American University of Beirut
Department of Electrical and Computer Engineering
EECE 210 - Electric Circuits
Spring 23-24
Quiz 1
Closed Book - No Programmable Calculators - 90 minutes
Mr. N. Ashkar
February 27, 2024
Name: ________________________
ID:___________________________
Solve the following problems;
Provide your answers on the attached Scantron card;
This question sheet must be returned with the Scantron card;
There is No penalty;
Mark with a pencil your LAST NAME, your First name Initial (FI) and your Middle name Initial (MI);
Mark your AUB ID Number in the box titled “ID NUMBER”;
Write the name of your course instructor on the Scantron card;
Use a pencil for marking your answers;
When using an eraser, make sure that you have erased well.
This exam has 9 pages.
1
Problem 1
In the circuit shown in Fig. 1, Vs={36, 24, 12, 18} V. Find 𝐼𝑠 .
(a) 1.5 A
(b) 1 A
(c) 0.5 A
(d) 0.75 A
(e) None of the above
Fig. 1: Circuit for Problem 1
Solution
𝑅𝑒𝑞 =
(8)(24)
15 (5)(20)
+5+
+
+4
32
3
25
𝑅𝑒𝑞 = 6 + 5 + 5 + 4 + 4 = 24 Ω
𝐼𝑠 =
𝑉𝑠
24
Problem 2
In the circuit shown in Fig. 2, R1= {6, 7, 12, 17} Ω and R2={24, 6, 12, 4} Ω.
Determine 𝑅𝐴𝐵 .
(a) 14 Ω
(b) 11 Ω
(c) 18 Ω
(d) 20 Ω
(e) None of the above
2
Fig. 2: Circuit for Problem 2
Solution
The 12 Ω resistor is short-circuited.
𝑅𝐴𝐵 = (((9 𝑝 18) + 6)𝑝𝑅2 ) + 𝑅1
𝑅𝐴𝐵 = ((6 + 6)𝑝𝑅2 ) + 𝑅1 =
12𝑅2
+ 𝑅1
12 + 𝑅2
Problem 3
In the circuit shown in Fig. 3, Vs= {10, 40, 20, 30} V. Find 𝐼𝑦 .
Fig. 3: Circuit for Problem 3
Answer
(a) 0.15 A
(b) 0.6 A
(c) 0.3 A
(d) 0.45 A
(e) None of the above
3
Solution
𝑅𝑎𝑏 = ((60 𝑝 30 𝑝 80) + (40 𝑝 120)) + 4
𝑅𝑎𝑏 = (16 + 30) + 4 = 50Ω
The current leaving the voltage source is:
𝐼𝑇 =
𝑉𝑠
𝐴
50
Using current divider rule,
𝐼𝑦 = 𝐼𝑇
120
𝑉𝑠 3 3𝑉𝑠
=
=
𝐴
120 + 40 50 4 200
Problem 4
In the circuit shown in Fig. 4, the voltage 𝑉𝑠 = {32, 48, 16, 8} V and R={100, 150,
200, 250} Ω . Find 𝑉0 .
Fig. 4: Circuit for Problem 4
(a) 20 V
(b) 30 V
(c) 10 V
(d) 5 V
(e) None of the above
Solution
Using Voltage divider rule, we obtain
𝑉0 = 𝑉𝑠
125
5𝑉𝑠
=
125 + 75
8
𝑉
4
Problem 5
In the circuit shown in Fig. 5, Is={6, 3, 12, 9} A. Determine 𝑉0 .
Fig. 5: Circuit for Problem 5
(a) -16 V
(b) -8 V
(c) -32 V
(d) -24 V
(e) None of the above
Solution
Using KCL, the current entering the parallel resistors branches is 2𝐼𝑠 A. Using current
divider rule, the current entering the 4Ω resistor is given by:
𝐼4 = 2𝐼𝑠
6
2𝐼𝑠
=
6+4+8
3
Using KVL,
𝑉0 = −4𝐼4 = −
8𝐼𝑠
3
Problem 6
In the circuit shown in Fig. 6, Vs={10, 5, 15, 25} V. Determine 𝑣0 .
Fig. 6: Circuit for Problem 6
5
(a) 16 V
(b) 8 V
(c) 24 V
(d) 40 V
(e) None of the above
Solution
Using KVL, we obtain
𝑦𝑖𝑒𝑙𝑑𝑠
−𝑉𝑠 − 20𝑖𝜎 + 5𝑖𝜎 + 40𝑖𝜎 = 0 →
𝑖𝜎 =
𝑉𝑠
25
Using Ohm’s law
𝑣0 = 40𝑖𝜎 =
40𝑉𝑠 8𝑉𝑠
=
25
5
Problem 7
In the circuit shown in Fig. 7, the voltage and current at the terminals of the circuit
element are zero for 𝑡 < 0. For 𝑡 ≥ 0 they are 𝑣 = 𝐴𝑒 −400𝑡 𝑉 and 𝑖 = 𝐴𝑒 −400𝑡 𝑚𝐴.
Calculate the total energy associated with the circuit element if A={20, 40, 80, 60}.
Fig. 7: Circuit for Problem 7
(a) 0.5 mJ
(b) 2 mJ
(c) 8 mJ
(d) 4.5 mJ
(e) None of the above
Solution
𝑝(𝑡) = 𝑣(𝑡)𝑖(𝑡) = 𝐴2 𝑒 −800𝑡 𝑚𝑊
The energy is given by:
6
∞
∞
2 −800𝑡
𝐸 = ∫ 𝑝(𝑡)𝑑𝑡 = ∫ 𝐴 𝑒
0
0
𝐴2
𝑑𝑡 =
𝑚𝐽
800
Problem 8
In the circuit shown in Fig. 8, R={1,2,5,3} Ω , Vs={2,8,60,24} V, P1=6 W, P2={5,2,174,42} W. Determine the current 𝐼𝑠 .
Fig. 8: Circuit for Problem 8
(a) 1 A
(b) 2 A
(c) 6 A
(d) 4 A
(e) None of the above
Solution
𝑃𝑇 = 𝑅𝐼𝑠2 + 𝑃1 + 𝑃2 = 𝑉𝑠 𝐼𝑠
Solving, we obtain
𝑅𝐼𝑠2 − 𝑉𝑠 𝐼𝑠 + (𝑃1 + 𝑃2) = 0
Problem 9
In the circuit shown in Fig. 9, R= {10, 5, 20,15} Ω. Determine 𝑅𝑎𝑏 .
Fig. 9: Circuit for Problem 9
7
(a) 6 Ω
(b) 3 Ω
(c) 12 Ω
(d) 9 Ω
(e) None of the above
Solution
Denote the node between the 2 upper resistors as node c, redraw the circuit, we obtain
𝑅𝑎𝑏 = ((𝑅 𝑝 𝑅) + 𝑅) 𝑝 𝑅 =
3𝑅
5
Problem 10
In the circuit shown in Fig. 10, Vs= {67, 134, 201, 268} V. Determine 𝐼𝑥
Fig. 10: Circuit for Problem 10
(a) 2 A
(b) 4 A
(c) 6 A
(d) 8 A
(e) None of the above
Solution
𝑦𝑖𝑒𝑙𝑑𝑠
It is clear that 12𝐼𝐴 = 18𝐼𝑋 →
5
2
𝐼𝑋 = 𝐼𝐴 . The current in the 8Ω resistor, using KCL,
3
is 𝐼𝐴 + 𝐼𝑋 = 𝐼𝐴 . Using KVL in the left loop, we obtain
3
8
𝑦𝑖𝑒𝑙𝑑𝑠
5
3𝑉𝑠 𝑦𝑖𝑒𝑙𝑑𝑠
2𝑉𝑠
−𝑉𝑠 + 8 ∗ 𝐼𝐴 − 3𝐼𝐴 + 12𝐼𝐴 = 0 →
𝐼𝐴 =
→
𝐼𝑋 =
3
67
67
Problem 11
In the circuit shown in Fig. 11, Vs={10, 20, 30, 40} V and R={2, 6, 4, 7} Ω.
Fig. 11: Circuit for Problem 11
The mesh-current equations are as follows:
Mesh 1 (corresponding to mesh-current 𝑖1 ): 3𝑖1 − 𝑎𝑖2 + 𝑉𝑠 = 0
Mesh 2 (corresponding to mesh-current 𝑖2 ): −3𝑖1 + 𝑏𝑖2 − 𝑐𝑖3 − 12 = 0
Mesh 3 (corresponding to mesh-current 𝑖3 ): −5𝑖2 + 𝑒𝑖3 + 12 = 0
where a, b, c, and e are constants having the unit ohm. Find the value of a+b+e.
(a) 23 Ω
(b) 27 Ω
(c) 25 Ω
(d) 28 Ω
(e) None of the above
Solution
𝑦𝑖𝑒𝑙𝑑𝑠
Mesh 1: 𝑉𝑠 + 3(𝑖1 − 𝑖2 ) = 0 →
𝑎=3
𝑦𝑖𝑒𝑙𝑑𝑠
Mesh 2: 𝑅𝑖2 + 3(𝑖2 − 𝑖1 ) + 5(𝑖2 − 𝑖3 ) − 12 = 0 →
12 = 0. Comparing, we get b = R+8, and c = 5.
𝑦𝑖𝑒𝑙𝑑𝑠
Mesh 3: 12+5(𝑖3 − 𝑖2 ) + 5𝑖3 = 0 →
e=10.
− 3𝑖1 + (𝑅 + 8)𝑖2 − 5𝑖3 −
−5𝑖2 + 10𝑖3 + 12 = 0. This implies that
9
Therefore,
𝑎 + 𝑏 + 𝑒 = 3 + 𝑅 + 8 + 10 = 𝑅 + 21
Problem 12
In the circuit shown in Fig. 12, Vs= {2, 3, 1, 7} V. Find 𝑉2 .
Fig. 12: Circuit for Problem 12
(a) 20 V.
(b) 30 V
(c) 10 V
(d) 70 V
(e) None of the above
Solution
In the left of the circuit and using KVL, we obtain
𝐼1 =
𝑉𝑠
3
From the right of the circuit,
𝑉2 = 30𝐼1 = 10𝑉𝑠
10
Problem 13
In the circuit shown in Fig. 13, Vs= {2,4,6,8} V and Is= {8,6,4,2} A. Find the value of
𝑉𝐴𝐵 .
Fig. 13: Circuit for Problem 13
(a) 25 V
(b) 20 V
(c) 15 V
(d) 10 V
(e) None of the above
Solution
Using KVL,
1
𝑉𝐴𝐵 = 𝑉𝑠 + 3𝐼𝑠
2
Problem 14
In the circuit shown in Fig. 14, Vs= {9,12,15,18} V. Find 𝐼𝑥 .
Fig. 14: Circuit for Problem 14
11
a) 1.5 A
b) 2 A
c) 2.5 A
d) 3 A
e) None of the above
Solution
At the node above 𝐼𝑋 , we have
𝑉𝑋 − 𝑉𝑠 𝑉𝑋 𝑉𝑋 − 4𝐼𝑋
+ +
=0
2
4
2
But 𝑉𝑋 = 4𝐼𝑋 . This implies
𝑦𝑖𝑒𝑙𝑑𝑠
4𝐼𝑋 − 𝑉𝑠
𝑉𝑠
+ 𝐼𝑋 = 0 →
𝐼𝑋 =
2
6
12