AEE1000 / MEC1151 - Electrical Circuits and Analysis
Dr Hong Fan, University of Glasgow, 2022
Tutorial – Topic 1 : Basic Concepts
[Charge and Current]
1.1 Determine the current flowing through an element if the charge flow is given by
(a) q (t ) = (3t + 8) mC
"ᵗ '
C) alt) 20 e cos 50T =
(b) q (t ) = (3e − t − 5e −2t ) nC :-/ It) :
dadt
(c) q (t ) = 20e −4t cos(50t ) µ C - ↳ᵗ
cos 50T
"+
Sin -50T
=
-80C -
1000 @
1. 12 ) qlt) =
(31-+8) t "
b) g. It ) ( ze b- e-
-
= -
j
"ᵗ
i. I / E) = d9_ i. 1 It) = DI
: -
e- ( 8005501-+1000 sin 50T)µA
dit #
dit
=
3mA # ᵗ "
1nA
-
=L-3@
-
+ toe
*
1.2 A current of 3.2 A flows through a conductor. Calculate how much charge passes through any
cross-section of the conductor in 20 s.
/ A :/ Cls
i.
Charge = 3.2×20
: 64C
#
1.3 The charge entering a certain element is shown in Figure 1.1. Find the current at:
(a) t = 1 ms (b) t = 6 ms (c) t = 10 ms b) @ t 6ms =
1- 32 ) @ t -2ms
-
91-11=80MC
91T) -80mL
-
:/ = = da
:@ t =
1mg at
91T) =
40mC = 80--80
6- 2
dqlt)
:-/ =
;÷n =
at =
0A
#
80 -
40
=
I
c) @ t -8ms -
= 40A
# qlt) = SOMC
Figure 1.1 i. @ t = 10ms
9. (t) = 40mC
.
'
.
/ =
rise
-
= dqct)
run It
-
40 -
80
- -
to -
8
= -
20 A
#
Dr Hong Fan, University of Glasgow, 2022
Tutorial – Topic 1 : Basic Concepts
[Charge and Current]
1.1 Determine the current flowing through an element if the charge flow is given by
(a) q (t ) = (3t + 8) mC
"ᵗ '
C) alt) 20 e cos 50T =
(b) q (t ) = (3e − t − 5e −2t ) nC :-/ It) :
dadt
(c) q (t ) = 20e −4t cos(50t ) µ C - ↳ᵗ
cos 50T
"+
Sin -50T
=
-80C -
1000 @
1. 12 ) qlt) =
(31-+8) t "
b) g. It ) ( ze b- e-
-
= -
j
"ᵗ
i. I / E) = d9_ i. 1 It) = DI
: -
e- ( 8005501-+1000 sin 50T)µA
dit #
dit
=
3mA # ᵗ "
1nA
-
=L-3@
-
+ toe
*
1.2 A current of 3.2 A flows through a conductor. Calculate how much charge passes through any
cross-section of the conductor in 20 s.
/ A :/ Cls
i.
Charge = 3.2×20
: 64C
#
1.3 The charge entering a certain element is shown in Figure 1.1. Find the current at:
(a) t = 1 ms (b) t = 6 ms (c) t = 10 ms b) @ t 6ms =
1- 32 ) @ t -2ms
-
91-11=80MC
91T) -80mL
-
:/ = = da
:@ t =
1mg at
91T) =
40mC = 80--80
6- 2
dqlt)
:-/ =
;÷n =
at =
0A
#
80 -
40
=
I
c) @ t -8ms -
= 40A
# qlt) = SOMC
Figure 1.1 i. @ t = 10ms
9. (t) = 40mC
.
'
.
/ =
rise
-
= dqct)
run It
-
40 -
80
- -
to -
8
= -
20 A
#