**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

This 4Ω is in parallel of 4Ω to lower side, so their equivalent is

This 2Ω is in series of 5Ω and 1Ω, so the sum of series resistances is

2Ω+ 5Ω+1Ω=8Ω

Now at upper side, 4Ω and 2Ω are in series

4Ω+ 2Ω=6Ω

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

This 3Ω is in series of 3Ω and 1Ω, so the sum of series is

3Ω+ 3Ω+1Ω=7Ω

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= R1*IS*

=12

=2V

V2=R2*Is*

=22

=4V

V3=R3*IS*

= 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 =12*2*

=24 W

PT=P1+P2+P3 or PT=I2RT

=4+8+12

=24 W