RL parallel circuit. No value given

[stepfanie]

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

Uploaded with ImageShack.us[/PLAIN]


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

 

[NascentOxygen]

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

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?

 

[stepfanie]

Hi :D

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

 

[stepfanie]

for Va

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

 

[NascentOxygen]

for Va

Va= R*V/R+R ?

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)

Vb= I am stuck :(

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

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

 

[stepfanie]

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

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

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

Vab= Va-Vb

 

[NascentOxygen]

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

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

You may be more familiar with calculations using jX_L 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.

 

[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· X<sub>L[/sub ∠+90°

im confused sorry</sub>

 

[stepfanie]

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

 

[NascentOxygen]

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· X<sub>L[/sub ∠+90°

im confused sorry</sub>

<sub>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°</sub>

 

[NascentOxygen]

So, using the potential divider relation:

V_B = V_L/(V_L + V_R)

=❪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?

=❪X_L∠+90°❫/❪X_L∠+90° + R∠0°❫[/color]

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., jX_L?

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

 

[stepfanie]

So, using the potential divider relation:

V_B = V_L/(V_L + V_R)

=❪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?

=❪X_L∠+90°❫/❪X_L∠+90° + R∠0°❫[/color]

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., jX_L?

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

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

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

 

[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

 

[NascentOxygen]

So, using the potential divider relation:

V_B = V∠0°[/color] · V_L/(V_L + V_R)

= V∠0°[/color] · ❪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?

= V∠0°[/color] · ❪X_L∠+90°❫/❪X_L∠+90° + R∠0°❫

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

 

[stepfanie]

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

Noted!

 

[stepfanie]

any can tell me

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

 

[NascentOxygen]

for jXL I am quite familiar.

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 * jX_L / (R + jX_L)

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

= V{½ - jX_L/(R + jX_L}

https://www.physicsforums.com/images/icons/icon2.gif 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.

 

[NascentOxygen]

any can tell me

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

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

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

 

[stepfanie]

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 * jX_L / (R + jX_L)

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

= V{½ - jX_L/(R + jX_L}

https://www.physicsforums.com/images/icons/icon2.gif 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.

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

I thought i already have the answer for Vab?

 

[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?

 

[NascentOxygen]

Sorry, I can't understand what you are doing because I think you are not using enough brackets. The "/" sign applies only to the next term unless you use brackets.

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

Is this with a denominator that is all REAL? There would not be a j anywhere in the denominator if the denominator was REAL.

I see you have j^2 a few times, you know j^2 = -1 do you? So replace j^2 with -1 and separate the real component from the imaginary part.

 

[stepfanie]

this is my final answer

= (v/2)*[ (R^2 - 2jRXL + XL^2) / ( R^2 - XL^2 ) ]

 

[NascentOxygen]

Almost there.

Fix it so j^2 = -1
and clearly separate the real from the imaginary.

 

[stepfanie]

so...

z= (v/2) * [ (R^2)/(R^2+XL^2) + j(-2RXL - XL^2) / (R^2+XL^2)

:D right?

 

[NascentOxygen]

No quite right. The only imaginery term in the numerator is 2jRXL

You don't set z to this. You set Va-Vb = ...

 

[NascentOxygen]

Once you have it as Va-Vb = Re + jIm
it can be viewed as a vector on a plane with vert axis the I am and horiz axis the Re part.

So you should be able to determine the magnitude of the vector, and the angle it makes with the horiz.

 

[stepfanie]

Va-Vb = (v/2) * [ (R^2 - XL^2)/(R^2+XL^2) + (-2jRXL) / (R^2+XL^2)

><

 

[Sam 007]

don't know if this is correct Vab= V/2*[ ( (R^2-XL^2)/(R^2+XL^2) ) + (-2jRXL/(R^2+XL^2) )

 

[NascentOxygen]

Va-Vb = (v/2) * [ (R^2 - XL^2)/(R^2+XL^2) + (-2jRXL) / (R^2+XL^2)

Looks right.

So now determine magnitude and angle.

 

[Sam 007]

Just out of curiosity how do you calculate the angle with such a complex equation, would you just assume (R^2-XL^2)=1 and for j would be -2 so would it come out as tan^-1(-2/1)= -63.43°? Got same the assignment