Parallel R-L circuits, modeled motor

AI Thread Summary
A 0.5hp motor draws 1000W and 10A from a 120V, 60 Hz source, and is modeled as a parallel R-L circuit. The phasor diagram indicates that voltage leads current by 90 degrees, but the user struggles to find the resistance (R) and inductance (L) values, which are given as 14.4 and 21.7 Ohms. The user attempts to apply power equations but finds it challenging to relate them to the circuit parameters. They calculate R using the formula R = V²/P, yielding a value consistent with the expected answer. Overall, the discussion revolves around understanding the relationship between power, resistance, and reactance in the context of a parallel R-L circuit.
Bradracer18
Messages
203
Reaction score
0

Homework Statement



A 0.5hp motor is observed to draw 1000W and 10 A when connected across a 120V, 60 Hz source.

The motor can be modeled as a parallel R-L circuit. Sketch the phasor diagram and find the values for R and L.

---There are more questions to this, but I might be able to do them, so want to start here. I can't understand how to get the answers in the back of the book(both in Ohms).

The answers are 14.4 and 21.7 Ohms.

Homework Equations




Not sure yet...


The Attempt at a Solution




My phasor diagram has voltage(E=120V) leading Current(I=10A) by 90 degrees. So, in other words, the voltage arrow points in the + x direction and the I points somewhere down at an angle between 0 and -90 degrees(on a conventional x-y axis).


I can't even get started...so any help to get me started would be great!

Brad
 
Physics news on Phys.org
For reference here is some info on a parallel R-L circuit.

Parallel resistor-inductor circuits
http://www.allaboutcircuits.com/vol_2/chpt_3/4.html


Is there a discussion in one's textbook regarding real, reactive and apparent power?
 
yeah, I have the equations for them.

Total Power VA = IE

Actual Power = P = IEcos(theta)

Reactive Power = Pr = IEsin(theta)I don't see how these help though, I tried to use them, and solved for theta...and got 33.6 degrees...but how does that help, and is that even correct??
 
Well how about 1000 W? This is the Real Power!

So P = V2/R => R = V2/P = (120 V)2 / (1000 W)

Then Apparent or Total Power = (120 V)(10A) = 1200 VA = S

and S2 = P2 + Q2, where Q is Reactive Power,

and Q = V2/X, where X is the Reactance = \omegaL for a pure inductive load, and \omega = 2\pif

http://www.allaboutcircuits.com/vol_2/chpt_3/2.html
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top