Mesh-Current Analysis of the Amplifier Circuit
4.9 Source Transformations
R
vs
a
b (a)
is
a
b (b)
R
Figure 4.36 Source transformations.
Example 4.8 Using Source Transformations to Solve a Circuit
a) For the circuit shown in Fig. 4.37, find the power associated with the 6 V source.
b)State whether the 6 V source is absorbing or delivering the power calculated in (a).
Solution
a) If we study the circuit shown in Fig. 4.37, know- ing that the power associated with the 6 V source is of interest, several approaches come to mind. The circuit has four essential nodes and six essential branches where the current is unknown. Thus we can find the current in the branch containing the 6 V source by solving either three mesh-current equa- tions or three node-voltage equations.
Choosing the mesh-current approach involves [4 - 1]
36 - (4 - 1)]
solving for the mesh current that corresponds to the branch current in the 6 V source.
Choosing the node-voltage approach involves solving for the voltage across the resistor, from which the branch current in the 6 V source can be calculated. But by focusing on just one branch current, we can first simplify the circuit by using source transformations.
30 Æ 20 30
4 6
10
5
6 V
40 V
Figure 4.37 The circuit for Example 4.8.
already on our list of simplifying techniques. We begin expanding this list with source transformations. A source transformation, shown in Fig. 4.36, allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa. The double-headed arrow emphasizes that a source transformation is bilateral; that is, we can start with either configuration and derive the other.
We need to find the relationship between and that guarantees the two configurations in Fig. 4.36 are equivalent with respect to nodes a,b.
Equivalence is achieved if any resistor experiences the same current flow, and thus the same voltage drop, whether connected between nodes a,b in Fig. 4.36(a) or Fig. 4.36(b).
Suppose is connected between nodes a,b in Fig. 4.36(a). Using Ohm’s law, the current in is
(4.52) Now suppose the same resistor is connected between nodes a,b in Fig. 4.36(b). Using current division, the current in is
(4.53) If the two circuits in Fig. 4.36 are equivalent, these resistor currents must be the same. Equating the right-hand sides of Eqs. 4.52 and 4.53 and simplifying, (4.54) When Eq. 4.54 is satisfied for the circuits in Fig. 4.36, the current in is the same for both circuits in the figure for all values of If the current through is the same in both circuits, then the voltage drop across is the same in both circuits, and the circuits are equivalent at nodes a,b.
If the polarity of is reversed, the orientation of must be reversed to maintain equivalence.
Example 4.8 illustrates the usefulness of making source transforma- tions to simplify a circuit-analysis problem.
is vs
RL RL
RL.
RL is =
vs
R. iL =
R R + RL is.
RL RL
iL = vs
R + RL
. RL
RL
RL
is
vs
4.9 Source Transformations 111
6 V 30
6
10 4
20 8 A
6 V 5
(a) First step
30
6
10
4 4
32 V
6 V
(b) Second step
20 30
4
1.6 A
6 V
(c) Third step
12 4
19.2 V
(d) Fourth step Figure 4.38 Step-by-step simplification of the circuit shown in Fig. 4.37.
Next, we can replace the parallel combination of the and resistors with a resistor.
This resistor is in parallel with the 8 A source and therefore can be replaced with a 32 V source in series with a resistor, as shown in Fig. 4.38(b). The 32 V source is in series with of resistance and, hence, can be replaced by a cur- rent source of 1.6 A in parallel with , as shown in Fig. 4.38(c). The and parallel resis- tors can be reduced to a single resistor. The parallel combination of the 1.6 A current source
12 Æ 30 Æ 20 Æ
20 Æ
20 Æ 4 Æ
4 Æ
4 Æ 5 Æ
20 Æ
and the resistor transforms into a voltage source of 19.2 V in series with . Figure 4.38(d) shows the result of this last transformation. The current in the direction of the voltage drop across
the 6 V source is or 0.825 A.
Therefore the power associated with the 6 V source is
b) The voltage source is absorbing power.
p6V = (0.825)(6) = 4.95W.
(19.2 - 6)>16, 12 Æ 12 Æ
A question that arises from use of the source transformation depicted in Fig. 4.38 is, “What happens if there is a resistance in parallel with the voltage source or a resistance in series with the current source?” In both cases, the resistance has no effect on the equivalent circuit that pre- dicts behavior with respect to terminals a,b. Figure 4.39 summarizes this observation.
The two circuits depicted in Fig. 4.39(a) are equivalent with respect to terminals a,b because they produce the same voltage and current in any resistor inserted between nodes a,b. The same can be said for the cir- cuits in Fig. 4.39(b). Example 4.9 illustrates an application of the equiva- lent circuits depicted in Fig. 4.39.
RL
Rs
Rp
(a)
is Rp
R
vs
a
b
R
vs
a
b
(b)
a
b Rs
R R
a
b is
Figure 4.39 Equivalent circuits containing a resistance in parallel with a voltage source or in series with a current source.
We must reduce the circuit in a way that pre- serves the identity of the branch containing the 6 V source. We have no reason to preserve the identity of the branch containing the 40 V source. Beginning with
this branch, we can transform the 40 V source in series with the resistor into an 8 A current source in parallel with a resistor, as shown in Fig. 4.38(a).
5 Æ 5 Æ
Example 4.9 Using Special Source Transformation Techniques
a) Use source transformations to find the voltage in the circuit shown in Fig. 4.40.
b) Find the power developed by the 250 V voltage source.
c) Find the power developed by the 8 A current source.
vo
15 5 25
250 V 125
10 8 A
100 vo
Figure 4.40 The circuit for Example 4.9.
20 25
250 V
8 A vo 100
Figure 4.41 A simplified version of the circuit shown in Fig. 4.40.
20 10 A 25 8 Avo 100
Figure 4.42 The circuit shown in Fig. 4.41 after a source transformation.
2 A vo 10
Figure 4.43 The circuit shown in Fig. 4.42 after combining sources and resistors.
b) The current supplied by the 250 V source equals the current in the resistor plus the current in the
resistor. Thus
Therefore the power developed by the voltage source is
c) To find the power developed by the 8 A current source, we first find the voltage across the source. If we let represent the voltage across the source, positive at the upper terminal of the source, we obtain
and the power developed by the 8 A source is 480 W.
Note that the and resistors do not affect the value of vobut do affect the power calculations.
10 Æ 125 Æ
vs + 8(10) = vo = 20, or vs = -60 V,
vs
p250V(developed) = (250)(11.2) = 2800W.
is = 250 125 +
250 - 20
25 = 11.2 A.
25 Æ
125 Æ
We now use a source transformation to replace the 250 V source and resistor with a 10 A source in parallel with the resistor, as shown in Fig. 4.42. We can now simplify the circuit shown in Fig. 4.42 by using Kirchhoff’s current law to com- bine the parallel current sources into a single source. The parallel resistors combine into a single resistor. Figure 4.43 shows the result. Hence vo = 20 V.
25 Æ 25 Æ
Solution
a) We begin by removing the and resis- tors, because the resistor is connected across the 250 V voltage source and the resistor is connected in series with the 8 A current source. We also combine the series-connected resistors into a single resistance of . Figure 4.41 shows the sim- plified circuit.
20 Æ
10 Æ 125 Æ
10 Æ 125 Æ