# RL parallel circuit. No value given!

1. Apr 21, 2012

### stepfanie

[URL=http://imageshack.us/photo/my-images/849/photo1zvk.jpg/][PLAIN]http://img849.imageshack.us/img849/9894/photo1zvk.jpg[/URL]

Guys I'm very new to this subject. This is 1 of my question form assignment.

a) find Vab(magnitude and angle)
b) plot magnitude of Vab against XL
c) find the active power dissipated in each resistor
d) find reactive power idssipated in XL
e) find total current I
f) find power factor of the circuit
g) find total apparent power delivered by voltage source
h) prove total generated apparent power is equal to total dissipated apparent power

Im kind of headache now. Can anyone help me with this? the due date is day after tomorrow

2. Apr 21, 2012

### Staff: Mentor

Hi stepfanie! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

① Calculate the voltage at A, in magnitude and angle. You aren't told a value for R, so keep it as R.

② How will you calculate the voltage at B?

Last edited by a moderator: May 5, 2017
3. Apr 21, 2012

### stepfanie

Hi :D

for Vab i think Vab=Va-Vb, use voltage divider rule? I'm not really sure. :(

4. Apr 21, 2012

### stepfanie

for Va

Va= R*V/R+R ?
Vb= im stuck :(

5. Apr 21, 2012

### Staff: Mentor

It's a voltage divider. You should write that with brackets so everyone can understand it.

Can you simplify this expression: R*V/(R+R)

Assume a current I is flowing through the series resistor and inductor. The voltage across the resistor is given by Ohm's Law, VR = I·R

What is the voltage across the inductor? What equation have you learnt that for AC relates V and I for an inductor?

6. Apr 21, 2012

### stepfanie

R*V/(R+R) = V/2

V across inductor : V=I*Z (am i right)?

if im right, Va= V/2 Vb= (I*R)+(I*Z)

Vab= Va-Vb

7. Apr 21, 2012

### Staff: Mentor

Z is a general term. For an inductor, it's usually written as jXL. There is a magnitude and an angle associated with it, viz., |XL| ∠+90°

So for an inductor, V = I∠0°· XL∠+90°
and to evaluate this you multiply the magnitudes and add the angles, giving
V = I· XL ∠+90°

You may be more familiar with calculations using jXL instead of the angle.

Either way, the voltage at B is found using a voltage divider comprising an inductor and a resistor, they each have impedance of some sort.

Last edited: Apr 21, 2012
8. Apr 21, 2012

### stepfanie

im sorry i can't really get you. :(

V = I∠0°· XL∠+90° is to find V for inductor?

what does the sub means? V = I· XL[/sub ∠+90°

im confused sorry

9. Apr 21, 2012

### stepfanie

because no value given, so basically have to show the ways to find the answer which make me confused.

10. Apr 21, 2012

### Staff: Mentor

Sorry, sub is a formatting command, it should be hidden but I mis-typed it. I'll fix that.

Should be:
So for an inductor, V = I∠0°· XL∠+90°
and to evaluate this you multiply the magnitudes and add the angles, giving
V = I· XL ∠+90°

11. Apr 21, 2012

### Staff: Mentor

So, using the potential divider relation:

VB = VL/(VL + VR)

=❪I∠0°· XL∠+90°❫/❪I∠0°· XL∠+90° + I∠0° · R❫

The (I∠0°) factor cancels from top and bottom, leaving the ratio of impedances. Maybe you knew it would turn out to be that?

=❪XL∠+90°❫/❪XL∠+90° + R∠0°❫

So you can see it's very much the same as the formula you memorized for a resistive divider. You can use reactances just the same, so long as you include their angle.(For a resistor, the angle is 0°.)

Can you express the denominator (shown in blue) as a magnitude and an angle? It's just like adding vectors, and here we need to sum two that are at right angles.

Also, you didn't answer my question: are you familiar with analysis using complex numbers, where there is a j, e.g., jXL?

I have to leave off now. Will be back tomorrow. :)

12. Apr 21, 2012

### stepfanie

for jXL im quite familiar.
ok i will try to understand what you trying teach me.

Thank you so much! Will see you tomorrow. ;)

13. Apr 22, 2012

### stepfanie

so far i got

a) Va= (R*∠0 / 2R)
Vb= ❪ XL∠+90° ❫ / ❪ XL∠+90° + R∠0° ❫
Vab= Va-Vb = (R*∠0 / 2R) - ❪ XL∠+90° ❫ / ❪ XL∠+90° + R∠0° ❫

b) how to plot the magnitude of Vab (y-axis) against XL (x-axis)

c) active power dissipated in each resisance
formula=(I^2)*R

[( V∠0 (3R+Xl∠90 )) / ( 3R+2RXl∠90 )]^2 *R

d) reactive power dissipated in XL
Q=I^2 * jXL
[( V∠0 (3R+Xl∠90 )) / ( 3R+2RXl∠90 )]^2 * Xl∠90

e) I total= I1 + I2

I1 = (V∠0)/2R
I2 = (V∠0)/(R+ XL∠90)

I total = [(V∠0)/2R] + [(V∠0)/(R+ XL∠90)]
= [V∠0(3R+Xl∠90)] / [3R + 2RXl∠90]

# correct me if i made any mistake :D

14. Apr 22, 2012

### Staff: Mentor

CORRECTION:: See where I have fixed my mistake of omitting input voltage in the above.

15. Apr 22, 2012

### stepfanie

Noted!

16. Apr 22, 2012

### stepfanie

any can tell me

I total = I1 + I2 or square root of [ (I1)^2 + (I2)^2 ] in parallel circuit ?

17. Apr 22, 2012

### Staff: Mentor

Well, if you are familiar with analysis using complex numbers, then of course we should be using that method‼

So far, you have established Va = V/2, where V is the input voltage.

To find Vb, use the potential divider approach with the impedance of each element as I showed previously, but this time we'll use their complex impedance.

Vb = V * jXL / (R + jXL)

Therefore, Va - Vb = V/2 - V*jXL/(R + jXL)

= V{½ - jXL/(R + jXL}

https://www.physicsforums.com/images/icons/icon2.gif [Broken] Can you simplify this by making the denominator real for the entire RHS?

Next, separate the RHS into REAL and IMAGINARY parts.

See how you go with that.

Last edited by a moderator: May 5, 2017
18. Apr 22, 2012

### Staff: Mentor

Total current is the vector sum of vector I₁ + vector I₂

Note: these vectors are more correctly termed "phasors".

19. Apr 22, 2012

### stepfanie

V( R- jXL / 2(R+ jXL) )

Last edited by a moderator: May 5, 2017
20. Apr 22, 2012

### stepfanie

complex impedance

z= v/2 [(R - jXL / R + jXL) x (R - jXL / R - jXL)]

= v/2 [ (R - jXL)*(R - jXL) / (R + jXL)*(R - jXL) ]

= v/2 ( R^2 - jRXL - jRXL + j^2XL^2 ) / ( R^2 - jRXL + jRXL - j^2XL^2 )

= v/2 (R^2 - 2jRXL + j^2XL^2) / ( R^2 - j^2XL^2 )

am i right?