Cleveland State University
DEPARTMENT of MECHANICAL ENGINEERING
MCE 315 – Electrical Systems
Fall 2024
Instructor: Reza Harirforoush, PhD
Assignment #3
DUE: Friday, October 18, 2024 (Submission to BlackBoard)
Solve five out of seven problems.
Circuit Theorems
1- Problem 4.41*
To find RTh, consider the circuit below
14 W
a
6W
5W
b
RTh = 5 //(14 + 6) = 4W = R N
Applying source transformation to the 1-A current source, we obtain the circuit below.
6W
- 14V +
14 W
VTh
a
+
6V
3A
5W
b
At node a,
1
14 + 6 - VTh
V
= 3 + Th
6 + 14
5
¾
¾®
VTh = -8 V
VTh
= (-8) / 4 = -2 A
RTh
IN =
Thus,
RTh = RN = 4W ,
VTh = -8V,
I N = -2 A
2- Problem 4.72*
Solution
(a)
RTh and VTh are calculated using the circuits shown in Fig. (a) and (b) respectively.
From Fig. (a),
RTh = 2 + 4 + 6 = 12 ohms
From Fig. (b),
-VTh + 12 + 8 + 20 = 0, or VTh = 40 V
4W
6W
2W
RTh
2W
4W
- +
+
+
- 8V
20V
(a)
(b)
12V
6W
VTh
+ -
-
(b)
i = VTh/(RTh + R) = 40/(12 + 8) = 2A
(c)
For maximum power transfer,
RL = RTh = 12 ohms
(d)
p = VTh2/(4RTh) = (40)2/(4x12) = 33.33 watts.
Energy Storage Elements
3- The circuit shown in Figure has reached steady state before the switch closes at time t = 0.
(a) Determine the values of iL(t), vC(t), and vR(t) immediately before the switch closes.
(b) Determine the value of vR(t) immediately after the switch closes.
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3
4- Problem 6-46*
Under dc conditions, the circuit is as shown below:
2W
iL
+
vC
4W
3A
-
By current division,
iL =
4
(3) = 2A, vc = 0V
4+2
wL =
1æ1ö
1
L i 2L = ç ÷(2) 2 = 1J
2è2ø
2
wc =
1
1
C v c2 = (2)( v) = 0J
2
2
The Complete Response of RL and RC Circuits
5- Problem 7-21*
Solution
The circuit can be replaced by its Thevenin equivalent shown below.
Rth
Vth
+
-
2H
80
(60) = 40 V
80 + 40
80
R th = 40 || 80 + R =
+R
3
Vth =
4
I = i(0) = i(¥) =
Vth
40
=
R th 80 3 + R
2
1
1 æ 40 ö
÷ =1
w = L I 2 = (2)ç
2
2 è R + 80 3 ø
40
40
=1 ¾
¾® R =
R + 80 3
3
R = 13.333 Ω
6- A security alarm for an office building door is modeled by the circuit of Figure. The switch
represents the door interlock, and v is the alarm indicator voltage. Find v(t) for t > 0 for the
circuit of Figure. The switch has been closed for a long time at t=0-.
Solution: First, use source transformations to obtain the equivalent circuit
for t < 0:
for t > 0:
1
L
1
So iL ( 0 ) = 2 A, I sc = 0, Rt = 3 + 9 = 12 W, t =
= 2 =
s
Rt 12
24
and iL ( t ) = 2e -24t
t >0
Finally v ( t ) = 9 iL ( t ) = 18 e -24t
t >0
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7- (Use Simulink)
The circuit shown in Figure is at steady state before the switch closes at time t=0. The input to
the circuit is the current of the current source, 4 mA. The output of this circuit is the current in
the inductor, i(t). Plot the output i(t) as a function of t. Use the plot to obtain an analytic
representation of i(t) for t > 0.
For star problems, please refer to the textbook Fundamentals of Electric Circuits by Charles Alexander
and Matthew Sadiku, 7th Edition (McGraw-Hill).
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