DC Networks - Electrical and Electronic Principles

DC Networks - Electrical and Electronic Principles
Physics
Student Name:
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Course Number:
19 April 2019

ASSIGNMENT
Q1: CIRCUIT A
i) For resistors R1 and R2 we get
100× 100
100+100 =50Ω (Product over sum)
For resistors R4 and R5 we get
330× 200
200+330 =¿124.53 Ω (Product over sum)
Total resistance=50 Ω+124.53 Ω+ 550 Ω=724.53 Ω
ii) Using V=IR (Ohm’s law)
12=724.53 I
I= 12
724.53=0.01656 A
iii) V1=IR
=0.01656×50
=0.828V
V2 =IR
=0.01656×550
=9.108V
V3=IR
=0.01656×124.53
=2.062V
iv) The total current in the circuit (i.e. 0.01656A) will be divided
Inversely between the two resistors R4 and R5.
Thus current through R5 will be 200
530 ×0.01656 A
=0.006249 A

CIRCUIT B
i) For resistors R1 ,R2 and R3
1/R=1/20 + 1/60 +1/10 = 10/60
R= 60/10 = 6 Ω
For R4 and R5
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) V1=IR
=0.001456×6
=0.008737 V
V2 =IR
=0.001456×54.55
=0.074248 V
V3=IR0.001456×2000
=2.912 V