Engineering RL parallel circuit. No value given

[stepfanie]

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.

So how can we determine the total active power? I have tried but only come out with that. :(

 

[stepfanie]

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

Is it like how I did in the previous question finding the real denominator and imn thingy ?

 

[Sam 007]

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.

If that's the case then the real power would be V^2/(R1+R2) wouldn't it?

for part d)

=V.( jXL/ (R+jXL) *(R-jXL/R-jXL)

=V.( (jRXL+XL^2)/(R^2+XL^2)

=V.( XL^2 /(R^2+XL^2) + jRXL/(R^2+XL^2) )

magnitudeV.√(( XL^2 /(R^2+XL^2))^2 + (RXL/(R^2+XL^2) )^2 )θ=tan^-1(RXL/XL^2)

Than the reactive power would be

Q=V^2/XL

Q=[V.√(( XL^2 /(R^2+XL^2))^2 + (RXL/(R^2+XL^2) )^2 )]^2/XL

Right?

 

[NascentOxygen]

If that's the case then the real power would be V^2/(R1+R2) wouldn't it?

For R1 and R2, yes.

for part d)

=V.( jXL/ (R+jXL) *(R-jXL/R-jXL)

=V.( (jRXL+XL^2)/(R^2+XL^2)

=V.( XL^2 /(R^2+XL^2) + jRXL/(R^2+XL^2) )

magnitude


V.√(( XL^2 /(R^2+XL^2))^2 + (RXL/(R^2+XL^2) )^2 )


θ=tan^-1(RXL/XL^2)

Than the reactive power would be

Q=V^2/XL

Q=[V.√(( XL^2 /(R^2+XL^2))^2 + (RXL/(R^2+XL^2) )^2 )]^2/XL

Right?

That looks right for inductor's reactive power. You could tidy up this expression.

 

[NascentOxygen]

So how can we determine the total active power? I have tried but only come out with that. :(

There are a number of ways. You could determine the magnitude of the current in each resistor and use I²R, then sum the powers. Or you could determine the magnitude of the voltage across each resistance and use E²/R.

 

[NascentOxygen]

for d)

Q=|V^2| / XL

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

v= (jVXL)/ (R+ jXL)

Right here would be a good place to convert it to a magnitude and angle, because from this point on in part (d) we are only concerned with the magnitude, and we need to square it to find power. The angle information is of no further interest here so it's just cluttering up your maths.[/color]

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? :/

You have the right idea. But power is a not a complex quantity, it won't have a "j" in it, certainly not in its denominator.

 

[Sam 007]

for r1 and r2, yes.

That looks right for inductor's reactive power. You could tidy up this expression.

But what happens to R3 is it not going to be included as real power? If it is, is it going to be calculated as v3= v*r3/(jxl+r3) using the voltage divider rule

than the real power of r3 = v3^2/r3
Does r3 have a real power?

tidy expression
q=[v^2*( xl^4 /(r^4+xl^4) + rxl^2/(r^4+xl^4) ]/xl

 

[NascentOxygen]

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

Not quite as easy as that, I'm afraid.

You have two parallel branches, so you have to determine the current in each and then sum them. Complex numbers will be needed.

 

[NascentOxygen]

But what happens to R3 is it not going to be included as real power? If it is, is it going to be calculated as v3= v*r3/(jxl+r3) using the voltage divider rule

than the real power of r3 = v3^2/r3
Does r3 have a real power?

Whenever current flows in a resistance, there is real power involved. Yes, yes, and yes.

tidy expression
q=[v^2*( xl^4 /(r^4+xl^4) + rxl^2/(r^4+xl^4) ]/xl

Try again.

 

[Sam 007]

For part E.
Branch one

R+R=2R

I=V/2R

Branch 2

R+jXL

I=V/(R+jxL)

than find the magnitude and angle than add Is suppose

V*(R/(R^2+XL^2)- jXL/(R^2+XL^2)

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


θ= tan^-1(XL/R)


Itotal = V*√((R/(R^2+XL^2)^2- (jXL/(R^2+XL^2))^2) + I=V/2R

 

[NascentOxygen]

For part E.
Branch one

R+R=2R

I=V/2R

Branch 2

R+jXL

I=V/(R+jxL)

than find the magnitude and angle than add Is suppose

That should work.

 

[Sam 007]

V*(R/(R^2+XL^2)- jXL/(R^2+XL^2)

V*√((R/(R^2+XL^2)^2- (XL/(R^2+XL^2))^2)θ= tan^-1(XL/R)Itotal = V*√((R/(R^2+XL^2)^2- (XL/(R^2+XL^2))^2) + V/2R

 

[Sam 007]

For part C... R3

V3=V*R3/(R3+jXL)
=V( R/(R+jXL)* (R-jXL)/(R-jXL)

= V*(R^2/(R^2+XL^2) - jRXL/(R^2+XL^2))mag=V*√(R^2/(R^2+XL^2))^2 - (RXL/(R^2+XL^2))^2)

θ=tan^-1(RXL/R^2)

Real power R3=(V*√(R^2/(R^2+XL^2))^2 - (RXL/(R^2+XL^2))^2))^2/R

Real power R1 and R2= V^2/2R

 

[stepfanie]

Real power R1 and R2= V/2R

isn't v^2 / 2R ?

 

[Sam 007]

isn't v^2 / 2R ?

yes ,sorry forgot to include ^2

 

[NascentOxygen]

V*(R/(R^2+XL^2)- jXL/(R^2+XL^2)

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


θ= tan^-1(XL/R)


Itotal = V*√((R/(R^2+XL^2)^2- (XL/(R^2+XL^2))^2) + V/2R

You can't just place a "+" sign between the two currents and sum their magnitudes because they don't have the same angle. You can add the currents by taking their different angles into account, or, alternatively, you can separately add their real parts, and add their imaginary parts, to find the real and imaginary part of the total current.

 

[stepfanie]

thank you nascent :D you help a lot..!