Solving Reactive Power for Purely Inductive and Capacitive Circuits

[ws0619]

Hello!
From what I have learned now is the topic of sinusoidal steady-state power.In this topic,the reactive power for purely inductive and purely capacitive confuse me for a long time.
Try check out the problem I faced from this bookhttp://books.google.com.my/books?id...&oi=book_result&resnum=1&ct=result#PPA395,M1".Reactive power for a purely inductive circuit is:p=-Qsin2wt or reactive power for a purely capacitive circuit is p=-Qsinwt?
Can anybody help me solve it?
thanks!

 

[MATLABdude]

I can't see the question, but in both cases, it's really dependent on what your power supply happens to be (and I can't ascertain that from what you've given). However, in complex power, power factor (for a sinusoidal system) is the ratio of the reactive power to real power, which happens to be the cosine of the angle between the voltage and current waveforms.

So given what you know about how current leads voltage in capacitors and inductors, what would be the power factor, and what portion of the apparent power in the circuit would be reactive?

HINT: http://en.wikipedia.org/wiki/Reactive_power#Power_factor

 

[ws0619]

thanks!I just looking for book error.By the way i have found out the answer.

 

[xcvxcvvc]

I can't see the question, but in both cases, it's really dependent on what your power supply happens to be (and I can't ascertain that from what you've given). However, in complex power, power factor (for a sinusoidal system) is the ratio of the reactive power to real power, which happens to be the cosine of the angle between the voltage and current waveforms.

So given what you know about how current leads voltage in capacitors and inductors, what would be the power factor, and what portion of the apparent power in the circuit would be reactive?

HINT: http://en.wikipedia.org/wiki/Reactive_power#Power_factor

ummmmmmmmmmmm dot dot dot dot dot

power factor is the ratio of average aka real power to total power aka complex power.

real power = cos(angle) * complex power

so the calculation becomes cos(angle) * complex power divided by complex power = cos (angle)

for op:

mathematically, reactive power is simply power with an imaginary component. Complex power, the vector sum of real power and reactive power, is V times I*. The star means I conjugate, which means you reverse the sign of the angle for the current.

Real power is from a resistor. You should know that a resistor's current and voltage are in phase. Intuitively, the equation begins to make sense: subtracting one angle from another angle of equal magnitude equals zero. Cos(0) = 1, so all of the power formed by a resistor is real. Sin(0) = 0, none of the power formed is reactive. Reactive power surfaces when the current and voltage through a component is not in phase like with inductors and capacitors. For a purely capacitive or inductive load, your current and voltage are either 90 degrees or -90 degrees out of phase - which result in zero for the cos element and 1 for the sin element. Notice, the reactive power for an inductor has a positive sign and a capacitor has a negative sign. If you have inductors in a load with a bad PF, you can add a parallel capacitor to increase the PF. This is so because the reactive powers literally cancel out, so the total complex power becomes less and nearer to 1 for the ratio of real power to complex power.Remember, power is additive. If you had 3 components and a source, the power the source is supplying equals the power of component 1 + component 2 + component 3. Thus, if you want to increase power factor, you can use the sine of complex power to see how much reactive power there is in the circuit. Then, plug in the PF you want and solve for how much complex power there should be(the real power won't change - only the complex power). So complex power becomes real power divided by PF. Then, you find the difference between the complex power you have and the complex power you want. This is how much complex power you want to get rid of. Adding a capacitor or inductor in parallel with your load won't change the load's power, because voltage across parallel devices is the same. So you use voltage squared divided by impedance to solve for the impedance of the capacitor. After that, you can use the impedance to solve for the farads needed for the capacitor - 1/(cw) = capacitive reactance

 

[MATLABdude]

ummmmmmmmmmmm dot dot dot dot dot
power factor is the ratio of average aka real power to total power aka complex power.

Duly corrected. Whoops!

 

[Dr.D]

xcvxcvvc said, "mathematically, reactive power is simply power with an imaginary component. Complex power, the vector sum of real power and reactive power, is V times I*. The star means I conjugate, which means you reverse the sign of the angle for the current."

All of this is just a wee bit awkward since voltage and current are in reality real variables in the physical world.