Download 1. Ohm`s law

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Chapter 2. Resistive circuits
2014. 9. 12.
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Contents
1. Ohm’s law
2. Kirchhoff’s laws
3. Series and parallel resistor combinations
4. Y-Δ transformation
5. Circuits with dependent sources
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Resistors : microscopic view
nucleus
electrons
Entering resistive material, charges are decelerated, which
decrease current flow.
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Types of resistors
(1), (2), and (3) are high power resistors.
(4) and (5) are high-wattage fixed resistors.
(6) is a high precision resistor.
(7)–(12) are fixed resistors with different power
ratings.
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1. Ohm’s law
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resistance
 (t )  i (t )  R
1
G
R
Power absorption :
( R  0)
; conductance
p (t )   (t )  i (t )  i 2 R 
2
R
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Example 2.1
Determine the current and the power absorbed by the resistor.
I 
12
 6 [mA]
2k
P  VI  (12)( 6  103 )  0.072 [W ]
 I 2 R  (6  103 ) 2 ( 2k )
 V 2 / R  (12) 2 / 2k
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Glossary
(1) Node
A node is simply a point of connection of two or more circuit elements.
node
Although one node can be spread out with perfect
conductors, it is still only one node
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(2) loop
(3) branch
A loop is simply any closed path through the circuit in which no node
is encountered more than once
a branch is a single or group of components such as resistors or a
source which are connected between two nodes
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2. Kirchhoff’s law
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(1) Kirchhoff ’s current law (KCL) :
the algebraic sum of the currents entering(out-going) any node is zero
→ the sum of incoming currents is equal to the sum of outgoing currents.
I1  I 2  (  I 3 )  (  I 4 )  (  I 5 )  0
 I1  I 2  I 3  I 4  I 5
(2) Kirchhoff’s voltage law (KVL),
the algebraic sum of the voltages around any loop is zero
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Kirchhoff’s Current law
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 I n (t )  0
n
1 (t )
i2 (t )
i1 (t )
 0 (t )
I
 2 (t )
i1 (t ) 
R1

 3 (t )
Current definition
i
0 - 1
i3 (t )
R3
a
(t )  i1 (t )  i2 (t )  i3 (t )  0
n
R2
R1
n
R
b
a  b
R
, i2 (t ) 
0 - 2
R2
0 - 1 0 -  2
R1

R2

, i3 (t ) 
0 - 3
R3
0 - 3
R3
0
•
The direction of a current can be chosen
arbitrarily.
•
The value of a current can be obtained from
a voltage drop along the direction of current
divided by a resistance met.
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Kirchhoff’s Voltage law
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 n ( t )  0
n
Sum of voltage drops along a closed loop should be equal to zero!
1 (t )   2 (t ) -  s (t )  0
1 (t) R1
 2 (t) C1
Voltage convention

Vab  Va  Vb
 s (t)
-
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Example 2.6
Find the unknown currents in the network.
I1  80[mA]
I 4  70[mA]
Node 1 :
I1  60m  20m  0
Node 2 :
I 4  I1  I 6  0
Node 3 :
 I 4  I 5  60m  40m  0
Node 4 :
 I 5  20m  30m  0
Node 5 :
I 6  40m  30m  0
I 5  50[mA]
I 6  10[mA]
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Example E2.6
Find the current ix in the circuits in the figure.
ix  10ix  44m  0
 ix  4[mA]
 ix  10ix  120m  12m  0
 ix  12[mA]
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Example E2.8
Find Vad and Veb in the network in the figure.
Vad  24  4  6  26[V ]
Veb  8  6  24  10[V ]
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Example 2.15
Given the following circuit, let us find I, Vbd and the power absorbed by the 30kΩ
resistor. Finally, let us use voltage division to find Vbc .
 6  I  10k  I  20k  12  I  30k  0
6
I 
 0.1[mA]
60k
Vbc 
20k
 ( 6)  2 [V ]
20k  40k
Vbd  I  20k  12  2  12  10 [V ]
P30 k  I 2  30k  0.01  106  30k  0.3 [mW ]
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Series resistors
equivalent
RS  R1  R2    RN
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Parallel resistors
equivalent
1
1
1
1
 

RP R1 R2
RN
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Example 2.19
Given the circuit, we wish to find the current in the 12-kΩ load resistor.
equivalent
IL 
1
12k
1
1

4k 12k
 ( 1m) 
1
 ( 1m)  0.25 [mA]
31
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Example 2.20
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We wish to determine the resistance at terminals A-B in the network in the figure.
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Y-Δ transformation
Δ
Y
equivalent
R ( R  R3 )
Ra  Rb  2 1
R1  R2  R3
Ra 
R1 R2
R1  R2  R3
R1 
Ra Rb  Rb Rc  Rc Ra
Rb
R3 ( R1  R2 )
R1  R2  R3
Rb 
R2 R3
R1  R2  R3
R2 
Ra Rb  Rb Rc  Rc Ra
Rc
R3 R1
Rc 
R1  R2  R3
R3 
Ra Rb  Rb Rc  Rc Ra
Ra
Rb  Rc 
Rc  Ra 
R1 ( R2  R3 )
R1  R2  R3
Y 
 Y
R  R1  R2  R3
 RY 
R
3
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Example 2.26
Given the network in Fig. 2.36a, let us find the source current IS .
Ra 
12k  18k
 6 [ k ]
12k  18k  6k
RP 
Rb 
18k  6k
 3 [ k ]
12k  18k  6k
 IS 
Rc 
12k  6k
 2 [ k ]
12k  18k  6k
12k  6k
 4 [ k ]
12k  6k
12
 1.2 [mA]
6k  4 k
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2.8 Circuits with dependent sources
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Example 2.27
Let us determine the voltage Vo in the circuit in the figure.
 12  I1  3k  2000  I1  I1  5k  0
I1 
12
 2 [mA]
3k  2k  5k
V0  I1  5k  10 [V ]
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Example 2.28
Given the circuit in the figure containing a current-controlled current source, let us
find the voltage Vo.
10m 
Vs
V
 I 0  4 I 0  0, I 0  s
6k
3k
10m 
Vs Vs
  0  60  5Vs  0
6k 1k
Vs  12 [V ], V0  8 [V ]
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Example 2.30
An equivalent circuit for a FET common-source amplifier or BJT common-emitter
amplifier can be modeled by the circuit shown in the figure. We wish to determine an
expression for the gain of the amplifier, which is the ratio of the output voltage to the
input voltage.
GND can be
arbitrarily set.
0V
RL  R3 || R4 || R5
 i  i1R1  i1R2  0
g 
R2
i
R1  R2
0V
G ( gain ) 
0   g m g RL
g R
 R2 
0

  m L g   g m RL 
i
i
R

R
2 
 1
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Transistor amplifier
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Transistor
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2.10 Application examples
Example 2.33 : The Wheatstone bridge circuit.
R3
Rx
R
R

 1 2
R1  R3 R2  Rx
R3 Rx
 Rx  R2
R3
R1
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