Adding phasers in a parallel circuit problem

Below is a practice question that I am having trouble progressing further with. The (abbreviated) question and progress as follows:

A circuit of 90V's supply is in parallel with two branches. Branch 1: Capacitive reactance of 3 Ohm in series with 6 Ohm resistor. Branch 2: Inductive reactance of 5 Ohm in series with 4 Ohm resistor. Find the total current and current across each branch. Confirm these results with a phasor diagram.

My progress:

B1 = 6 + 3J and B2 = 4 + 5J
B1 into polar: Root(6^2 + 3^2) = 6.7082 , Tan^-1(3/6) = 26.565.
B2 into Polar: Root(4^2 + 5^2) = 6.4031 , Tan^-1(5/4) = 54.340.


I=V/r or I=E/Z

B1I = (90/6.7082 angle 26.565 = 13.4164 angle 26.565) and
B1I Rect = (13.4164 cos 26.565) + (13.4164 sin 26.565) gives 5.9978 + 12.00100J

B2I = 90/6.403 angle 51.340 = 14.0559 angle 51.340
B2I Rect = (14.0559 cos 51.340) + (14.0559 sin 51.340) gives 8.7806 + 10.97578J


Kirchoffs law and common sense states the two currents in parallel added will give total current:

Polar:(13.4164+14.0559) ang (26.555 + 51.340) =  27.4723 angle 77.895


Now converting to rectangular:

27.4723 cos 77.895 =  5.760 and
27.4723 sin 77.895 = 26.861J


Therefore total current should be 5.760 + 26.861J

Assuming my calculations are correct how would I then prove this using phasers ? I have only done phaser calculations using non-imaginary ; I am told this method is easier.

I have attempted to prove this using further polor to rect calculation and they don't appear to add up ;although this could be due to a lack of sleep.

Progress update Thanks for edits and info thus far.

Update given capacitive reactance is negative:

B1 = 6 - 3J
Into polar: 6.7082 , Tan^-1(-3/6) = 333.433 (-26.565)


Calculating Current rectangularly first:

Total current would be 90/((6 - 3j) + (4 + 5j)) = 90/ 10 + 2j

 =(90/10 + 2j) * (10 - 2j)/(10 - 2j)
= (900 + 180j) / (100 -20j + 20j -2j^2)
=  (900 + 180j) / (102)
= 8.8235 + 1.7647j


To polar :

Sqrt(8.8235^2 + 1.7647^2) = 8.99824 , Tan^-1(1.7647/8.8235) = angle 11.3099
`

__ Assuming this is correct how would I prove with phasers ; again I am unsure due to only using phasers with real values.

• Wouldn't a capacitive reactance result in a negative imaginary term? So $Z_{B1} = 6-3j$ Feb 4 '15 at 9:08
• Hmm I believe you are correct Feb 5 '15 at 12:17

1. The impedance offered by a capacitor with reactance $X_C$ is $-jX_C$. So the impedance offered by branch1 is 6-3j.
2. $A_1\angle\theta_1 + A_2 \angle \theta_2 \ne (A_1 + A_2)\angle (\theta_1 + \theta_2)$