Chapter
1
Basic
concepts
Fund
quest o
S
of
Electric
circuits
Notes
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Basic Concepts of Electric Circuits: Exercises and Questions, Exercises of Electrical Circuit Analysis
A collection of exercises and questions related to basic concepts of electric circuits. It covers topics such as current, voltage, power, and energy. Suitable for students studying introductory electrical engineering or physics courses. It can be used as a supplementary resource for practice and self-assessment.
Typology: Exercises
2024/2025
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Chapter
Basic
concepts
Fund
quest o
S
of
Electric
circuits
Notes
More on second page SAMPLE t 11
xion) = 6. 24x10.8 elections = 10 , 000 , 000 , 000 ( . 60 2x10) = 1 . 602x10C ,
q
(
15t
10
3)m) t = 1. Os i = = 0-15-10(3) e to = 0
15
30 bt =
- 15 +
302 3 q(t) =
15 + 30e 3() E
. 506mA
100 %
a)a = 5 P = U : I = v
T
b) P = VI I= v I 100 #lov = v 4 . W = Pt P= VI p = 115 (12) V= 115( = 13805 I = 12 A w= 1380(24) = 33 , 120 W/n/
12 kw/hr, 45A zV j P , =
5(45)
225W P = 2 (45) -Pow P = V I 43 = ( . 12(20)] (25) I bow Pu = 3(25) = 75W
15
.
18
·
p
= Go
v= 1200
12
=2010-
= 6 - 24x10' elections
26 ·
E
= PE
5axin
= 0. 0833
loin
= 0. 10
v(t) = locos (2t)
W
=
Sid
= 5(
. 0833) + 4(0. 0833)
3(
.
- 8(
.
- 4(
.
i(t)
=
20(
- e - 0.
5t)nA P=
VI 86400
= 2. 33
MWh
Gar
I Q
8
a)q(x)
= Side u =
a
27 .
= /20(
e
- 0 . 5
-)de
du
=
- 5d
0
. 5A
= I
o
=
de
=
5(86,
50
43
,
200(
.
24x108)
=
#43,
200 L E20x
elections
=
40(
fe"]
b)
I ma 1600h
=
yo(u
e4]
=
666
= - 40(
. 5 + -
e
- 0
.
14
i(f)
= Ge ma
21 .
days
[(t)
=
40(
. 5 + -
e
- 0
.
5 )
i(
- )
=
Et
A.
q(1)
=
40(
.
5(1)
2
V(t)
=
lo W=
PE
P= V :
=
4 .26(
a =
Sid
b)
p
= Vi
a)a
S
: d
=
2
W
=
viE
2
V
(1)
= 10 cos(2(1) =
/Gede
u =
- 2t w =
12
(10)
(00) (60)
= 4
. 49395 O
du =
- 2dt
=
If"-
↓
t
- 728m
i(l)
= 20(l
e
5xi)
de
=
8684 A =
-Ebe
-I
W = PE
P =
9439(
. 8604)
:
be
7460(1800)
= loypo 1000 78
. 65W
=
3e"
= 13428000 [
120-
2
30(4) : 120
30t
=
- 32
zt
lo So
.
P
= V :
16
A
30(2)
=
602460
120
=
3e
(
↓Foto
Esm(
!
book for
uurs
O
500(4)
= 2
. 4 Kn
x oses
a
v(V)
b)
P = VI
22 .
=
24 cents
I
50
- (
=
10
↑
=
gar
=
10 (6) ide
= dy
Q
= 15
2524624 =
10(6(
-2)
= 2t)
dt
= 4
P
= VI
= 10FFze2t)
=
v(t)
= - 120
- 25
= 7 .
5x
his
30th)
=
150 + Oct(2 P= f)
(bei)
↑
Every
(
- 30 + )
= 600
150 + 2
74
i =
2000 A
4
2
- 600
- 150/4)
: 0
C) B
=
t
=
3m/x
P(t)
150(2)
= 300
300
- 600
- 150(2) =
- 300
=
3x10-3s
·
so
Y
- 600
- 150(3)
== 150
W=
NPat
4t
·
(s)
=
-10e
dt
n
=
4t
idei
=
- 300 -
D
I
·
dh
=-
=
se
b)
W=
150
- 150
=
180e")
=
2000
/No
Toj
= 180e
- "
3
=
2000(3x
- To
Fo:2A P
=
(80e
(
"()
El
·
=
17p. T
a)
E
= PE
- Fo
= PorA
=
12
se
= Book
- 200X
800x
200x10 + 1200x + 200X
↓
3A
: )
Q =?
= 10 ,
000
W
Q
= Side 210kwh
&
P
,
=
30(-0) =
43 ,
20t
=
- 180 - -
30
72
28
- 56 + 43
O
b)
HORNY
= 0
.
42
42
= 6(12)
- 210 = 156
P
= 10 +
/
,
200
25
=. 72
W
43
=?
P
=
210
- 156
I
C =
- 24x10" electrons
py
=
28(1)
=
54W
Pool
a) 6
. 482x
Elections
x
eas
= 0
.
1038
=
28W
P = VT
45
= 28(2)
=
u
iid E = pe
P = Vi
= 564
a
= fit
vit
↑o
=
5(
- 3)(2)
= Qu
b)
1
.
24x10" ele
yes
necess
= -0. 108
=
- 30
w =
3
.888x
.
5
. 832x
%
m
- 2
. 46x10"
electors
I
10e ers
=
- 042
d)
- 628x e
exons
'Beeo
= - .
080c.