# Phasors

Discussion in 'Homework Help' started by regexp, Nov 28, 2010.

1. ### regexp Thread Starter New Member

Nov 20, 2010
24
0
Hi,

I have this simple circuit

i Just need to find the phasor current I. But i must be doing something wrong,
because my answer differs a bit from the answer in the book.

1. Replace the capacitance with it's complex impedance
$-\frac{1}{333\cdot10^{-5}} = -J300.3$

2. find $Z_eq = \frac{1}{1/115 + 1/-300.3} = 107.39$

3. 1585/107.39 = 14.76, the correct answer is supposed to be 15.11.

What am i doing wrong?

2. ### mik3 Senior Member

Feb 4, 2008
4,846
67
You forgot to include the j in the Zeq equation.

3. ### regexp Thread Starter New Member

Nov 20, 2010
24
0
How do you mean? The angle?

4. ### mik3 Senior Member

Feb 4, 2008
4,846
67
It should be:

Zeq=1/[1/115-1/j300.3]=1/[1/115+j/300.3]

which equals:

Zeq=1/(0.0093<20.95)=107.53<-20.95

5. ### debjit625 Well-Known Member

Apr 17, 2010
790
186
On the picture the voltage source will be 1585 * 1.4142 (root 2) i.e 2241.19 Volt and the polar angle is 12 deg.

Good luck

6. ### TweedleDee New Member

Nov 9, 2010
5
0
Hmm, I'm getting the same answer as the OP.

7. ### TweedleDee New Member

Nov 9, 2010
5
0

But you still use 1585 rms for the calculation right?

8. ### mik3 Senior Member

Feb 4, 2008
4,846
67
Accurately,

Zeq=107.39<-20.95 Ω

and if the RMS value for voltage is used,

I=14.759<32.95 A

9. ### regexp Thread Starter New Member

Nov 20, 2010
24
0

What am i doing wrong?

Feb 4, 2008
4,846
67

11. ### regexp Thread Starter New Member

Nov 20, 2010
24
0
Ok, thanks for clearing that up.
I f#&%? hate my book.

12. ### mik3 Senior Member

Feb 4, 2008
4,846
67
Well, do not always trust me too.

13. ### regexp Thread Starter New Member

Nov 20, 2010
24
0
Hi,

I have one last question.

If i have parallel RLC phasors, how do i get the angle of the resulting phasor.

For example, i would like to find the equivalent of these two parallel phasors $37.94\angle71.5$ and
$133.33\angle-90$

I know how to get |Z| , just not how to get the angle.

$\frac{1}{\frac{1}{37.94} + \frac{1}{133.33}} = 29.53$