RL parallel circuit. No value given

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just wonder is 1 or -1 ?

if -1 then 63.43°
 
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magnitude is (-2jRXL) / (R^2+XL^2) Vertical = imaginary

angle is (v/2) * [ (R^2 - XL^2)/(R^2+XL^2) Horizontal = real
 
stepfanie said:
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.
 
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 :)
 
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sqrt of ( (R^2-XL^2)^2+ (-2jRXL)^2)

but the angle.
 
Sam 007 said:
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.
 
NascentOxygen said:
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) ?
 
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 :)
 
√(((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 )

? :)
 
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Cool thanks so much for the help Nascent :)
 
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
 
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 ) ?
 
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 ?
 
stepfanie said:
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 ? :)
 
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stepfanie said:
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-XL branch in order to determine the power in that resistor, and reactive power in that XL. Use complex algebra (real + j imaginary) to do this, since you've learned that method.
 
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?
 
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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? :/
 
Sam 007 said:
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 ? :rolleyes:

and for b) we just use the magnitude of Vab and how about the XL ?
 
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
 
Sam 007 said:
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.
 
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
 
Sam 007 said:
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.
 
Sam 007 said:
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.
 
stepfanie said:
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
 
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