Three impedances are connected in parallel across 60 sin (wt) voltage source V1, Z1 = 500 rms, Z2= j400 rms and Z3= -j800 rms. Find current in each impedances show that total current = voltage admittance, Also draw phase diagram. -

$Current = \\ Vs = 60 \lt 0^o \\ Z1= 50 \lt 0^o Ω \\ ∴ I1 = Vs/Z1 = 60 \lt 0^o/ 50 \lt 0^o = 1.2 \lt 0^o A \\ Vs= 60 \lt 0^o \\ Z2 = 40 \lt 90^o Ω (0 +j40) \\ ∴ I2= 60 \lt 0^o/ 40 \lt 90^o = 1.5 \lt 90^o A \\ Vs= 60 \lt 0o \\ Z3= (0-j80) = 80 \lt - 90^o Ω \\ ∴ I3= 60 \lt 0o/ 80 \lt -90^o \\ = 0.75 \lt 90o A \\ \text{Total current-} I= I1 + I2 + I3 \\ = (1.2 + j0) + (0-j1.5) + (0+ j0.75) \\ ∴ I= (1.2 – j0.75) = 1.42\lt -32o Amp$

Proof: I= V * Y

Y1 $= 1/Z1 = 1/ 50 \lt 0o = 0.02 \lt 0^o \\ = 0.02+j0) s$

Y2 $= 1/ Z2 = 1/ 40\lt90o =0.025 \lt-90^o \\ = (0-j0.025)S$

Y3 $= 1/Z3 = 1/80\lt-20o = 0.0125 \lt 90o S \\ = (0+j0.0125)S$

∴ Y= Y1 + Y2 +Y3

= 0.02 – j0.025 + j0.0125

= 0.02 – j0.0125

=0.0236 < $-32^oS$

I= V* Y

I = 60<0o * 0.236 <-32o

I = 1.42< - 32o Amp