Engineering RL parallel circuit. No value given

[stepfanie]

just wonder is 1 or -1 ?

if -1 then 63.43°

 

[NascentOxygen]

just wonder is 1 or -1 ?

What?

if -1 then 63.43°

The answer won't be a number, it will be an expression in R and X_L.

See https://www.physicsforums.com/showpost.php?p=3877493&postcount=26

 

[stepfanie]

magnitude is (-2jRXL) / (R^2+XL^2) Vertical = imaginary

angle is (v/2) * [ (R^2 - XL^2)/(R^2+XL^2) Horizontal = real

 

[NascentOxygen]

magnitude is (-2jRXL) / (R^2+XL^2) Vertical = imaginary

angle is (v/2) * [ (R^2 - XL^2)/(R^2+XL^2) Horizontal = real

It seems that you don't know what is meant by magnitude or angle of a vector?

A complex number can be regarded as a vector with its real part drawn horizontally and the imaginary part drawn vertically. The magnitude is the length of that vector.

 

[Sam 007]

hi Nascent ,would the magnitude be √( (R^2-XL^2)^2+ (-2jRXL)^2)?

while the angle be θ=(tan^-1(-2jRXL/(R^2-XL^2)))° ?

cheers :)

 

[stepfanie]

sqrt of ( (R^2-XL^2)^2+ (-2jRXL)^2)

but the angle.

 

[NascentOxygen]

hi Nascent ,would the magnitude be √( (R^2-XL^2)^2+ (-2jRXL)^2)?

No, you have left out something major.

while the angle be θ=tan^-1(-2jRXL/(R^2-XL^2))° ?

There is no j in either magnitude or angle. You drop the j. There won't be a degree symbol where you wrote it, either.

 

[stepfanie]

No, you have left out something major.



There is no j in either magnitude or angle. You drop the j. There won't be a degree symbol where you wrote it, either.

is it the denominator (R^2 + XL^2) ?

 

[Sam 007]

ok so would the magnitude be
√( ((R^2-XL^2)/(R^2+XL^2))^2 + (-2RXL/(R^2+XL^2)) ^2 )?while the angle be θ=tan^-1(-2RXL/(R^2-XL^2)) ?

Thanks :)

 

[NascentOxygen]

ok so would the magnitude be
√( ((R^2-XL^2)/(R^2+XL^2))^2 + (-2RXL/(R^2+XL^2)) ^2 )?

Wasn't there a factor of 2 in the denominator?

while the angle be θ=tan^-1(-2RXL/(R^2-XL^2)) ? ✔[/size][/color]

 

[Sam 007]

√(((R^2-XL^2)/(2R^2+2XL^2))^2 + (-2RXL/(2R^2+2XL^2)) ^2 ) or

√(((R^2-XL^2)/(2R^2+2XL^2))^2 + (-RXL/(R^2+XL^2)) ^2 )

? :)

 

[NascentOxygen]

√(((R^2-XL^2)/(2R^2+2XL^2))^2 + (-2RXL/(2R^2+2XL^2)) ^2 ) ? how about now :)

That looks right. You could take out the common denominator.

 

[Sam 007]

Cool thanks so much for the help Nascent :)

 

[NascentOxygen]

There is supposed to be a V in front of all that, too, if this is Va-Vb.

 

[stepfanie]

so the magnitude = V sqrt of ((R^2-XL^2)/(2R^2+2XL^2))^2 + (-RXL/(R^2+XL^2)) ^2 )

that's it right? :D

 

[Sam 007]

hmm should it be since its the magnitude of Vab

V*√(((R^2-XL^2)/(2R^2+2XL^2))^2 + (-2RXL/(2R^2+2XL^2)) ^2 ) or


√(V*((R^2-XL^2)/(2R^2+2XL^2))^2 + (-2RXL/(2R^2+2XL^2)) ^2 ) or


√(V^2*((R^2-XL^2)/(2R^2+2XL^2))^2 + (-2RXL/(2R^2+2XL^2)) ^2 ) ?

 

[NascentOxygen]

so the magnitude = V sqrt of ((R^2-XL^2)/(2R^2+2XL^2))^2 + (-RXL/(R^2+XL^2)) ^2 )

that's it right? :D

Yes.

 

[stepfanie]

for f) calculate the power factor of the circuit.

power factor = true power / apparent power which is (I^2 *R) / (I^2 *Z)

so i can use the equation in c) which is the active power dissipated in each resistance divide by d) which is the reactive power dissipated in XL by using power factor formula.

am i right ?

 

[NascentOxygen]

V·√(((R^2-XL^2)/(2R^2+2XL^2))^2 + (-2RXL/(2R^2+2XL^2)) ^2 )

Something seems wrong.

That whole expression simplifies to V/2. That can't be right? But I think it is right!

That makes things easier!

 

[stepfanie]

for f) calculate the power factor of the circuit.

power factor = true power / apparent power which is (I^2 *R) / (I^2 *Z)

so i can use the equation in c) which is the active power dissipated in each resistance divide by d) which is the reactive power dissipated in XL by using power factor formula.

am i right ?

c) active power dissipated in each resisance
formula p=(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

so for Power Factor

P/Q =

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

is it correct ? :)

 

[NascentOxygen]

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

yes

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

But I don't like the look of that.

d) reactive power dissipated in XL
Q=I^2 * jXL

Yes, without the j.

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

I don't think so.

You will have to work out the current through the R-X_L branch in order to determine the power in that resistor, and reactive power in that X_L. Use complex algebra (real + j imaginary) to do this, since you've learned that method.

 

[Sam 007]

C) active power in each resistor?
P=V^2/R1
P=V^2/R2
P=V^2/R3
XL has no active power
D) Reactive power?

Q= V^2/XL

E) Total current
It=V/R1+V/R2+V/R3+V/XL

not sure?

 

[stepfanie]

for d)

Q=|V^2| / XL

v= (jXL / (R + jXL)) *v -> voltage divider rule.

v= (jVXL)/ (R+ jXL)

v^2 = [ (jVXL)/ (R+ jXL) ] ^2

v = (j^2 V^2 XL^2) / ((R + jXL)^2)

v= (-V^2 XL^2) / ((R + jXL)^2)

so sub V into Q=|V^2| / XL

Q= [(-V^2 XL^2) / ((R + jXL)^2)] / XL

Q= (-V^2 XL^2) / (XL*((R + jXL)^2)) ->VAR

am i right? :/

 

[stepfanie]

C) active power in each resistor?
P=V^2/R1
P=V^2/R2
P=V^2/R3
XL has no active power
D) Reactive power?

Q= V^2/XL

E) Total current
It=V/R1+V/R2+V/R3+V/XL

not sure?

for c) we not need to calculate the total P ? just P in each resistance ?

and for b) we just use the magnitude of Vab and how about the XL ?

 

[Sam 007]

I don't really think you have to use Vab from part B for part C just the voltage V∠0°

Yeah for part C. just calculate the power in each resistance, XL is not a resistor its an inductor has no resistance.

But than again I am not sure will have to ask nascent

 

[stepfanie]

I don't really think you have to use Vab from part B for part C just the voltage V∠0°

Yeah for part C. just calculate the power in each resistance, XL is not a resistor its an inductor has no resistance.

But than again I am not sure will have to ask nascent

thank you.

but if i want to find the total active power ? what will it be..
and can u solve d)? I am not sure I am doing correctly.

 

[Sam 007]

well this is the way I would do it but stephanie I think it may not be right


P=V^2/R1+V^2/R2+V^2/R3

 

[stepfanie]

well this is the way I would do it but stephanie I think it may not be right


P=V^2/R1+V^2/R2+V^2/R3

this is what i got as well.

 

[NascentOxygen]

well this is the way I would do it but stephanie I think it may not be right

P=V^2/R1+V^2/R2+V^2/R3

That could only be right iff there is a voltage of amplitude V across each individual element, R1, R2, and R3, and there isn't. None has V volts across itself.

But for the series pair R-R you can say V volts exists across that pair.

 

[NascentOxygen]

for d)

Q=|V^2| / XL

v= (jXL / (R + jXL)) *v -> voltage divider rule.

https://www.physicsforums.com/images/icons/icon14.gif That's the right way for finding the voltage across the inductance. Now convert this to a vector having a magnitude and an angle. Power calculations involve the square the magnitude of the voltage across any element. https://www.physicsforums.com/images/icons/icon6.gif