Question

6. Two impedances = ( 4 + j7 )Ω and z2 = ( 4 - 3j )Ω are connected in parallel determine:

a. the current flowing

b. its phase relative to the 120V supply voltage.

Total impedance for a parallel circuit is given by 1/Z = 1/Z + 1/Z2 and the current is given by I = V/Z

Answer 1

$\frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2}$

$\frac{1}{Z} = \frac{1}{4+j7} + \frac{1}{4-3j}$

$\frac{ 1}{Z} = \frac{4-3j + 4+7j}{(4+7j)(4-3j)}$

$\frac{1}{Z} = 0.22+0.012j$

V = 120 + 0j

I = $\frac{V}{Z}$ = V × $\frac{1}{Z}$

I = (120+0j) × (0.22 + 0.012j)

I = 26.4 + 1.44j