EE301
Exam 1 Review
1) If 85 Joules of Energy is required to move 368 Coulombs of charge, what is the Voltage Potential
created?
𝑉𝑉 =
𝑊𝑒
85
=
= 𝟎. 𝟐𝟐𝟑𝟑𝟏𝟏 𝑽𝑽
𝑄
368
2) Through a given surface of a conductor’s cross section area, 15 mA of current passes at a rate of
200 µs. What is the amount of charge?
𝑄 = 𝐼 ∙ 𝑡 = (15𝑥10−3 ) ∙ (200𝑥10−6 ) = 3𝑥10−6 = 𝟑𝟑 𝝁𝑪
3)
A 6 V battery has a rating of 52 A-h. What is the average amount of current if the battery is
drained after operating for 145.8 minutes?
𝐼𝑎𝑣𝑒 =
52 𝐴 − ℎ
= 𝟐𝟐𝟏𝟏. 𝟒𝟒 𝑨
145.8 𝑚𝑖𝑛
�
�
60 𝑚𝑖𝑛/ℎ
4) What is the Voltage across a 220 Ω resistor if 15 mA passes through it?
15 mA
+
V
220 Ω
-
𝑉𝑉 = 𝐼𝑅𝑅 = (15𝑥10−3 ) ∙ (220) = 𝟑𝟑. 𝟑𝟑 𝑽𝑽
5) According to Ohm’s Law, what is proportional to Current if Resistance is constant?
𝑽𝑽𝒐𝒐𝒍𝒕𝒂𝒈𝒆 ∝ 𝑪𝒖𝒓𝒓𝒆𝒏𝒕
6) Find 𝑹𝑹𝑶𝑶 for the circuits below.
a.
60 Ω
15 Ω
28 Ω
19 Ω
6Ω
𝑅𝑅 ′ =
𝑅𝑅𝑇𝑇 =
𝑅𝑅𝑇𝑇
(6)(28)
+ 15 = 19.94 Ω
6 + 28
(19.94)(19)
+ 60 = 𝟔𝟗. 𝟕𝟑𝟑 Ω
19.94 + 19
b.
10 Ω
25 Ω
50 Ω
20 Ω
16 Ω
28 Ω
22 Ω
𝑅𝑅𝑇𝑇
𝑅𝑅 ′ =
𝑅𝑅 ′′ =
(10 + 20)(28)
+ 16 = 30.48 Ω
10 + 20 + 28
𝑅𝑅 ′′′ =
32 Ω
(30.48)(50)
= 18.94 Ω
30.48 + 50
(22)(32)
= 13.04 Ω
22 + 32
𝑅𝑅𝑇𝑇 =
(13.04 + 18.94)(25)
= 𝟏𝟏𝟒𝟒. 𝟎𝟑𝟑 Ω
13.04 + 18.94 + 25
𝑅𝑅 ′ =
(9 + 14 + 7)(18)
+ 12 = 23.25 Ω
9 + 14 + 7 + 18
𝑅𝑅𝑇𝑇 =
(46)(23.25)
= 𝟏𝟏𝟓𝟓. 𝟒𝟒𝟒𝟒 Ω
46 + 23.25
c.
24 Ω
18 Ω
4Ω
12 Ω
𝑅𝑅𝑇𝑇
9Ω
18 Ω
14 Ω
7Ω
𝑅𝑅 ′′ = 24 + 18 + 4 = 46 Ω
7) Find the quantities for the circuit below.
Supply (𝑷𝑷𝑰𝑰𝑰𝑰 )
+
-
30 V
ƞ𝒐𝒐
Stage 1
80 Ω
Stage 2
Stage 3
Load (𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶 )
10 Ω
𝑰𝑰𝟏𝟏
𝑰𝑰𝑺𝑺
ƞ𝟏𝟏
100 Ω
𝑰𝑰𝟐𝟐
ƞ𝟐𝟐
200 Ω
𝑰𝑰𝟑𝟑
ƞ𝟑𝟑
𝑰𝑰𝑳𝑳
200 Ω
25 Ω
ƞ𝑳𝑳
a. Find 𝑰𝑰𝑺𝑺
1
1
1
1 −1
+
+
+ � + 10 + 80 = 106.67 Ω
𝑅𝑅𝑇𝑇 = �
100 200 200 25
𝐼𝑆 =
𝑉𝑉𝑆
30
=
= 𝟎. 𝟐𝟐𝟖 𝑨
𝑅𝑅𝑇𝑇 106.67
b. Find 𝑷𝑷𝑰𝑰𝑰𝑰
𝑃𝐼𝑁 = 𝐼𝑆 ∙ 𝑉𝑉𝑆 = (0.28)(30) = 𝟖. 𝟒𝟒 𝑾
c. Find 𝑰𝑰𝟏𝟏 , 𝑰𝑰𝟐𝟐 , 𝑰𝑰𝟑𝟑 , and 𝑰𝑰𝑳𝑳 (Hint: Use Current divider)
𝑅𝑅𝑇𝑇′ = �
1
1
1
1 −1
+
+
+ � = 16.67 Ω
100 200 200 25
𝑅𝑅𝑇𝑇′
16.67
(𝐼𝑆 ) =
(0.28) = 0.04668 = 𝟒𝟒𝟔. 𝟔𝟖 𝒎𝑨
𝐼1 =
𝑅𝑅1
100
𝑅𝑅𝑇𝑇′
16.67
(𝐼𝑆 ) =
(0.28) = 0.02334 = 𝟐𝟐𝟑𝟑. 𝟑𝟑𝟒𝟒 𝒎𝑨
𝐼2 =
𝑅𝑅2
200
𝐼3 =
𝐼𝐿 =
𝑅𝑅𝑇𝑇′
16.67
(𝐼𝑆 ) =
(0.28) = 0.02334 = 𝟐𝟐𝟑𝟑. 𝟑𝟑𝟒𝟒 𝒎𝑨
𝑅𝑅3
200
𝑅𝑅𝑇𝑇′
16.67
(𝐼𝑆 ) =
(0.28) = 0.1867 = 𝟏𝟏𝟖𝟔. 𝟕 𝒎𝑨
𝑅𝑅𝐿
25
d. Find 𝑷𝑷𝑳𝑳𝒐𝒐𝒔𝒔 𝟏𝟏 at Stage 1, 𝑷𝑷𝑳𝑳𝒐𝒐𝒔𝒔 𝟐𝟐 at Stage 2, and 𝑷𝑷𝑳𝑳𝒐𝒐𝒔𝒔 𝟑𝟑 at Stage 3
𝑃𝐿𝑜𝑠𝑠 1 = (𝐼𝑆 )2 ∙ (𝑅𝑅80 ) = (0.28)2 (80) = 𝟔. 𝟐𝟐𝟕𝟐𝟐 𝑾
𝑃𝐿𝑜𝑠𝑠 2 = (𝐼𝑆 )2 ∙ (𝑅𝑅10 ) + (𝐼1 )2 ∙ (𝑅𝑅1 ) = (0.28)2 (10) + (0.04668)2 (100) = 𝟏𝟏. 𝟎𝟎𝟐𝟐 𝑾
𝑃𝐿𝑜𝑠𝑠 3 = (𝐼2 )2 ∙ (𝑅𝑅2 ) + (𝐼3 )2 ∙ (𝑅𝑅3 ) = (0.02334)2 (200) + (0.02334)2 (200)
𝑃𝐿𝑜𝑠𝑠 3 = 𝟎. 𝟐𝟐𝟏𝟏𝟕𝟗 𝑾
e. Find 𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶
f.
𝑃𝑂𝑈𝑇𝑇 = (𝐼𝐿 )2 ∙ (𝑅𝑅𝐿 ) = (0.1867)2 (25) = 𝟎. 𝟖𝟕𝟏𝟏𝟒𝟒 𝑾
Calculate Total Efficiency (ƞ𝑶𝑶𝒐𝒐𝒕𝒂𝒍 ) of the circuit
ƞ𝑇𝑇𝑜𝑡𝑎𝑙 =
𝑃𝑂𝑈𝑇𝑇
0.8714
(100) =
(100) = 𝟏𝟏𝟎. 𝟑𝟑𝟕%
𝑃𝐼𝑁
8.4
g. Calculate the Efficiency at each Stage (ƞ𝟏𝟏 , ƞ𝟐𝟐 , ƞ𝟑𝟑 , 𝑎𝑛𝑑 ƞ𝑳𝑳 ). Hint: for each stage 𝑷𝑷𝑰𝑰𝑰𝑰 is
the 𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶 from the previous stage. 𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶 is 𝑷𝑷𝑰𝑰𝑰𝑰 − 𝑷𝑷𝑳𝑳𝒐𝒐𝒔𝒔 at that stage.
𝑃𝑂𝑈𝑇𝑇 1
𝑃𝐼𝑁 1 − 𝑃𝐿𝑜𝑠𝑠 1
8.4 − 6.272
2.128
(100) =
(100) =
(100) =
(100)
𝑃𝐼𝑁
𝑃𝐼𝑁
8.4
8.4
ƞ1 = 𝟐𝟐𝟓𝟓. 𝟑𝟑𝟑𝟑%
ƞ1 =
𝑃𝑂𝑈𝑇𝑇 2
𝑃𝐼𝑁 2 − 𝑃𝐿𝑜𝑠𝑠 2
2.128 − 1.002
1.126
(100) =
(100) =
(100) =
(100)
𝑃𝐼𝑁 2
𝑃𝑂𝑈𝑇𝑇 1
2.128
2.128
ƞ2 = 𝟓𝟓𝟐𝟐. 𝟗𝟏𝟏%
ƞ2 =
𝑃𝑂𝑈𝑇𝑇 3
𝑃𝐼𝑁 3 − 𝑃𝐿𝑜𝑠𝑠 3
1.126 − 0.2179
0.9081
(100) =
(100) =
(100) =
(100)
𝑃𝐼𝑁 3
𝑃𝑂𝑈𝑇𝑇 2
1.126
1.126
ƞ3 = 𝟖𝟎. 𝟔𝟓𝟓%
ƞ3 =
ƞ𝐿 =
𝑃𝑂𝑈𝑇𝑇 𝐿𝑜𝑎𝑑
𝑃𝑂𝑈𝑇𝑇 𝐿𝑜𝑎𝑑
0.8714
(100) =
(100) =
(100) = 𝟗𝟓𝟓. 𝟗𝟔%
𝑃𝐼𝑁 𝐿𝑜𝑎𝑑
𝑃𝑂𝑈𝑇𝑇 3
0.9081
h. Does ƞ𝑶𝑶𝒐𝒐𝒕𝒂𝒍 = ƞ𝟏𝟏 ∙ ƞ𝟐𝟐 ∙ ƞ𝟑𝟑 ∙ ƞ𝑳𝑳 ?
ƞ𝑇𝑇𝑜𝑡𝑎𝑙 = (0.2533)(0.5291)(0.8065)(0.9596) = 0.1037 = 𝟏𝟏𝟎. 𝟑𝟑𝟕%
𝟏𝟏𝟎. 𝟑𝟑𝟕% = 𝟏𝟏𝟎. 𝟑𝟑𝟕%
8) Use Voltage Divider ONLY to find all the voltages for the circuit below. Do not simplify the
circuit.
𝑉𝑉1 =
𝑉𝑉2 =
17
(60) = 𝟐𝟐𝟒𝟒. 𝟐𝟐𝟗 𝑽𝑽
25 + 17
8
(60) = 𝟏𝟏𝟎. 𝟒𝟒𝟑𝟑 𝑽𝑽
𝑉𝑉3 =
8 + 26 + 12
𝑉𝑉4 =
+
25
(60) = 𝟑𝟑𝟓𝟓. 𝟕𝟏𝟏 𝑽𝑽
25 + 17
+
25 Ω
60 V
𝑽𝑽𝟏𝟏
-
+
-
8Ω
+
26 Ω
17 Ω
𝑽𝑽𝟐𝟐
-
12
(60) = 𝟏𝟏𝟓𝟓. 𝟔𝟓𝟓 𝑽𝑽
𝑉𝑉5 =
8 + 26 + 12
12 Ω
9) Perform a Source Transformation on the Voltage Source and Current Source and then find 𝑰𝑰𝑿 .
(Hint: Use Nodal Analysis to solve for 𝑰𝑰𝑿 after Source Transformation)
12 Ω
48 V
𝐼=
𝑰𝑰𝒙𝒙
+
-
𝑉𝑉 48
=
=4𝐴
𝑅𝑅 12
4A
12 Ω
−4 +
𝑰𝑰𝒙𝒙
𝑉𝑉
+
6Ω
-
8Ω
1.875 A
𝑉𝑉 = 𝐼𝑅𝑅 = (1.875)(8) = 15 𝑉𝑉
8Ω
+
6Ω
-
𝑉𝑉 𝑉𝑉 𝑉𝑉 − 15
+ +
= 0 {𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑏𝑦 48}
12 6
8
−192 + 4𝑉𝑉 + 8𝑉𝑉 + 6(𝑉𝑉 − 15) = 0
−192 + 12𝑉𝑉 + 6𝑉𝑉 − 90 = 0
18𝑉𝑉 = 282
𝑉𝑉 = 15.67 𝑉𝑉
+
-
𝐼𝑋 =
15 V
𝑉𝑉
15.67
=
= 𝟐𝟐. 𝟔𝟏𝟏 𝑨
𝑅𝑅6
6
𝑽𝑽𝟒𝟒
+
+
26
(60) = 𝟑𝟑𝟑𝟑. 𝟗𝟏𝟏 𝑽𝑽
8 + 26 + 12
𝑽𝑽𝟑𝟑
𝑽𝑽𝟓𝟓
-
10) Find the Thevenin Equivalent from the 𝑹𝑹𝑳𝑳 perspective and draw the New Circuit.
20 Ω
6Ω
5Ω
4Ω
𝑹𝑹𝑳𝑳
2A
20 Ω
6Ω
5Ω
4Ω
𝑅𝑅𝑇𝑇𝐻 = 4+ 6 = 10 Ω
𝑽𝑽𝒐𝒐
20 Ω
5Ω
𝑽𝑽𝑶𝑶𝑻𝑻
𝑹𝑹𝑻𝑻𝑻𝑻
6Ω
4Ω
2A
Note the 5 and 20 Ω
Resistor are in an open
branch therefore they are
not included in the 𝑹𝑹
calculation.
Note the 6 Ω Resistor is on
an open branch therefore
no current passes through it
and no voltage drop occurs.
+
𝑽𝑽𝑶𝑶𝑻𝑻
-
10 Ω
𝑉𝑉𝑇𝑇𝐻 = IR4 = 8 V
FINAL THEVENIN CIRCUIT
8V
+
-
𝑹𝑹𝑳𝑳