6Ω
15Ω
10
75V+
$12Ω
60Ω
1.6V0
2
3
BB KVL
-75+ Vd +12 (I1-I2)-7Id=0
Vd+12 (I1-I2)-7Id=75
3V1+5I1=75
By KVL at Loop 2
7Id+ 12 (I2-I1)+15I2+60 (I2-I3)=0
7Id+87I2-12I1-60I3=0
7Id-87I1-2Vd-96Vd=0
-81d=98Vd
Id=-$\frac{98}{80}$ Vd=-1.225Vd
After solving we get the Value Vd=-24V and Id = -29.4 mA
$I_1 = \frac{V_d}{6} = -4mA$
I2-Id=-29.4mA
I3= 1.6 V1=1.6x24=38.4 A
Pab=75x4=300W
7I1(I1-I2)=7x284x25.4= 5,227,32w
1.6 Vd is givn by 60 (I2-I3)= 60 (-29.4+ 38.4)=60x9=540V
540Vx 39.4= 20, 736 W