Question

A voltage of 200 ∠ 53º8′ is applied across two impedances in parallel. The values of impedances are (12 + j 16) and (10 − j 20). Determine the kVA, kVAR and kW in each branch and the power factor of the whole circuit. 

Answer 1

The circuit is shown in Fig.. 

Y_A = 1/(12 + j16) = (12 −j16)/[(12 + j16) (12 −j16)] 

= (12 −j16)/400 = 0.03 −j0.04 mho 

Y_B = 1/(10 −j20) = 10 + j20/[(10 −j20) (10 + j20)]

= (10 + j20)/500 = 0.02 + j0.04mho

Now V = 200 ∠53º8′ = 200 (cos 53º8′ + j sin 53º8′ ) 

= 2000 (0.6 + j0.8) = 120 + j160 volt 

I_A = VY_A = (120 + j160) (0.03 −j0.04) 

= (10 + j0) ampere (along the reference axis) 

∴ I_B = VY_B = (120 + j160) (0.02 + j0.04) 

= − 4.0 + j8 ampere (leading)