Download Example Equivalent Circuit Problem

ECE 2300 Circuit Analysis
Dr. Dave Shattuck
Example Equivalent Circuit Problem
37[k]
2.2[k]
4.7[k]
+
8.2[k]
5[mA]
6.2[k]
iX
27[k]
3.3[k]
vW
-
Find iX and vW.
Solution:
We begin this problem by replacing the 3.3[k] resistor and the 4.7[k] resistor with
their equivalent resistor value, which is 8.0[k]. Then, we take that 8.0[k] resistor, which is in
parallel with the 6.2[k] resistor, and replace them by their equivalent resistance, which is
3.49[k]. We redraw the circuit that results, and get the following circuit diagram.
37[k]
2.2[k]
+
8.2[k]
5[mA]
3.49[k]
vX
27[k]
There are several things to notice about this diagram. First, we note that vW is no longer present
in this circuit. There is no place to label vW in this diagram. The voltage vW is inside the
equivalent circuit, and cannot be labeled outside the equivalent circuit in the original circuit. We
redraw the original circuit in the figure that follows, enclosing the part of the circuit that we have
replaced within a dashed line.
37[k]
2.2[k]
4.7[k]
+
8.2[k]
+
6.2[k]
5[mA]
vX iX
27[k]
+
-
vX
-
3.3[k]
vW
-
Note that because the upper terminal for the voltage vW is inside the equivalent circuit, and not
present outside the equivalent circuit, it is not possible to label vW on the diagram after replacing
the equivalent circuit with the 3.49[kW] resistor. Note, however, that since the voltage vX can be
labeled outside the equivalent circuit, it can be labeled both before and after replacing the
equivalent circuit with the 3.49[kW] resistor. We have shown vX both inside and outside the
equivalent circuit, in this diagram. It should be clear that both of these versions of vX are the
same. Let’s again show the simplified circuit, and work from there.
37[k]
2.2[k]
+
8.2[k]
5[mA]
3.49[k]
vX
27[k]
We continue to simplify the circuit, by replacing the 3.49[k] resistor and the 2.2[k]
resistor with their equivalent resistor, which is 5.69[k]. Then, we take that 5.69[k] resistor,
which is in parallel with the 8.2[k] resistor, and replace them by their equivalent resistance,
which is 3.36[k]. We redraw the circuit that results, and get the following circuit diagram.
37[k]
+
5[mA]
3.36[k]
vZ
27[k]
-
It is now pretty straightforward to solve the voltage vZ, using Ohm’s Law. Note that the 5[mA]
current is now flowing up through the 3.36[k] resistor, in the active sign convention with
respect to the voltage vZ. Thus, using Ohm’s Law, we can write that
vZ    5  mA   3.36  k  16.8  V .
Now that we have vZ, we go back to our earlier diagram, and noting that vZ is also present
in this diagram, we use it to solve for vX with the VDR. For this example, we reproduce the
diagram following this paragraph, adding the definition of vZ. Of course, if we were solving this
by hand, we would simply have added vZ to our existing diagram, from earlier in the problem.
37[k]
2.2[k]
+
8.2[k]
5[mA]
27[k]
vZ
-
+
3.49[k]
vX
-
In this circuit, the 2.2[k] resistor and the 3.49[k] resistor are in series, and so we can use
VDR to write
3.49  k 
v X  vZ

3.49  k   2.2  k 


3.49  k 
v X   16.8  V  

 3.49  k   2.2  k  


v X  10.3 V  .
Next, we can use the original diagram to find iX, using Ohm’s Law. We reproduce the original
diagram here, just for clarity.
37[k]
2.2[k]
4.7[k]
+
8.2[k]
+
6.2[k]
5[mA]
iX
27[k]
vX
-
3.3[k]
vW
-
Using Ohm’s Law, we write
vX
10.3[V]
iX 


6.2[k] 6.2[k]
iX  1.66  mA.
In addition, we can use VDR to solve for vW, and get
3.3 k
vW  v X

3.3 k  4.7  k 


3.3 k
vW   10.3 V  

 3.3 k  4.7  k 


vW  4.25 V.
Alternative Solution:
There are several good ways to solve this problem. Another choice would have been to
have defined the current through the 2.2[k] resistor. We will define it as iQ, as shown in the
circuit that follows.
37[k]
2.2[k]
+
8.2[k] iQ
5[mA]
27[k]
vZ
-
+
3.49[k]
vX
-
Again, note that we did not have to redraw the circuit, as we have done here. This definition
could have been completed in the diagrams already given.
In this diagram, iQ can be found from the CDR. Note that the series combination of the
2.2[k] resistor and the 3.49[k] resistor is in parallel with the 8.2[k] resistor. Note also that
the 5[mA] current feeds this parallel combination. Thus, we can use CDR, to get


8.2  k
iQ  5[mA] 

 8.2  k  5.69  k 


iQ  2.95[mA].
We can now use this value of iQ to find iX, using the CDR again. The diagram is reproduced here
for clarity.
37[k]
2.2[k]
4.7[k]
+
8.2[k] iQ
5[mA]
27[k]
+
6.2[k]
iX
vX
-
3.3[k]
vW
-
Using CDR again, we can write


8.0  k 
iX  iQ 

 8.0  k   6.2  k  




8.0  k 
iX   2.95[mA] 

 8.0  k   6.2  k  


iX   1.66[mA] .
This is the same result that we got earlier, which should be no surprise. It is simply a
different way to get the same answer. It is left as an exercise for the student to solve for the
current through the 3.3[k] resistor, and find vW in this way. The answer should not change.