RL parallel circuit & vector problem

In summary, in this problem, an amp meter reads 5 amps and its watt meter reads 500 watts. The wattmeter measures real power only, and so I used the following equation to find the inductance: (1/jWL)^2 = 11 / 20736. I think this is close to being right if somebody could check this.
  • #1
commelion
40
0
Hi I have been working on this problem and would like to know if I am close to being correct, I am 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.

thanks in advance.
 
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  • #2
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.
 
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  • #3
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
commelion said:
so in the circuit problem the amp meter reads 5 amps, this is the magnitude because you do not know the angle. ?
Ammeters always read magnitude. That's all they can do; they read the magnitude of the current flowing through them.

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|.?
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 possesses 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. :smile:

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.
 
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  • #5
so when doing these problems for instance in the question, I've 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, I've plotted these and the do indeed make sense, the also make 5 amps when (a^2+b^2=c^2) is used
im particularly interested 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
Examples:
z=34.5 ohms @ angle = +23 degrees
|z| = 34.5 ohms
 
  • #7
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 ?

thanks in advance
 
  • #8
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
hi

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

thanks
 
  • #10
commelion said:
can you show me how you got 1.2 ohms please, i understand now the vectors and magnitude
oops. Looks like a finger fumble when using the calculator. :eek:

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.
 

1. What is a RL parallel circuit?

A RL parallel circuit is an electrical circuit that consists of a resistor (R) and an inductor (L) connected in parallel. In this type of circuit, the voltage across both components is the same, but the currents through each component can differ.

2. How do you calculate the total resistance in a RL parallel circuit?

The total resistance (RT) in a RL parallel circuit is calculated using the formula: RT = (R^-1 + XL^-1)^-1, where R is the resistance in ohms and XL is the reactance of the inductor in ohms.

3. Can you explain the concept of reactance in a RL parallel circuit?

Reactance is the measure of opposition to the flow of alternating current (AC) in a circuit due to the presence of inductance. In a RL parallel circuit, the inductor creates a reactance that opposes the flow of current, causing a phase shift between the voltage and current.

4. How do you solve a vector problem in a RL parallel circuit?

To solve a vector problem in a RL parallel circuit, you first need to convert all the values (resistance, inductance, and voltage) into their polar form, which includes both magnitude and phase angle. Then, you can use vector addition and subtraction to find the total voltage and current in the circuit.

5. What is the difference between a series and parallel RL circuit?

In a series RL circuit, the resistor and inductor are connected in a series, meaning that the same current flows through both components. In contrast, a parallel RL circuit has the resistor and inductor connected in parallel, so the voltage across both components is the same, but the currents through each component may differ.

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