# RL parallel circuit & vector problem

1. Sep 10, 2011

### commelion

Hi I have been working on this problem and would like to know if I am close to being correct, im also having difficulty understanding the difference between knowing when to use (vector and magnitude of a vector) relationships correctly, I give an example of this below.

A inductor and resistor (unknown values) are connected in parallel and supplied with a voltage of 120 volts 60 hz. Find the inductance in milli henries if the circuits amp meter reads 5 amps and its watt meter reads 500 watts

Suggested solution

The watt meter measures real power only, therefore I used the following
R = V^2/P and got 28.8 ohms

Magnitude Z = Magnitude V/ Magnitude I
And got Magnitude Z = 24 ohms

Here why is it important to use the magnitude here and not just Z = V/I from (ohms law) ?
I am having difficulty knowing when the correct notation is needed.

I then tried the following for parallel circuits |Z|^-1 = |R^-1+ jWL^-1|

With some algebra I got the following

(1/jWL)^2 = 11 / 20736
jWL = 43.41 ohms

L = 115 milli henries using 2pi*fL

I think this is close to being right if somebody could check this i would be grateful, for the question about the vectors and magnitude notation I have searched on line for an explanation to no avail, hopefully some one here can help.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 10, 2011

### fleem

Everything is vectors, so always use vector math. (Of course, some problems will all be one-dimensional vectors and have angles and unreal components of value zero, but you're still using vectors!). It is never useful to neglect a non-zero angle or ignore a non-zero unreal component at any point in your calculations, except in the final answer when only the magnitude or real component is specifically all you've been asked for. Only in casual conversation might one use the word "impedance" to imply only the magnitude of the impedance. Technically, and in your test answers, its best to show the vector notation in the answer unless its clear only the magnitude should be specified. For example, it ~might~ be OK to give only magnitudes in answers if your textbook or your teacher regularly uses the word "impedance" to refer only to the magnitude of impedance--but personally I think that wouldn't be a good way to teach, even though admittedly its sometimes used that way in casual conversation.

Last edited: Sep 10, 2011
3. Sep 12, 2011

### commelion

thanks for the help

correct me i i get this wrong please !

so in the circuit problem the amp meter reads 5 amps, this is the magnitude because you do not know the angle. ?

likewise say i was to put the amp meter in series with the resistor this would be entirely real as it has no angle and therefore does not need the magnitude notation ie. |I|.?

4. Sep 12, 2011

### Staff: Mentor

Ammeters always read magnitude. That's all they can do; they read the magnitude of the current flowing through them.

Ammeters always read magnitude. That's all they can do; they read the magnitude of the current flowing through them. If you put the ammeter in series with anything, or a combination of things, the ammeter always reads the magnitude of the current flowing through the ammeter. It can't do anything else.

It is instructive to imagine you possess lots of meters, so you can connect ammeters in series with every element in a circuit simultaneously. You might see 3 amps in the resistor, 4 amps in the parallel inductor, yet at the same time only 5 amps total into the circuit. (I just made those figures up, for illustration.) It is up to you to draw the vectors on a sheet of paper; the meters won't do that part of the task for you.

Ammeters and voltmeters always measure magnitude. * They have only two input terminals, so know nothing about real or reactive. They have no way of referencing one sinusoid with another to see a phase difference.

Wattmeters are clever gadgets. They measure real power, and that tells you how much power is being dissipated in the resistances in your circuit. To manage this feat, wattmeters have more than two input terminals so they can compare the phase of one wave with that of another.

Last edited: Sep 12, 2011
5. Sep 12, 2011

### commelion

so when doing these problems for instance in the question, ive worked out that the (inductor current is 2.76<-90deg) and the (resistor current is 4.17< 0deg)

these can be repersented by Il = 2.76<-90deg and Ir = 4.17<0deg, there is no need to put the magnitude notation ie || on these as the are just vectors, however the tolal amps in the circuit is 5 amps, this should use the magnitude notation ie |It| = 5 amps, ive plotted these and the do indeed make sense, the also make 5 amps when (a^2+b^2=c^2) is used
im particularly intrested in getting the notation of magnitude and vectors correct.

for example when faced with the magnitude of impedance (|z|) = 24 ohms from question
whats the difference between |z| and just z anybody ?

thanks for the help so far

6. Sep 12, 2011

### Staff: Mentor

Examples:
z=34.5 ohms @ angle = +23 degrees
|z| = 34.5 ohms

7. Sep 12, 2011

### commelion

alarm !!!

sq root of (34.5)^2 + (23)^2 = 41.46 not 34.5

i do understand that my questions here are not of universal importance but everything hinges on me understanding the above ?????????????

8. Sep 12, 2011

### Staff: Mentor

Taking my example of an impedance with magnitude 34.5 at angle +23 deg.
It resolves into a real component of 34.5 cos 23 = 31.75 ohms
and a reactive component of 34.5 sin 23 = 1.2 ohms

Since the reactive component is positive we see that it's in the 1st quadrant, hence due to inductance.

So we have a Pythagorean triangle with base 31.75, height 1.2, and hypotenuse 34.5 The magnitude of the impedance is 34.5 ohms.

9. Sep 13, 2011

### commelion

hi

can you show me how you got 1.2 ohms please, i understand now the vectors and magnitude

thanks

10. Sep 14, 2011

### Staff: Mentor

oops. Looks like a finger fumble when using the calculator.

Should be:
a reactive component of 34.5 sin 23 = 13.48 ohms

So we have a Pythagorean triangle with base 31.75, height 13.48, and hypotenuse 34.5 The magnitude of the impedance is 34.5 ohms.