Download PEQWS_Mod01_Prob07_v04

Dave Shattuck
University of Houston
Practice Examination Questions With Solutions
Module 1 – Problem 7
Filename: PEQWS_Mod01_Prob07.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 30 minutes*.
Problem Statement:
In the circuit shown below, find the voltages vX and vO.
+
vX
+
iS=
5[mA]
R2 =
330[W]
R3 =
220[W]
-
R1 =
560[W]
+
vQ
R4 =
100[W]
-
-
+
R5 =
750[W]
-
vS=
25vQ
* This “time allowed” figure is based on the student’s circuit analysis skills
at this point. Once the student has learned the material in the second module, this
problem should be completed in significantly less time than what is shown here.
Page 1.7.1
vO
Dave Shattuck
University of Houston
Problem Statement:
In the circuit shown below, find the voltages vX and vO.
+
vX
R1 =
560[W]
+
iS=
5[mA]
R2 =
330[W]
R3 =
220[W]
-
+
vQ
R4 =
100[W]
vO
-
-
+
R5 =
750[W]
-
vS=
25vQ
Solution:
The first step in the solution is to define the variables that we are going to
need. I know that I need to do this, because I expect to have to write a KVL
around the loop in the middle of the circuit, and some of these voltages are not yet
defined. Specifically, I want to define the voltage across the R1 resistor, and across
the R5 resistor. Let’s do that first. I have chosen to name the voltages as v1 and v5,
but this choice is arbitrary. I have chosen a particular polarity for each of these
voltages, but again this choice is arbitrary. How did I know that I needed to define
these two voltages? The key is that I looked at this circuit, and decided that I was
going to need to write KVL around the loop that is marked with a red dashed line
in the figure that follows. Notice that all of the rest of the voltages in this loop are
already defined. Note that they may not yet be known, but they are defined.
Having definitions for variables is an important first step, which is sometimes
skipped by beginning students.
Page 1.7.2
Dave Shattuck
University of Houston
+
vX
+
iS=
5[mA]
R2 =
330[W]
R3 =
220[W]
v1
R1 =
560[W]
+
vQ
R4 =
100[W]
-
vO
-
-
+
R5 =
750[W]
+
+
v5
vS=
25vQ
-
-
Let’s next address solving for v5. We would like to find the current through
R5, which will then give us v5 by Ohm’s Law. We can find the current, which we
will call i5, by writing KCL for the closed surface shown with a dashed red line in
the figure that follows.
+
vX
+
iS=
5[mA]
R2 =
330[W]
R3 =
220[W]
v1
R1 =
560[W]
+
vQ
+
R4 =
100[W]
-
vO
-
i5
+
+
R5 =
750[W]
v5
Page 1.7.3
-
-
vS=
25vQ
Dave Shattuck
University of Houston
When we apply KCL to this closed surface, we get
i5  0.
This follows because this is the only current crossing this surface. Stated another
way, the current i5 is zero because current has no way of returning to the left part of
the circuit. In fact, we can say that there is no “circuit” for this current.
With this, we can say from Ohm’s law that
v5  i5 R5  0.
Next, we recognize that we can write KCL for the upper left hand node. The
currents through the resistors can be written in terms of the voltage vQ, which gives
us
iS 1 
vQ
R2

5[mA] 
vQ
R3
 0, or
vQ
330[W]

vQ
220[W]
 0.
Solving this for vQ, we have
vQ
vQ

 5[mA], or
330[W] 220[W]
vQ 7.58[mS]  5[mA], which when solved yields
vQ  660[mV].
Now we are part of the way towards our solution. We have vQ, and we also
now know vQ which is 25 times this, or
vS  25vQ  16.5[V].
Once we can find v1, we will then be able to find vQ. Let’s solve for v1.
To get v1, we write KVL around the right hand loop, in terms of the current
in that loop. We need to define that current first, which means labeling it in the
circuit diagram. I have called it i1, and then I can write KVL around the loop
indicated in the following diagram with a dashed red line, using that current. Of
course, I get the voltage across each resistor using Ohm’s Law.
Page 1.7.4
Dave Shattuck
University of Houston
+
vX
+
i1
iS=
5[mA]
v1
R1 =
560[W]
+
i1
R2 =
330[W]
R3 =
220[W]
vQ
+
vO
-
i5
R4 =
100[W]
-
+
+
R5 =
750[W]
v5
-
vS=
25vQ
-
Writing KVL, I have
vS  v1  vO  0, and then using Ohm's Law, this becomes
vS  (i1 560[W])  (i1100[W])  0.
It is very important to recognize in this equation that
vO  (i1100[W]),
because the voltage and current for this resistor have been defined in the active
sign convention. Since we know vS, we can solve for i1, and get
16.5[V]  (i1 560[W])  (i1100[W])  0, or
16.5[V]
 25[mA].
660[W]
Now, it is clear that
vO  (i1100[W])  ( 25[mA]100[W])
i1  
vO  2.5[V].
Finally, we can write KVL for the loop drawn with a dashed red line in the
circuit that follows.
Page 1.7.5
Dave Shattuck
University of Houston
+
vX
+
iS=
5[mA]
R2 =
330[W]
R3 =
220[W]
v1
R1 =
560[W]
+
vQ
R4 =
100[W]
-
vO
-
-
+
R5 =
750[W]
+
+
v5
-
The KVL will be
v1  vS  v5  vQ  vX  0, or
 25[mA] 560[W]  16.5[V]  0  660[mV]  vX
We solve this for vX, and we get
vX  1.84[V].
Page 1.7.6
-
 0.
vS=
25vQ