Chapter 9 - The Complete Response of Circuits With Two Energy Storage Elements
EMLAB 1 Chapter 2. Resistive circuits 2014. 9. 12.
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Transcript of EMLAB 1 Chapter 2. Resistive circuits 2014. 9. 12.
EMLAB
1
Chapter 2. Resistive circuits
2014. 9. 12.
EMLAB
2Contents
1. Ohm’s law
2. Kirchhoff’s laws
3. Series and parallel resistor combinations
4. Y-Δ transformation
5. Circuits with dependent sources
EMLAB
3
Entering resistive material, charges are decelerated, which decrease current flow.
Resistors : microscopic view
nucleus
electrons
EMLAB
4Types 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.
EMLAB
5
Rtit )()(
1. Ohm’s law
)0( R
RRitittp
22)()()(
Power absorption :
resistance
RG
1 ; conductance
EMLAB
6Example 2.1
Determine the current and the power absorbed by the resistor.
][62
12mA
kI
kRV
kRI
WVIP
2/)12(/
)2()106(
][072.0)106)(12(
22
232
3
EMLAB
7Glossary
(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 con-ductors, it is still only one node
EMLAB
8
(3) branch
(2) loop A loop is simply any closed path through the circuit in which no nodeis 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
EMLAB
92. Kirchhoff’s law
(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.
(2) Kirchhoff’s voltage law (KVL), the algebraic sum of the voltages around any loop is zero
0)()()( 54321 IIIII
54321 IIIII
EMLAB
10Kirchhoff’s Current law
0)( n
n tI
)(0 t
)(2 ti)(1 ti
)(3 ti3
303
2
202
1
101
321
-)(,
-)(,
-)(
0)()()()(
Rti
Rti
Rti
titititIn
n
Ri ba
ab
R
Current definition • The direction of a current can be chosen ar-bitrarily.
• The value of a current can be obtained from a voltage drop along the direction of current divided by a resistance met.
R2
)(1 t )(2 t
)(3 t
R1
R3
0---
3
30
2
20
1
10 RRR
EMLAB
11Kirchhoff’s Voltage law
-(t)1 -(t)2
-
(t)s
0)( n
n t
Sum of voltage drops along a closed loop should be equal to zero!
0)(-)()( 21 ttt s
R1 C1
Voltage convention
baab VVV
EMLAB
12Example 2.6
Find the unknown currents in the network.
020601 mmINode 1 :
Node 2 : 0614 III
Node 3 : 0406054 mmII
030205 mmINode 4 :
Node 5 : 030406 mmI
][801 mAI ][505 mAI ][106 mAI ][704 mAI
EMLAB
13Example E2.6
Find the current ix in the circuits in the figure.
04410 mii xx
][4 mAix
01212010 mmii xx
][12 mAix
EMLAB
14
Find Vad and Veb in the network in the figure.
Example E2.8
][266424 VVad
][102468 VVeb
EMLAB
15Example 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 .
0301220106 kIkIkI
][1.060
6mA
kI
][101221220 VkIVbd
][3.0301001.030 6230 mWkkIP k
][2)6(4020
20V
kk
kVbc
EMLAB
16Series resistors
equivalent
NS RRRR 21
EMLAB
17Parallel resistors
NP RRRR
1111
21
equivalent
EMLAB
18Example 2.19
Given the circuit, we wish to find the current in the 12-kΩ load resistor.
equivalent
][25.0)1(13
1)1(
121
41
121
mAmm
kk
kIL
EMLAB
19Example 2.20We wish to determine the resistance at terminals A-B in the network in the figure.
EMLAB
20Y-Δ transformation
equivalent
321
312 )(
RRR
RRRRR ba
321
213 )(
RRR
RRRRR cb
321
321 )(
RRR
RRRRR ac
Y
RRR
RRR
RRR
RRR
RRR
RRR
c
b
a
321
13
321
32
321
21
Y
R
RRRRRRR
R
RRRRRRR
R
RRRRRRR
a
accbba
c
accbba
b
accbba
3
2
1
ΔY
321 RRRR 3
RRY
EMLAB
21Example 2.26
Given the network in Fig. 2.36a, let us find the source current IS .
][661812
1812
kkkk
kkRa
][361812
618
kkkk
kkRb
][261812
612
kkkk
kkRc
][4612
612
kkk
kkRP
][2.146
12mA
kkIS
EMLAB
222.8 Circuits with dependent sources
Example 2.27
052000312 111 kIIkI
][2523
121 mA
kkkI
][10510 VkIV
Let us determine the voltage Vo in the circuit in the figure.
EMLAB
23Example 2.28
Given the circuit in the figure containing a current-controlled current source, let us find the voltage Vo.
k
VIII
k
Vm ss
3,04
610 000
0560016
10 sss Vk
V
k
Vm
][8],[12 0 VVVVs
EMLAB
24Example 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.
543 |||| RRRRL
02111 RiRii
ig RR
R
21
2
Lgm Rg 0
21
20)(RR
RRg
RggainG Lm
i
gLm
i
GND can be arbitrarily set.
V0
V0
EMLAB
25
Transistor
Transistor amplifier
EMLAB
262.10 Application examples
Example 2.33 : The Wheatstone bridge circuit.
xx
x
R
R
R
R
RR
R
RR
R 2
3
1
231
3
1
32 R
RRRx