5. General Circuit Analysis with
Phasors
The previous sections discuss basic analysis methods based on
equivalence, reduction, and circuit theorems. These methods are
valuable because we work directly with element impedances and
thereby gain insight into steady-state circuit behavior. We also need
general methods, such as node and mesh analysis, to deal with
more complicated circuits than the basic methods can easily
handle. These general methods use node - voltage or mesh-current
variables to reduce the number of equations that must be solved
simultaneously.
Node-voltage equations involve selecting a reference node and
assigning a node-to-datum voltage to each of the remaining
nonreference nodes. Because of KVL, the voltage between any two
nodes equals the difference of the two node voltages. This
fundamental property of node voltages plus the element
impedances allow us to write KCL constraints at each of the nonreference nodes.
Fig. 8-28: An example node
For example, consider node A in Fig. 8-28. The sum of currents
leaving this node can be written as
Rewriting this equation with unknowns grouped on the left and
known inputs on the right yields
Expressing this result in terms of admittances produces the
following equation.
This equation has a familiar pattern. The unknowns VA, VB and VC
are the node-voltage phasors. The coefficient [Y1+Y2+Y3] of VA is
the sum of the admittances of all of the elements connected to
node A. The coefficient Y2 of VB is admittance of the elements
connected between nodes A and B, while Y3 is the admittance of
the elements connected between nodes A and C. Finally, IS1 and IS2
are the phasor current sources connected to node A, with IS1
directed into IS2 directed away from the node. These observations
suggest that we can write node-voltage equations for phasor
circuits by inspection, just as we did with resistive circuits.
Circuits that can be drawn on a flat surface with no crossovers are
called planar circuits. The mesh-current variables are the loop
currents assigned to each mesh in a planar circuit. Because of
KCL, the current through any two-terminal element is equal to the
difference of the two adjacent meshes. This fundamental property
of mesh currents together with the element impedances allow us to
write KVL constraints around each of the meshes.
Fig. 8-29: An example mesh
For example, the sum of voltages around mesh A in Fig. 8-29 is
The mesh A equation is obtained by equating this sum to the sum
of the source voltages produced in mesh A. Arranging this equation
in standard form yields
This equation also displays a familiar pattern. The unknowns IA, IB,
and IC are mesh-current phasors. The coefficient [Z1+Z2+Z3] of IA is
the sum of the impedances in mesh A. The coefficient [Z2 ] of IB is
the impedance in both mesh A and mesh B, while [Z3 ] is the
impedance common to meshes A and C. Finally, VS1 and VS2 are
the phasor voltage sources in mesh A. These observations allow us
to write mesh-current equations for phasor circuits by inspection.
The preceding discussion assumes that the circuit contains only
current sources in the case of node analysis and voltage sources in
mesh analysis . If there is a mixture of sources, we may be able to
use the source transformations discussed in Sect. 8-4 to convert
from voltage to current sources, or vice versa. A source
transformation is possible only when there is an impedance
connected in series with a voltage source or an admittance in
parallel with a current source.
EXAMPLE 8-20
Use node analysis to find the node voltages VA and VB in Fig. 8-30
(a).
Fig. 8-30
SOLUTION:
The voltage source in Fig. 8-30(a) is connected in series with an
impedance consisting of a resistor and inductor connected in
parallel. The equivalent impedance of this parallel combination is
Applying a source transformation produces an equivalent current
source of
Fig. 8-30(b) shows the circuit produced by the source
transformation eliminates node B. The node-voltage equation at the
remaining non-reference node in Fig. 8-30(b) is
Solving for VA yields
Referring to Fig. 8-30(a), we see that KVL requires VB=VA+10
90 . Therefore, VB is found to be
EXAMPLE 8-21
Use node analysis to find the current IX in Fig. 8-31.
-
Fig. 8-31
SOLUTION:
In this example we use node analysis on a problem solved in
Example 8-12 using a
-to-Y transformation and solved again in
Example 8-18 using a Thevenin equivalent circuit. The voltage
source cannot be replaced by source transformation because it is
not connected in series with an impedance. By inspection, the node
equations at nodes A and B are
A node equation at node C is not required because the voltage
source forces the condition VC=75
0 . Substituting this constraint
into the equations of nodes A and B and arranging the equations in
standard form yields two equations in two unknowns:
Solving these equations for VA and VB yields
Using these values for VA and VB, the unknown current is found to
be
This value of IX is the same as the answer obtained in Example 812 and again in Example 8-18. Review these three examples
together to gain perspective on different approaches to phasor
circuit analysis.
EXAMPLE 8-24
The circuit in Fig. 8-32 is an equivalent circuit of an ac induction
motor. The current IS is called the stator current, IR the rotor current,
and IM the magnetizing current. Use the mesh-current method to
solve for the branch currents IS, IR and IM.
Fig. 8-32
SOLUTION:
Applying KVL to the sum of voltages around each mesh in Fig. 832 yields
Arranging these equations in standard form yields
Solving these equations for IA and IB produces The required stator,
rotor, and magnetizing currents are related to these mesh current,
as follows:
EXAMPLE 8-25
Use the mesh-current method to solve for output voltage V2 and
input impedance ZIN of the circuit in Fig. 8-33.
Fig. 8-33
SOLUTION:
The circuit contains a voltage-controlled voltage source. We initially
treat the dependent source as an independent source and use KCL
to write the sum of voltages around each mesh:
Arranging these equations in standard form produces
Using Ohm's law, the control voltage VX is
Eliminating VX from the mesh equations yields
Solving for the two mesh currents produces
Using these values of the mesh currents, the output voltage and
input impedance are