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DC Networks - Electrical and Electronic Principles.
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DC Networks - Electrical and Electronic Principles
Physics
Student Name:
Instructor Name:
Course Number:
19 April 2019
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Question 3
i)
For resistors R
1
,R
2
and R
3
1/R=1/20 + 1/60 +1/10 = 10/60
R= 60/10 = 6
Ω
For R
4
and R
5
60
×
660
660
=
54.55
Ω
(Product over sum)
Total resistance is 6
Ω+
54.55
Ω
+
(
2
×
1000
)
Ω
=2060.55 Ω
ii)
V=IR (Ohm’s law)
3=2060.55 I
I=
3
2060.
.
=0.001456 A
iii)
V
1
=IR
=0.001456
×
6
=0.008737 V
V
2
=IR
=0.001456
×
54.55
=0.074248 V
V
3
=IR0.001456
×
2000
=2.912 V
iv)
Current through R
5
=
60
660
×
0.001456 A
=0.0001324 A
Circuit B
Circuit A (1N4148 Diode) in forward bias
V
1
0.1V
0.2V
0.3V
0.4V
0.5V
0.6V
0.7V
0.8V
0.9V
1.0V
1.1V
1.2V
V
D
0.1V
0.2V
0.3V
0.4V
0.5V
0.6V
0.7V
0.8V
0.9V
1.0V
1.1V
1.2V
I
D
0.0001
V
0.0002V
0.0003V
0.0004V
0.0005V
0.000
6V
0.000
7V
0.000
8V
0.000
85V
0.000
96V
0.000
96V
0.000
98V
Circuit A (1N4148 Diode) in reverse bias
V
1
2V
4V
6V
8V
10V
12 V
V
D
-2 V
-4V
-6V
-8V
-10V
-12V
I
D
-0.002 A
-0.004A
-0.006A
-0.008A
-0.0086A
-0.0088A
Circuit B (BAT54 Diode) in forward bias
V
1
0.1V
0.2V
0.3V
0.4V
0.5V
0.6V
0.7V
0.8V
0.9V
1.0V
1.1V
1.2V
V
D
0.1V
0.2V
0.3V
0.4V
0.5V
0.6V
0.7V
0.8V
0.9V
1.0V
1.1V
1.2V
I
D
0.0001V
0.0002V
0.00
03V
0.00
04V
0.00
05V
0.00
06V
0.00
07V
0.00
08V
0.00
09V
0.00
095V
0.00
095V
0.00
096V
Circuit B (BAT54 Diode) in forward bias
V
1
2V
4V
6V
8V
10V
12 V
V
D
-2V
-4V
-6V
-8V
-10V
-12 V
I
D
-0.002 A
-0.004A
-0.006A
-0.008A
-0.0082A
-0.084A
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