ELECTRICAL PRINCIPLES.

Can you please make sure all working is shown for Task 1 & 2. For task 3&4 can you please use the iciruit app for apple mac and take screenshots of the circuits and fill the tables. http://icircuitapp.com Can you please make sure a graph is put together from the completed tables as well. If you need any further information please let me know but all the requirements for completion are in the assignment document attached. Many thanks, Steven

Running head: ELECTRICAL PRINCIPLES 1
Electrical and Electronic Principles
Name
Institution

ELECTRICAL PRINCIPLES 2
Electrical and Electronic Principles
Task 1 Circuit A
Part (i)
R1=R2=100 Ω , R3 =550 Ω , R4=330 Ω , R5=200 Ω , supply voltage=12V
Total Resistance , R=R1∨|R2+R3+R4|∨R5
¿ R1 R2
R1+ R2
+ R3 + R4 R5
R4 + R5
R1 R2
R1+ R2
= 100 ×100
100+100 =50 Ω
R4 R5
R4 + R5
= 330 ×200
330+200 =120 Ω
Total Resistance , R=50+550+120=720 Ω
Part (ii)
Current drawn ¿ supply , I= Supply Voltage
Total Resistance
¿ 12
720 A= 1
60 A=0.016667 A=16.667 mA
Part (iii)
Voltage across V 1=I R1 R2
R1+ R2
=0.016667 A ×50 Ω=0.8333 V
Voltage across V 2 =I R3=0.016667 A ×550 Ω=9.1667 V

ELECTRICAL PRINCIPLES 3
Voltage across V 3 =I R4 R5
R4 + R5
=0.016667 A ×120 Ω=2V
Part (iv)
Current through R4=I R5
R4 + R5
=0.016667 A × 200 Ω
(330+200)Ω
¿ 0.016667 A × 200
550 =0.00606 A=6.06 mA
Task 1 Circuit B
Part (i)
R1=20 Ω, R2=60 Ω, R3=10 Ω, R4 =60 Ω , R5=600 Ω , R6=2000 Ω supply voltage=3V
Total Resistance , R=R1∨¿ R2∨¿ R3+R4∨¿ R5+ R6
¿ R1 R2
R1+ R2
∨¿ R3+ R4 R5
R4+ R5
+ R6
R1 R2
R1+ R2
= 20 ×60
20+60 =15 Ω
R1 R2
R1+ R2
∨¿ R3= 15× 10
15+10 =6 Ω
R4 R5
R4 + R5
= 60 ×600
60+ 600 = 600
11 Ω
Total Resistance , R=6+ 600
11 +2000= 22 66 6
11 Ω=20 60.545 5 Ω
Part (ii)

ELECTRICAL PRINCIPLES 4
Current drawn ¿ supply , I= Supply Voltage
Total Resistance
¿ 3V
2060.5455 Ω=1.4 55925× 10−3 A=1.455925 mA
Part (iii)
Voltage across V 1=I R1 R2
R1+ R2
∨¿ R3 =1.455925× 10−3 A × 6 Ω=0.008736 V
Voltage across V 2 =I R4 R5
R4 + R5
=1.455925 ×10−3 A × 600
11 Ω=0.0 79414 2V
Voltage across V 3 =I R6=1.455925 ×10−3 A × 2000 Ω=2.9 118503V
Part (iv)
Current through R5=I R4
R4 + R5
=1.455925 ×10−3 A × 60 Ω
(60+600) Ω
¿ 1.491456 ×103 A × 60
660 =0.00013 2356 A
Task 2

ELECTRICAL PRINCIPLES 5
Applying Kirchhoff’s Current Law at junction C we get:
I 1+I2=I3 … Equation 1
Then, we apply Kirchhoff’s Voltage Law to loop 1 to get the equation:
V 1−R1 I1 −R2 I3=0
Substituting the values of voltage and resistance and simplifying yields:
6−2 I 1−4 I3=0
2 I1 +4 I3 =6
I 1+2 I3=3 … Equation2
Then, we apply Kirchhoff’s Voltage Law to loop 2 to get the equation:
V 2−R3 I2−R2 I 3=0 … Equation 3
Substituting the values of voltage and resistance and simplifying yields:

ELECTRICAL PRINCIPLES 6
12−2 I 2−4 I3=0
2 I2 +4 I3 =12
I 2+ 2 I3=6 … Equation3
Substituting the value of I 3 in equation 1 into equation 2 we get:
I 1+2 (I1 + I2)=3
I 1+2 I 1+2 I2=3
3 I1 +2 I 2=3 … equation 4
Substituting the value of I 3 in equation 1 into equation 3 we get:
I 2+ 2(I1 + I2)=6
I 2+2 I 1+2 I2=6
2 I1 +3 I 2=6 … equation 5
Then solve the simultaneous equations 4 and 5 as follows:
Multiply equation 4 with 2 and equation 5 with 3 to get:
2 ( 3 I1 +2 I2=3 ) … equation 4 ⇒6 I1 +4 I 2=6
( 2 I1 +3 I2=6 ) 3 … equation 5 ⇒6 I1 +9 I2=18
Then subtract the resulting equations from each other to get:
( 6 I1+ 9 I2 ) − ( 6 I1 + 4 I 2 ) =18−6
5 I2 =12

ELECTRICAL PRINCIPLES 7
I 2=12
5 =2.4 A
Substituting I2 =2.4 A into equation 4 yields
3 I1 +2 ( 2.4 )=3 … equation 4
3 I1 =3−4.8=−1.8
I 1=−1.8
3 =−0.6 A
Then from equation 1,
I 3=I 1+I2=−0.6 +2.4=1.8 A
Therefore , I1=−0.6 A , I 2=2.4 A ,∧I3=1.8 A
The power dissipated by R2=I 3
2 R2=1.82 ×4=12.96 W
Task 3 Circuit A: 1N4148 diode

ELECTRICAL PRINCIPLES 8
Table 1: 1N4148 diode Forward bias
V 1 (Volts) V D ( mV ) , 1 N 4148 diode I D (μA )
0.1 0 0
0.2 0 0
0.3 0 0
0.4 399.996 0.003981
0.5 499.9 0.099679
0.6 597.67 2.33
0.7 673.31 26.69
0.8 710.756 89.244
0.9 730.631 169.369
1.0 743.506 256.494
1.1 752.89 347.11 μA

ELECTRICAL PRINCIPLES 9
1.2 760.23 439.77μA
Table 2: 1N4148 diode Reverse bias
V 1 (Volts) V D ( Volts ) ,1 N 4148 diode I D (mA )
2 0 0
4 0 0
6 0 0
8 0 0
10 0 0
12 0 0
Task 3 Circuit B (BAT54 diode)

ELECTRICAL PRINCIPLES 10
Table 3: BAT54 diode Forward bias
V 1 (Volts) V D ( mV ) , BAT 54 diode I D (μA )
0.1 0 0
0.2 0 0
0.3 0 0
0.4 399.911 0.088547
0.5 495.885 4.115
0.6 555.439 44.561
0.7 580.169 119.831
0.8 593.745 206.255
0.9 602.871 297.129
1.0 609.691 330.309

ELECTRICAL PRINCIPLES 11
1.1 615.115 484.885
1.2 619.61 580.39
Table 4: BAT54 diode Reverse bias
Task 4
V 1 (Volts) V D ( Volts ) , BAT 54 diode I D (mA )
2 0 0
4 0 0
6 0 0
8 0 0
10 0 0
12 0 0

ELECTRICAL PRINCIPLES 12
0 100 200 300 400 500 600 700 800
0
50
100
150
200
250
300
350
400
450
500
1N4148 diode Forward bias
Voltage VD(MilliVolts)
Current ID(Micro Amperes)
0 100 200 300 400 500 600 700
0
100
200
300
400
500
600
700
BAT54 diode Forward bias
Voltage VD(MilliVolts)
Current ID(Micro Amperes)
From the graphs shown above, it is evident that the 1N4148 diode has a steeper slope as
compared to the BAT54 diode. On the other hand, no current flows when the diode is connected
in reverse bias.