Answer:
a, R eq = 0.5 ohms
b, I source = V/Req = 3/0.5 = 6A
c, I1 = 3A
I2 = 1.5A
I3 = 1.5A
d, P1 = 9W
P2 = 4.5W
P3 = 4.5W
e, P source = 18W
Explanation:
a, In a parallel circuit, [tex]\frac{1}{Req}[/tex] = [tex]\frac{1}{R1}[/tex] + [tex]\frac{1}{R2}[/tex] + [tex]\frac{1}{R3}[/tex]
=> Req = 0.5 ohms
b, I source = V/Req = 6A
c, I1 = V/R1 = 3/1 = 3A
I2 = V/R2 = 3/2 =1.5A
I3 = V/R3 = 3/2 = 1.5A
I1+I2+I3 = 6A = I source
d, P1 = [tex](I1)^{2}[/tex]R1 = 9x1 = 9W
P2 = [tex](I2)^{2}[/tex]R2= 2.25x2 = 4.5W
P3 = [tex](I3)^{2}[/tex]R3 = 2.25x2 = 4.5W
e, P source = IV = 6x3 = 18W
P1+P2+P3 = P source = 18W
