Q. 1 Solution:
Starting from lower side of circuit network, we can see that 2Ω and 2Ω are in series, so their combined effect is
2+2=4Ω
97b5d636-b0d8-40d1-8f33-9266b3779585-image.png
This 4Ω is in parallel of 4Ω to lower side, so their equivalent is
1b08526d-0c95-43b2-9f73-3927652469d8-image.png
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This 2Ω is in series of 5Ω and 1Ω, so the sum of series resistances is
2Ω+ 5Ω+1Ω=8Ω
f1ffb19b-69a3-4eb0-8d8d-4eb1316289f4-image.png
Now at upper side, 4Ω and 2Ω are in series
4Ω+ 2Ω=6Ω
b3a9072f-b02e-4881-aaf3-9b389aab13e0-image.png
This 6Ω is in parallel of 6Ω, so their equivalent is
df9ccb62-3a78-4fec-a9d8-9044a372be75-image.png
cac80a33-5165-4e48-b124-810aaad24d2e-image.png
This 3Ω is in series of 3Ω and 1Ω, so the sum of series is
3Ω+ 3Ω+1Ω=7Ω
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Now we can see 7Ω becomes in series of 8Ω, so equivalent resistance is
Req =7Ω + 8Ω
Req=15Ω
Q.2 Solution:
1)
To calculate the source current Is, Firstly, we calculate the total resistance
RT= R1+R2+R3
=6Ω
V=IR
IS=VS/RT
= 12/6
=2A
2)
Since all resistances are in series, same 2A current pass through each resistance.
V1= R1IS
=12
=2V
V2=R2Is
=22
=4V
V3=R3IS
= 32
=6V
3)
P1 =I2R1
= (2)2 *1
=4W
P2 =I2R2
= (2)2 *2
= 8W
P3 =I2R3
=(2)2 *3
= 12W
= 12W
4)-
Ps = VsIs
P =122
=24 W
PT=P1+P2+P3 or PT=I2RT
=4+8+12
=24 W