Chapter 2 Resistive Circuits Notes/Chapter 02.pdf · Chapter 2 Resistive Circuits Chapter 2 Resistive Circuits 1. Solve circuits (i.e., find currents and voltages of interest) by
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ELECTRICAL
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Chapter 2Resistive Circuits
Chapter 2Resistive Circuits
1. Solve circuits (i.e., find currents and voltages of interest) by combining resistances in series and parallel.
2. Apply the voltage-division and current-divisionprinciples.
3. Solve circuits by the node-voltage technique.
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Chapter 2Resistive Circuits
4. Find Thévenin and Norton equivalents.
5. Apply the superposition principle.
6. Draw the circuit diagram and state the principlesof operation for the Wheatstone bridge.
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
Circuit Analysis using Series/Parallel Equivalents
1. Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source.
2. Redraw the circuit with the equivalent resistance for the combination found in step 1.
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3. Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with a single source and a single resistance.
4. Solve for the currents and voltages in the final equivalent circuit.
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
Voltage Division
total321
111 v
RRRRiRv
++==
total321
222 v
RRRRiRv
++==
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Chapter 2Resistive Circuits
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Application of the Voltage-Division Principle
V5.1
156000200010001000
1000
total4321
11
=
×+++
=
+++= v
RRRRRv
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
Current Division
total21
2
11 i
RRR
Rvi
+==
total21
1
22 i
RRR
Rvi
+==
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
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Application of the Current-Division Principle
Ω=+×=
+= 20
60306030
32
32eq RR
RRR
A10152010
20
eq1
eq1 =
+=
+= siRR
Ri
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
Although they are veryimportant concepts,
series/parallel equivalents andthe current/voltage division
principles are not sufficient tosolve all circuits.
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Chapter 2Resistive Circuits
Node Voltage Analysis
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Writing KCL Equations in Terms of the Node Voltages for
Figure 2.16
svv =1
03
32
4
2
2
12 =−++−R
vvRv
Rvv
03
23
5
3
1
13 =−++−R
vvRv
Rvv
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02
21
1
1 =+−+ siRvv
Rv
04
32
3
2
2
12 =−++−R
vvRv
Rvv
siRvv
Rv =−+
4
23
5
3
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Chapter 2Resistive Circuits
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Chapter 2Resistive Circuits
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Circuits with Voltage Sources
We obtain dependentequations if we use all of thenodes in a network to writeKCL equations.
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( ) ( ) 01515
3
2
4
2
1
1
2
1 =−−++−−+R
vRv
Rv
Rv
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010 21 =++− vv
13
32
2
31
1
1 =−+−+R
vvR
vvRv
04
3
3
23
2
13 =+−+−Rv
Rvv
Rvv
14
3
1
1 =+Rv
Rv
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Node-Voltage Analysis with a Dependent Source
First, we write KCL equations at each node, including the current of the controlled source just as if
it were an ordinary current source.
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xs iiR
vv 21
21 +=−
03
32
2
2
1
12 =−++−R
vvRv
Rvv
024
3
3
23 =++−xiR
vR
vv
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Next, we find an expression for the controlling variable ix in terms of the node voltages.
3
23
Rvvix
−=
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Substitution yields
3
23
1
21 2R
vviR
vvs
−+=−
03
32
2
2
1
12 =−++−R
vvRv
Rvv
023
23
4
3
3
23 =−++−R
vvRv
Rvv
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Node-Voltage Analysis1. Select a reference node and assign variables for the unknown node voltages. If the reference node is chosen at one end of an independent voltage source, one node voltage is known at the start, and fewer need to be computed.
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2. Write network equations. First, use KCL to write current equations for nodesand supernodes. Write as many current equations as you can without using all ofthe nodes. Then if you do not have enough equations because of voltage sourcesconnected between nodes, use KVL to write additional equations.
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3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the node voltages. Substitute into the network equations, and obtain equations having only the node voltages as unknowns.
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4. Put the equations into standard form and solve for the node voltages.
5. Use the values found for the node voltages to calculate any other currents or voltages of interest.
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Mesh Current Analysis
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Choosing the Mesh Currents
When several mesh currents flow through one element, we consider the current in that element to be the algebraic sum of the mesh currents.
Sometimes it is said that the mesh currents are defined by “soaping the window panes.”
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Writing Equations to Solve for Mesh Currents
If a network contains only resistances and independent voltage sources, we can writethe required equations by following each current around its mesh and applying KVL.
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Using this pattern for mesh 1 of Figure 2.32a, we have
For mesh 2, we obtain( ) 024123 =++− BviRiiR
For mesh 3, we have
( ) 031132 =−+− BviRiiR
( ) ( ) 021312 =−−+− As viiRiiR
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In Figure 2.32b
( ) ( ) 021441211 =−−+−+ AviiRiiRiR
( ) ( ) 032612425 =−+−+ iiRiiRiR
( ) ( ) 043823637 =−+−+ iiRiiRiR
( ) ( ) 034814243 =−+−+ iiRiiRiR
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Mesh Currents in Circuits Containing Current Sources
A common mistake made by beginning students is to assume that the voltages across current sources are zero. In Figure 2.35, we have:
A21 =i0105)(10 212 =++− iii
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Combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined.
( ) ( ) 01042 32311 =+−+−+ iiiii
Mesh 3:
( ) ( ) 0243 13233 =−+−+ iiiii
512 =− ii
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026420 221 =+++− iii
124iivx −=
22ivx =
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Mesh-Current Analysis
1. If necessary, redraw the network without crossing conductors or elements. Then define the mesh currents flowing around each of the open areas defined by the network. For consistency, we usually select a clockwise direction for each of the mesh currents, but this is not a requirement.
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2. Write network equations, stopping after the number of equations is equal to the number of mesh currents. First, use KVL to write voltage equations for meshes that do not contain current sources. Next, if any current sources are present, write expressions for their currents in terms of the mesh currents. Finally, if a current source is common to two meshes, write a KVL equation for the supermesh.
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3. If the circuit contains dependent sources, find expressions for the controllingvariables in terms of the mesh currents. Substitute into the network equations,and obtain equations having only the mesh currents as unknowns.
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4. Put the equations into standard form. Solve for the mesh currents by use of determinants or other means.
5. Use the values found for the mesh currents to calculate any other currents or voltages of interest.
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Thévenin Equivalent Circuits
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Thévenin Equivalent Circuits
ocvVt =
sc
oc
ivRt =
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Finding the Thévenin Resistance Directly
When zeroing a voltage source, it becomes an open circuit. When zeroing a current source, it becomes a short circuit.
We can find the Thévenin resistance by zeroing the sources in the original network and then computing the resistance between the terminals.
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Step-by-step Thévenin/Norton-Equivalent-Circuit Analysis
1. Perform two of these:a. Determine the open-circuit voltage Vt = voc.
b. Determine the short-circuit current In = isc.
c. Zero the sources and find the Thévenin resistance Rt looking back into the terminals.
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2. Use the equation Vt = Rt In to compute the remaining value.
3. The Thévenin equivalent consists of a voltage source Vt in series with Rt .
4. The Norton equivalent consists of a current source In in parallel with Rt .
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Source Transformations
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Maximum Power Transfer
The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance.
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SUPERPOSITION PRINCIPLE
The superposition principle states that the total response is the sum of the responses to each of the independent sources acting individually. In equation form, this is
nT rrrr +++= L21
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WHEATSTONE BRIDGEThe Wheatstone bridge is used by mechanical and civil engineers to measure the resistances of strain gauges in experimental stress studies of machines and buildings.
31
2 RRRRx =
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