EGR271
Midterm Exam
1. For the circuit shown in Fig.1, find:
a. Vx
b. i3
Fig. 1
Using KCL we have
𝑖1 = 𝑖2 + 𝑖3
Let the voltage at the node be V
12 − 𝑉 𝑉 𝑉
= +
2
4 2
12 𝑉 𝑉 𝑉
− = +
2
2 4 2
12 𝑉 𝑉 𝑉
= + +
2
2 2 4
6=
2𝑉 + 2𝑉 + 𝑉
4
24 = 5𝑉
𝑉 = 𝑉𝑥 =
24
= 4.8
5
Current (i3) will be
𝑖3 =
𝑣
𝑅3
𝑖3 =
4.8
2
𝑖3 = 2.4 𝐴
(1/2)
2. For the circuit of Fig. 2, determine i
Fig. 2
Witing an appopiate Kcl equation fo node 1,
𝑉1 𝑉1 − 𝑉2
6=
+
10
12
And fo node 2.
𝑉2 𝑉2 − 𝑉1
4= +
5
12
The equation becomes
𝑉1 𝑉1 − 𝑉2
+
10
12
6𝑉1 + 5𝑉1 − 5𝑉2
6=
60
360 = 11𝑉1 − 5𝑉2
6=
𝑉2 𝑉2 − 𝑉1
+
5
12
12𝑉2 + 5𝑉2 − 5𝑉1
4=
12
4=
48 = −5𝑉1 + 17𝑉2
[
360
11 −5 𝑉1
][ ] = [
]
48
−5 17 𝑉2
V1=39.2593
V2 =14.3704
3. For the circuit shown in Fig.3, using supernode analysis write the KCL equations and the voltage
source equation.
Fig. 3
We stat by writing the KCL equation
𝑉1 − 𝑉2 𝑉1 − 𝑉3
6+6=
+
3
4
𝑉2 − 𝑉1 𝑉3 − 𝑉1 𝑉3 𝑉2
−6 − 12 =
+
+ +
3
4
5
1
𝑉2 − 𝑉1 = −18𝑉
We now consider the super node .2 current source are connected, and four
resistors. In order to continue we reduce the above equations while replacing V2=18+V1
12 =
12 =
𝑉1 − 𝑉2 𝑉1 − 𝑉3
+
3
4
𝑉1 − (−18 + 𝑉1 ) 𝑉1 − 𝑉3
+
3
4
𝑉1 + 18 − 𝑉1 𝑉1 − 𝑉3
+
3
4
18 𝑉1 − 𝑉3
12 =
+
3
4
12 =
6=
𝑉1 − 𝑉3
4
24 = 𝑉1 − 𝑉3
For the second equation,
−6 − 12 =
𝑉2 − 𝑉1 𝑉3 − 𝑉1 𝑉3 𝑉2
+
+ +
3
4
5
1
−18 =
(−18 + 𝑉1 ) − 𝑉1 𝑉3 − 𝑉1 𝑉3 (−18 + 𝑉1 )
+
+ +
3
4
5
1
−18 𝑉3 − 𝑉1 𝑉3 −18 + 𝑉1
−18 =
+
+ +
3
4
5
1
𝑉3 − 𝑉1 𝑉3 −18 + 𝑉1
+ +
4
5
1
𝑉3 − 𝑉1 𝑉3 𝑉1
−3 =
+ +
4
5
1
5𝑉3 − 5𝑉1 + 4𝑉3 + 20𝑉!
−3 =
20
9𝑉3 + 15𝑉1
−3 =
20
−21 =
−60 = 9𝑉3 + 15𝑉1
We now have two linear equations to solve (highlighted in yellow).
24 = 𝑉1 − 𝑉3
−60 = 15𝑉1 + 9𝑉3
To solve these two linear equations, we follow the following step
[
1
15
−1 𝑉1
24
][ ] = [
]
9 𝑉3
−60
𝑉1 = 6.5𝑉
𝑉3 = −17.5𝑉
𝑉2 = −18𝑉 + 6.5
𝑉2 = −11.5𝑉
4. For the circuit shown in Fig. 4, determine the power associated with the 4V source.
Fig. 4
We fist define two clockwise mesh currents. we write the KVL equation as
following.
−8 + 4𝑖1 + 2(𝑖1 − 𝑖2 ) − 4 = 0
−12 + 4𝑖1 + 2𝑖1 − 2𝑖2 = 0
6𝑖1 − 2𝑖2 = 12
doing the same for mesh 2
2 + 4 + 2(𝑖2 − 𝑖1 ) + 5𝑖2 = 0
6 + 2𝑖2 − 2𝑖1 + 5𝑖2 = 0
7𝑖2 − 2𝑖1 = −6
2𝑖1 − 7𝑖2 = 6
Now we have 2 linear equations to solve.
6𝑖1 − 2𝑖2 = 12
2𝑖1 − 7𝑖2 = 6
To solve these two linear equations, we follow the following step
[
6 −2 𝑖1
12
][ ] = [ ]
6
2 −7 𝑖2
𝑖1 = 1.895𝐴
𝑖2 = −0.316 𝐴
5. For the circuit of Fig. 5, using supermesh analysis write the KVL equations and add the current
source equation.
Fig. 5
A super mesh(i1,i3)
14 + 1(𝑖1 − 𝑖2 ) + 3(𝑖3 − 𝑖2 ) + 1𝑖2 = 0
Mesh i2
1(𝑖2 − 𝑖1 ) + 3(𝑖2 − 𝑖3 ) + 2𝑖2 = 0
𝑖1 − 𝑖3 = 6
𝑖1 = 6 + 𝑖3
We now replace in each i1 each equation
14 + 1(6 + 𝑖3 − 𝑖2 ) + 3(𝑖3 − 𝑖2 ) + 1𝑖2 = 0
20 + 4𝑖3 − 1𝑖2 = 0
20 = −4𝑖3 + 1𝑖2
1(𝑖2 − 6 − 𝑖3 ) + 3(𝑖2 − 𝑖3 ) + 2𝑖2 = 0
−6 − 4𝑖3 + 6𝑖2 = 0
−6 = 4𝑖3 − 6𝑖2
[
I3 =4.0714
−4 1 𝑖3
20
][ ] = [ ]
4 −6 𝑖2
−6
I2 =3.7143