UC Davis 1 Hussain Al-Asaad
UNIVERSITY OF CALIFORNIA—DAVIS
DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING
ENG17 — CIRCUITS I — SUMMER SESSION I — 2010
MIDTERM EXAM — SOLUTION
STUDENT INFORMATION
INSTRUCTIONS
• The exam is closed book and closed notes.
• One double-sided cheat sheet is allowed.
• Calculators are allowed.
• Print your name and your ID number.
• There are six problems in the exam. Choose five problems and solve them.
• Cross out the problem that you did not choose.
• Show your work for partial credit.
• If you need more space for your solution, use the back of the sheets.
EXAM GRADE
Name Hussain Al-Asaad
ID Number xxx-xx-xxxx
Problem Maximum Points Student Score
1 20 20
2 20 20
3 20 20
4 20 20
5 20 20
6 20 20
Total 100 100
UC Davis 2 Hussain Al-Asaad
1. EQUIVALENT RESISTANCE (20 POINTS)
Consider the circuit shown below. Let Req be the equivalent resistance of the circuit.
1. If R = 0 , find the equivalent resistance Req. (3 points)

2. If , find the equivalent resistance Req. (3 points)

3. Find the equivalent resistance Req (in terms of R). (14 points)
Since none of the resistors are in series or parallel, then it is not possible to combine the resistors to get the equiv-
alent resistance. Instead, we can insert a voltage source Vs and measure Req as:
Applying KCL at node a, we get .
Applying KVL at loop a-b-c-d, we get .
Applying KVL at loop a-c-b-d, we get .
Rewriting the above two equations and using the equation , we get:
Multiplying the above equations by (R+3) and (R+1) respectively, we get:
Adding the above two equations, we get:
.
Req R
1 
1 
3 
3 
R
1 
1 
3 
3 
–
+
Vs
is
i1 i2
i
i1 – i
i2 + i
a
b c
d
Req 1 3  3 1 +
1 3
1 3+
----------- 1 3
1 3+
-----------+ 3
4
-- 3
4
--+ 3
2
-- = = = =
R  =
Req 1 3+  3 1+  4 4
4 4
4 4+
----------- 2 = = = =
Req
Vs
is
----=
is i1 i2+=
1 i1 Ri 1  i2 i+  Vs–+ + 0=
3 i2 Ri– 3  i1 i–  Vs–+ 0=
is i1 i2+=
is R 1+ i+ Vs=
3is R 3+ – i Vs=
R 3+ is R 3+  R 1+ i+ R 3+ Vs=
3 R 1+ is R 1+  R 3+ – i R 1+ Vs=
4R 6+ is 2R 4+ Vs= 
Vs
is
---- 4R 6+
2R 4+
--------------- 2R 3+
R 2+
---------------= =  Req
2R 3+
R 2+
---------------=
UC Davis 3 Hussain Al-Asaad
2. NODE VOLTAGE ANALYSIS (20 POINTS)
Consider the circuit shown below.
1. Calculate the values of node voltages va, vb, and vc in terms of Vs. (13 points)
Since a voltage source Vs is placed between nodes c and a, then we have:

Applying KCL at node b, we get:
.
Applying KCL at ground node, we get:

Solving the above equations, we get:
and
2. If the total power delivered by the voltage source is 8W, determine the value of Vs. (7
points)
Applying KCL at node c, we get:

The power delivered by the voltage sources is given as:
.
1 
1 
1 
a c
1 A1 A
b
+–
Vs is
vc va– Vs=
vb va–
1
---------------
vb
1
----
vb vc–
1
---------------+ + 0=  3vb va– vc– 0=
vb
1
---- 1 1+=  vb 2=
vc
Vs 6+
2
--------------= va
6 Vs–
2
-------------=
vc vb–
1
--------------- is 1+= 
Vs 6+
2
-------------- 2– is 1+=  is
Vs
2
-----=
Pdelivered Vs is
Vs
2
2
----- 8= = =  Vs 4V=
UC Davis 4 Hussain Al-Asaad
3. MESH CURRENT ANALYSIS (20 POINTS)
Consider the circuit shown below.
1. Calculate the mesh currents in terms of Vs, Is, and R. (10 points)
By inspection of the circuit, we get: and .
Applying KVL for meshes 3 and 4, we get:
and
Rewriting the above two equations we get:
and
Solving the above two equations we get:
and
2. Assume that Vs = 1 V and Is = 1 A. If the total power absorbed by all resistors is 6 W,
determine the value of R. (10 points)
Substituting the values of Vs and Is, we get:
, , , and
The total power absorbed by the resistors is given by:

After substituting the values of the mesh currents in the above equation we get:

R
+–+ –
R
R
R
Vs
i1 i2
i3 i4
Vs
Is
Is
i1 Is= i2 Is–=
Vs– R i3 i1–  R i3 i4– + + 0= Vs R i4 i3–  R i4 i2– + + 0=
2Ri3 Ri4– Vs RIs+= R– i3 2Ri4+ V– s R– Is=
i3
Vs RIs+
3R
-------------------= i4
Vs RIs+
3R
-------------------–=
i1 1= i2 1–= i3
R 1+
3R
------------= i4
R 1+
3R
------------–=
Ptotal R i1 i2– 
2 R i2 i4– 
2 R i4 i3– 
2 R i3 i1– 
2+ + + 6= =
8R2 9R– 1+ 0=  8R 1–  R 1–  0=  R 1 1
8
--
 
 
 

UC Davis 5 Hussain Al-Asaad
4. SUPERPOSITION (20 POINTS)
By using superposition, determine the value of the voltage v in the circuit shown below.
Using superposition, , where vcs is the voltage due to the 4A current source and vvs is the voltage
due to 4V voltage source. To find the voltage vcs, the above circuit is reduced to the circuit below:
Applying KCL at bottom node c, we get:

Applying KCL at the right top node b, we get:

Applying KVL at the a-b-c-a loop, we get:

Hence, we can compute vcs as:
To find the voltage vvs, the original circuit is reduced to the circuit below:
Applying KVL at the a-b-c-a loop, we get:
.
Therefore, the value for v is calculated as follows:
1 
–
+ 4 V
1 
–
+
2 i
1 
4 A 2 va
+ va - + v -
i
v vcs vvs+=
1  1 
–
+
2 i
1 
4 A 2 va
+ va - + vcs -
i
ix
ima b
c
va 1  4  4= =
4 ix 2va i+ + + 0=  ix i+ 12–=
im i 2va+ + 0=  im i– 8–=
2i– 1  i– 8–  1  i– + + 0=  i 2–=
vcs 1  im  1  6–  6V–= = =
1 
–
+ 4 V
1 
–
+
2 i
1 
2 va
+ va - + vvs -
i
a b
c
va 1  0  0= =
2i– 1  i–  1  i–  4+ + + 0=  i 1=  vvs 1  i–  1V–= =
v vcs vvs+ 6– 1– 7V–= = =
UC Davis 6 Hussain Al-Asaad
5. THEVENIN CIRCUIT EQUIVALENT (20 POINTS)
Find the Thevenin circuit equivalent to the left of terminals a and b in the circuit shown below.
Note that the Thevenin voltage and resistance may be dependent on R.
Since the circuit has no independent voltage or current sources, then its thevenin equivalent is a resistor with a
resistance Rth that is computed by applying a 1A current source to the terminals a and b and solving for vab in order to
get Rth = vab .
Applying KCL at node c, we get:
Ohm’s law yields:
.
Now we can compute vab as follows:
Consequently,
R
a
b
R
2 vx
R
+ vx -
R
a
b
R
2 vx
R
+ vx - 1 A
c
ix
ix 1 2vx+=
vx Rix R 1 2vx+ = =  vx
R
1 2R–
---------------=
vab Rix Rix+ 2Rix 2R
vx
R
---- 
  2vx
2R
1 2R–
---------------= = = = =
Rth
2R
1 2R–
---------------=
a
b
Rth
2R
1 2R–
---------------=
UC Davis 7 Hussain Al-Asaad
6. NORTON CIRCUIT EQUIVALENT (20 POINTS)
Find the Norton circuit equivalent to the left of terminals a and b in the circuit below. Note that
the Norton current and resistance may be dependent on R.
It is very easy to compute RN as follows:
Applying KCL at node b, we get:
Applying KVL for loop a-c-b-a, we get:

So, the overall Norton circuit equivalent becomes as follows:
a
b
1 AR
1 VR
1 AR
1 A +–
a
b
R
R
R
RN
isc
ix
c
RN R 0  R+ R= =
ix 1+ isc=
1– Rix+ 0=  ix
1
R
--=  IN isc ix 1+
1
R
--- 1+ R 1+
R
------------= = = =
a
b
RR 1+R
------------