistors is equal to the product of their
resistances divided by their sum
3
VOLTAGE AND CURRENT LAWS
3.4.2
14
Wye-Delta
1
Rc
1
3
Rb
R1
Ra
2
3
R3
4
wye-delta
R2
4
2
wye-delta
It is required to have
1. R1−2 = R1 + R3 = Rb ||(Ra + Rc )
2. R1−3 = R1 + R2 = Rc ||(Ra + Rb )
3. R3−4 = R2 + R3 = Ra ||(Rb + Rc )
Here we used superposition theorem (linearity) to derive the above relations as
follows: suppose that three generators are connected to terminals 1-2, 1-3, and 2-3.
If only the first generator is working and the other two removed from the circuit,
then the generator will see R1−2 = R1 +R3 = Rb ||(Ra +Rc ). Similarly for the second
and third generators to produce the other two relations.
the following Maxima code defines the above three equations:
R11R(R1,R2):=R1*R2/(R1+R2);
eq1: R1+R3=R11R(Rb,Ra+Rc);
eq2: R1+R2=R11R(Rc,Ra+Rb);
eq3: R2+R3=R11R(Ra,Rc+Rb);
and the following Maxima command will solve for R1 , R2, R3 in terms of Ra , Rb, Rc.
linsolve([eq1,eq2,eq3],[R1,R2,R3]);
which gives
Rb Rc
Ra Rc
Ra Rb
R1 =
, R2 =
, R3 =
Rc + Rb + Ra
Rc + Rb + Ra
Rc + Rb + Ra
and the following Maxima command will solve for Ra , Rb , Rc in terms of R1 , R2, R3.
linsolve([eq1,eq2,eq3],[Ra,Rb,Rc]);
which gives
(R2 + R1) R3 + R1 R2
(R2 + R1 ) R3 + R1 R2
(R2 + R1) R3 + R1 R2
Ra =
, Rb =
, Rc =
R1
R2
R3
3
VOLTAGE AND CURRENT LAWS
15
3.4.3
The Single Loop Circuit
3.4.4
The Single Node-Pair Circuit
3.4.5
Series and Parallel Connected Independent Sources
• Ideal voltage sources in parallel are permissible only when each has the same
terminal voltage at every instant.
• Ideal current sources in series are permissible only when each has the same
current, including sign, at every instant of time.
3.4.6
Resistors in Series and Parallel
3.4.7
Voltage and Current Division
3.4.8
Practical Application
Three grounds: Earth ground, Signal ground, and Chassis ground.
Do not connect chassis ground to the Neutral wire coming from the 220V supply
since the voltage drop across the Neutral wire might not be too small.
4.7kΩ
4.7kΩ
9V +
−
4.7kΩ
+
vo
9V +
−
−
4.7kΩ
hkd7ech3earth1.m4
+
vo
−
hkd7ech3earth.m4
+
Earth ... Signal ... Chassis
Different Grounds
hkd7ech3earth2.m4
RLiveW ire
− vL +
vE Requipment +
− 220V
−
+ vN −
RN eutralW ire
vL + vE + vN = 220V
hkd7ech3earth3.m4
3
VOLTAGE AND CURRENT LAWS
3.4.9
16
Summary and Review
• KCL: sum of entering (+) and leaving (-) currents at any node=0
• KVL: sum of voltage rises and voltage drops in any loop=0
• Series elements
• Parallel elements
• Series resistors: Rs = R1 + R2
• Parallel resistors: Rp = R1 ||R2 =
R1 R2
R1 +R2
• Voltage sources in series : can be of any magnitude or polarity. voltage sources
in parallel must have the same magnitude and polarity
• Current sources in parallel: can be of any magnitude or direction. current
sources in series must have the same magnitude and direction
• Voltage division: v is the voltage across two series resistors R1 , R2,
i
+
R1
+ v1 −
+
v2 R2
−
v
−
series2r.m4
−v + v1 + v2 = 0, v1 = iR1, v2 = iR2. KVL: −v + iR1 + iR2 Substitute to get
2
1
v and v2 = R1R+R
v.
v1 = R1R+R
2
2
• Current division: i is the current flowing through two parallel resistors R1, R2 ,
i
i1
+
v
R1
i2
R2
−
parallel2r.m4
the current flowing in R1 equals i1 =
R2
R1 +R2 i
and i2 =
R1
R1 +R2 i
3
VOLTAGE AND CURRENT LAWS
3.5
17
Bridge Measurement
Here we have one resistor which we need to measure its value R1, connect the circuits as shown, where R2 , R3 have fixed values while R2 is adjustable. Vary R2 until
the Ammeter reads zero current.
Note that an Ammeter is equivalent to an ideal voltage source (zero internal resistance) having zero voltage.
1 Rs
2
R1
+
V1 +
− 3
−
R3
R2
V =0
ix 2
+−
4
R4
bridge.m4
0
when i32 = 0, then i1 = i2 and i3 = i4.
v12 = R1 i1 = R3 i3 and v20 = R2 i2 = R4 i4 which gives
R3
R1
=
R2
R4
3.5.1
Example
Verify using PSPICE that i32 = 0 for the above circuit given that V1 = 10, R1 =
10, R2 = 20, R3 = 7, R4 = 14.
bridge.cir: check balance in a bridge
v1
1
0
dc
10
rs
1
2
0.01
r1
2
3
10
r2
2
4
20
r3
3
0
7
r4
4
0
14
v2
3
4
dc
0
.op
.end
3
VOLTAGE AND CURRENT LAWS
18
and the output of the simulation
bridge.cir: check balance in a bridge
NODE
VOLTAGE
NODE
VOLTAGE
NODE
VOLTAGE
(
1)
10.0000 (
2)
9.9912 (
3)
4.1140
NODE
VOLTAGE
(
4)
4.1140
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
v1
v2
-8.816E-01
0.000E+00
TOTAL POWER DISSIPATION
8.82E+00 WATTS
As a homework exercise, type the above Spice file and check the bridge for the
values: r1 = 12, r2 = 35, r3 = 1.5k, r4 = 4.4k, v1 = 12.5 for being balanced or not?
If the bridge is not balanced, what is the current flowing in the Ammeter?
3
VOLTAGE AND CURRENT LAWS
3.6
19
Using PSpice to find the value of a resistor that satisfies some criterion
If we address the problem as follows:
Given V1 = 10, R1 = 10, R2 = 20, R4 = 17, use PSpice to find R3 that balances the
bridge?, i.e. the current i32 = 0.
Here we need to define R3 as a variable parameter. The parameter needs to be
assigned an initial value which can be used with the command .op, then we need to
.dc sweep this parameter from vale, to value, step value
bridge_param.cir
*use PSpice find R that balances the bridge
*vary R: first define it as variable resistor, then use
*.dc param from to step
v1 1 0 dc 10
r1 1 2 10
r2 2 0 20
r3 1 3 {R}
r4 3 0 14
v2 3 2 dc 0 ;act as Ammeter
.param R=1 ;needed for the .op
.dc param R 2 10 0.1 ; from ... to ... step
.probe
.end
and using PROBE, Trace, Add, I(V2) we get
bridge_param.cir
Date/Time run: 03/06/107 23:13:19
Temperature: 27.0
(A) bridge_param
300mA
200mA
100mA
0A
-100mA
2
3
4
5
6
7
8
9
10
I(v2)
R
Date: March 06, 2007
Page 1
Figure 4: Bridge Measurement via PSpice
From the plot, i23 = 0 at R = 7Ω.
Time: 23:15:10