HELP with AC calculations

Discussion in 'Homework Help' started by hihptsi, Jul 24, 2011.

1. hihptsi Thread Starter New Member

Jul 24, 2011
2
0
A circuit of inductor and capacitor in parallel is connected to resistor R in series. Applied ac power is E = 26V/0o . XL= 6 ohms XC= 12 ohms, R = 5 ohms. Find the total impedance Z, the currents IR, IL, and IC.
Determine powers PR, QL, QC, and S (magnitude only). All answers must be in phasor form except the power.

can someone run me thru the steps to calculate each of these, i am very confused and will be tested on this in a week. id really appreciate it!
Walt

2. blah2222 Distinguished Member

May 3, 2010
581
38
You would solve this the same way as if this were a DC circuit, except you are now dealing with complex numbers. What is giving you trouble specifically?

Go through it first and we will help you along the way. We are not going to do this for you.

3. hihptsi Thread Starter New Member

Jul 24, 2011
2
0
im having trouble with everything. my teacher explains things very poorly and goes extremly fast.

i understand that to find total Z you would

5+j6-j12
5-j6
sqrt 5^2+12^2 tan^-1 6/5

im lost after this how do i calculate the ohms and the phase angle???? if someone could explain this better for me id be very grateful.

4. blah2222 Distinguished Member

May 3, 2010
581
38
Okay so yes, you are correct for the series case the total impedance can be found simply by adding all of the impedance elements together:

Zs = R + jXl - jXc = 5 + j6 - j12 = 5 - j6 *Rectangular form

To get it in the phasor form you need to find the magnitude and phase of the complex number. It might help if you realize that this can be viewed on a 2-D complex plane where the x-axis is the real axis and y-axis is the imaginary axis, something like this:

In our case, Zs is +5 units (to the right) on the real plane and -6 units (downward) on the imaginary plane.

The magnitude of Zs is simply the distance from the origin to the point that these coordinates mark out. So using what we know from the Pythagorean Theorem:

$|Zs| = \sqrt{Re(Zs)^{2} + Im(Zs)^{2}}$

The phase is simply the angle that it takes to get from the positive real axis to Zs.

$Phase(Zs) = \arctan{\frac{Im(Zs)}{Re(Zs)}}$

Zs = |Zs|<Phase(Zs) *Phasor form

We already determined that the real part of Zs, Re(Zs), is +5 and the imaginary part of Zs, Im(Zs), is -6.

Finding the parallel total impedance is a bit trickier, but very doable. You just have to be careful with your complex algebra.

Zp = R || jXl || jXc = 5 || j6 || -j12

Try finishing off this part and we'll go from there to help.

Last edited: Jul 24, 2011