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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.

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    \$\begingroup\$ Wouldn't a capacitive reactance result in a negative imaginary term? So \$Z_{B1} = 6-3j\$ \$\endgroup\$
    – caveman
    Feb 4 '15 at 9:08
  • \$\begingroup\$ Hmm I believe you are correct \$\endgroup\$ Feb 5 '15 at 12:17
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I think the question was to find the currents using rectangular calculation only and then to verify using the polar form (phasors). You need to do the calculations using rectangular coordinates and polar separately.

There are a few mistakes in the calculations shown in your post.

  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)\$

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  • \$\begingroup\$ Great so it's all wrong /0\. Well thanks for pointing that out. \$\endgroup\$ Feb 5 '15 at 12:18
  • \$\begingroup\$ Could you perhaps explain your second point; I added these with respect the formulae supplied from my tutor :s. \$\endgroup\$ Feb 5 '15 at 12:20
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    \$\begingroup\$ To add phasors, you convert to rectangular form, i.e. x + jy, then add the components. Then you convert back to polar form. To multiply phasors, you use polar form, multiply the amplitudes and add the angles. \$\endgroup\$
    – caveman
    Feb 5 '15 at 22:43
  • \$\begingroup\$ Or else you can draw the vectors (phasors) on a graph and find out the resultant vector using the parallelogram method. \$\endgroup\$
    – nidhin
    Feb 6 '15 at 7:52

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