**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

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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

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This 6Ω is in parallel of 6Ω, so their equivalent is

df9ccb62-3a78-4fec-a9d8-9044a372be75-image.png

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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