AC Circuits confirmation

In summary: The expression Z = 120/(3.12481-3.01183j) is a complex number that can be expressed as Z = X + jY, where X is the real (resistive) component and Y is the imaginary (reactive) component. In this case, X = 19.9079 and Y = 19.1882. This is because the magnitude of complex numbers is given by |Z| = sqrt(X^2 + Y^2). Therefore, the expression Z = 19.9079 + 19.1882j is just a different way of writing the same complex number.
  • #1
charger9198
60
0
For the circuit given in the power factor is 0.72 lagging and
the power dissipated is 375 W.

12.jpg


Determine the:
(1) apparent power
(2) reactive power
(3) the magnitude of the current flowing in the circuit
(4) the value of the impedance Z and state whether circuit is inductive or
capacitive.

(1) Apparent power = True power / Power Factor (S=P/pf)

= 375 / 0.72

= 520.8333 VA

(2) Reactive Power = SQRT (apparent power^2) - (True Power^2)

= SQRT (520.8333^2)-(375^2)

= SQRT (130642.3264)

= 361.4448 VAR

(3) Magnitude of Current

Power= Voltage*Current *Power factor (P=V*I*pf)

375 = 120*I*0.72
I = P/(V*pf)
I = 375/(120*0.72)
I = 4.34 A

(4) Total ohms = Voltage/Current
= 120/4.34
=27.6498 ohms (minus 10)
z = 17.65 ohms

The power factor is lagging within the circuit therefore the the virvuit is inductive



Am i on the right lines here? or is there a better way of calculating the above results
 
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  • #2
I find your calculation of total ohms (part 4) to be a bit dubious. Where you've made the assumption that you can just subtract the resistor value from the impedance magnitude to leave the impedance Z of the "mystery component" is also suspect. Impedance will have real and imaginary parts.

I'd be more inclined to use the current magnitude and its known phase angle obtained from the power factor to construct the complex current: [itex] i = I (cos(\phi) - j\;sin(\phi)) [/itex], and then find the total (complex) impedance of the circuit:

[itex] Z_{tot} = \frac{120 V}{i} = (19.91 + j 19.19) \Omega [/itex]

Then subtract the given resistor from the real part of that to leave the impedance Z.
 
  • #3
Ok thanks gneill

so ill use

ϕ=acos(power factor)
=acos (.72)
= 0.76699

then;

i = 4.34(cos(0.76699)-jsin(0.76699))
i = 4.339566
Zt=120/4.339566
=27.28723 (minus 10)
=17.28723
 
  • #4
charger9198 said:
Ok thanks gneill

so ill use

ϕ=acos(power factor)
=acos (.72)
= 0.76699

then;

i = 4.34(cos(0.76699)-jsin(0.76699))
i = 4.339566 <---- This should be a complex value (real + imaginary components)
Zt=120/4.339566
=27.28723 (minus 10)
=17.28723

The complex current should retain real and imaginary components so that the impedance value obtained will also have real and imaginary components. These correspond to the resistance and reactance of the circuit.
 
  • #5
Think I've got it;

i=4.34(cos(0.76699)-jsin(0.76699)

i = 4.34(0.720003-0.693971)
i = (3.12481-3.01183j)

So

Z=120/(3.12481-3.01183j)

Z=19.9079 + 19.1882j

Minus ten from real gives

9.9079 ohms

This better? Are my answers ok for the other parts of the question?
 
  • #6
Yes, that looks better. So the "mystery component" has both resistance and inductance.

The other answers look fine to me; nicely calculated. :smile:
 
  • #7
Thanks for you help as always :)
 
  • #8
charger9198 said:
Thanks for you help as always :)

You're very welcome!
 
  • #9
gneill said - I'd be more inclined to use the current magnitude and its known phase angle obtained from the power factor to construct the complex current: , and then find the total (complex) impedance of the circuit:



Charger said - ϕ=acos(power factor)
=acos (.72)
= 0.76699

I work the phase angle out at 43.95, where does this 0.76699 come from? I'm having a real problem getting my head round this 'imaginary' thing.
 
  • #10
stemurdo said:
Charger said - ϕ=acos(power factor)
=acos (.72)
= 0.76699

I work the phase angle out at 43.95, where does this 0.76699 come from? I'm having a real problem getting my head round this 'imaginary' thing.

0.76699 is the angle in radians. You can work in degrees or radians as long as you're consistent.
 
  • #11
ok so i use cos-1 x power factor, gives me a phase angle of 43.95 degrees.

i=I(cos(43.95)-jsin(43.95))
=4.34(cos(43.95)-jsin(43.95))
=4.34(0.7199-j0.694)
=(3.1244-j3.012)

is this right so far? this is where I'm getting a bit lost and not sure of the next step
 
  • #12
am i right in saying that the jvalue is a negative value?
so it would actually be (3.1244 - (-3.012))
giving 120/6.134 = 19.555 for the real value? then how do you get the second jvalue?
 
  • #13
stemurdo said:
am i right in saying that the jvalue is a negative value?
so it would actually be (3.1244 - (-3.012))
giving 120/6.134 = 19.555 for the real value? then how do you get the second jvalue?

You don't add the real and imaginary components of a complex number, they are separate entities. Take a look at post #5 in this thread. The impedance will have real and imaginary components, too.
 
  • #14
sorry gneill, i can only ask for patience here. i appreciate the time.
i'm obviously missing something here, or just not grasping the concept at all!
if you don't add or subtract the jvalue then what purpose does it serve? how do you divide 120 by 3.124 and come up with 19.9 or whatever? my brain has packed up and gone home. :-)
 
  • #15
stemurdo said:
sorry gneill, i can only ask for patience here. i appreciate the time.
i'm obviously missing something here, or just not grasping the concept at all!
if you don't add or subtract the jvalue then what purpose does it serve? how do you divide 120 by 3.124 and come up with 19.9 or whatever? my brain has packed up and gone home. :-)

You should review the manipulation of complex numbers; how to add, subtract, multiply, and divide them. http://www.allaboutcircuits.com/vol_2/chpt_2/6.html to get you started.

If you enter "Complex number arithmetic" into a Google search you'll get more information, and even links to online complex number calculators.
 
  • #16
Should the impedance be Vs/I which is 27.63 ohms. If there is an Inductance or Capacitance in the circuit the Z will be higher than R?
 
  • #17
terrytibbs said:
Should the impedance be Vs/I which is 27.63 ohms. If there is an Inductance or Capacitance in the circuit the Z will be higher than R?

27.63 Ω represents the magnitude of the impedance. Impedance is a complex value comprised of a real (resistive) and an imaginary (reactive) part. In this sense, yes, the magnitude will be greater than the size of the individual components: ##|Z| = \sqrt{R^2 + X^2}## . So for X nonzero, |Z| > R.
 
  • #18
charger9198 said:
Think I've got it;

i=4.34(cos(0.76699)-jsin(0.76699)

i = 4.34(0.720003-0.693971)
i = (3.12481-3.01183j)

So

Z=120/(3.12481-3.01183j)

Z=19.9079 + 19.1882j

Minus ten from real gives

9.9079 ohms

This better? Are my answers ok for the other parts of the question?

i don't understand how get from Z=120/(3.12481-3.01183j) to Z=19.9079 + 19.1882j

can anyone explain this please if possible followed the link at the end but can understand from that how it get to Z=19.9079 + 19.1882j
 
  • #19
mattyh3 said:
i don't understand how get from Z=120/(3.12481-3.01183j) to Z=19.9079 + 19.1882j

can anyone explain this please if possible followed the link at the end but can understand from that how it get to Z=19.9079 + 19.1882j

It's basic complex arithmetic. You can convert the denominator to polar form (magnitude + angle) then do the division, convert the result back to rectangular form. Or, clear the imaginary term from the denominator by multiplying numerator and denominator by the complex conjugate of the denominator.

A web search on "complex arithmetic" will likely turn up a tutorial.
 
  • #20
gneill said:
It's basic complex arithmetic. You can convert the denominator to polar form (magnitude + angle) then do the division, convert the result back to rectangular form. Or, clear the imaginary term from the denominator by multiplying numerator and denominator by the complex conjugate of the denominator.

A web search on "complex arithmetic" will likely turn up a tutorial.

you see none of this is in the lesson book for this course the only way i can find is z=v/i

there is polar form but nothing like what's on this page.. i still can't understand what your trying to tell me

just tried a complex arithmetic calculator online and i don't get them answers
 
  • #21
mattyh3 said:
you see none of this is in the lesson book for this course the only way i can find is z=v/i

there is polar form but nothing like what's on this page.. i still can't understand what your trying to tell me

just tried a complex arithmetic calculator online and i don't get them answers

Your course may assume that you've studied complex arithmetic in another course; It's generally taught in a mathematics course. If you haven't taken such a class, then you should find an online tutorial on complex numbers.
 
  • #22
terrytibbs said:
Should the impedance be Vs/I which is 27.63 ohms. If there is an Inductance or Capacitance in the circuit the Z will be higher than R?

well i thought it would be z=v/i but looks different on here and inductance becuase p.f is lagging
 
  • #23
gneill said:
Your course may assume that you've studied complex arithmetic in another course; It's generally taught in a mathematics course. If you haven't taken such a class, then you should find an online tutorial on complex numbers.

i did a maths course before been aloud on this and there is nothing in there on this or how to do it
 
  • #24
mattyh3 said:
well i thought it would be z=v/i but looks different on here and inductance becuase p.f is lagging

Z = V/I is correct, assuming that V and I are complex values (well, here V is assumed to be the reference for the phasor angles, so V's imaginary component will be zero making V a real value). The I is determined from the given information using the power factor to determine the apparent current in complex form.
 
  • #25
still can't find anything of any use that i can understand where the 120 goes and you end up with 19.9079+19.1882j
 
  • #26
mattyh3 said:
still can't find anything of any use that i can understand where the 120 goes and you end up with 19.9079+19.1882j

It's just division. A real number, 120, is being divided by the complex number 3.12481-3.01183j .
 
  • #27
gneill said:
It's just division. A real number, 120, is being divided by the complex number 3.12481-3.01183j .

if i divide them into 120 it get 38.402+39.843j i don't understand any other way of doing it as am not great with maths
 
  • #28
mattyh3 said:
if i divide them into 120 it get 38.402+39.843j i don't understand any other way of doing it as am not great with maths

You'll have to find a complex number / complex arithmetic tutorial to learn about it. Or make use of an online complex arithmetic calculator (although that won't help you in tests or exams!).

Here's an example online calculator (found with via google with a couple of seconds effort):

http://www.1728.org/compnumb.htm
 
  • #29
gneill said:
You'll have to find a complex number / complex arithmetic tutorial to learn about it. Or make use of an online complex arithmetic calculator (although that won't help you in tests or exams!).

Here's an example online calculator (found with via google with a couple of seconds effort):

http://www.1728.org/compnumb.htm

thanks for the help but i can't even use the calculator it gives me weird answers nothing like what's on here

this is what i get 19.408277013161786 + i*19.70643553618951
 
  • #30
mattyh3 said:
thanks for the help but i can't even use the calculator it gives me weird answers nothing like what's on here

this is what i get 19.408277013161786 + i*19.70643553618951

Yes, and that's the correct result for the impedance. It's a complex number with real and imaginary parts.
 
  • #31
gneill said:
Yes, and that's the correct result for the impedance. It's a complex number with real and imaginary parts.

o great i thought i was wrong lol... so now i just need to take off the 10 ohm resistance from the real value of 19.40828158597505 to get z= 9.408 ohms
 
  • #32
mattyh3 said:
o great i thought i was wrong lol... so now i just need to take off the 10 ohm resistance from the real value of 19.40828158597505 to get z= 9.408 ohms

Yes, that's the idea.
 
  • #33
gneill said:
Yes, that's the idea.

thanks for the help just need to find how to do complex calcs on my calculator now
 
  • #34
May I ask gneill why we subtract 10 ohms as I can't find anything in my course about this? Just a quick question really so I know why this happens...
 
  • #35
Big Jock said:
May I ask gneill why we subtract 10 ohms as I can't find anything in my course about this? Just a quick question really so I know why this happens...

If you look at the image attached to the first post you'll see that the load consists of some unknown impedance Z in series with a 10 Ω resistor. The given information was used to calculated the total impedance of the circuit, so Z + 10 Ω. By subtracting the 10 Ω we're left with Z alone.
 
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<h2>1. What is an AC circuit?</h2><p>An AC (alternating current) circuit is a type of electrical circuit that has a source of alternating current, such as a wall outlet, and contains components like resistors, capacitors, and inductors. The current in an AC circuit changes direction periodically, unlike a DC (direct current) circuit where the current flows in one direction.</p><h2>2. How do you confirm an AC circuit?</h2><p>An AC circuit can be confirmed by using a multimeter to measure the voltage and current at different points in the circuit. The readings should show a sinusoidal waveform, indicating that the current is alternating. Additionally, the phase relationship between voltage and current should be consistent with the type of circuit (series or parallel) and the components used.</p><h2>3. What is the difference between series and parallel AC circuits?</h2><p>In a series AC circuit, the components are connected in a single loop, and the current flows through each component in sequence. In a parallel AC circuit, the components are connected in branches, and the current splits between them. The voltage across components in a series circuit is additive, while in a parallel circuit, the voltage across each component is the same.</p><h2>4. What is the role of impedance in AC circuits?</h2><p>Impedance is the total opposition to the flow of current in an AC circuit, similar to resistance in a DC circuit. It is a combination of resistance, capacitance, and inductance and is measured in ohms. Impedance affects the flow of current and the distribution of voltage in a circuit, and it can be calculated using Ohm's law.</p><h2>5. How does an AC circuit differ from a DC circuit in terms of power?</h2><p>In an AC circuit, the power is constantly changing due to the alternating current. The power is calculated using the root mean square (RMS) values of voltage and current. In a DC circuit, the power remains constant as the current flows in one direction. The power in a DC circuit is calculated using the product of voltage and current.</p>

1. What is an AC circuit?

An AC (alternating current) circuit is a type of electrical circuit that has a source of alternating current, such as a wall outlet, and contains components like resistors, capacitors, and inductors. The current in an AC circuit changes direction periodically, unlike a DC (direct current) circuit where the current flows in one direction.

2. How do you confirm an AC circuit?

An AC circuit can be confirmed by using a multimeter to measure the voltage and current at different points in the circuit. The readings should show a sinusoidal waveform, indicating that the current is alternating. Additionally, the phase relationship between voltage and current should be consistent with the type of circuit (series or parallel) and the components used.

3. What is the difference between series and parallel AC circuits?

In a series AC circuit, the components are connected in a single loop, and the current flows through each component in sequence. In a parallel AC circuit, the components are connected in branches, and the current splits between them. The voltage across components in a series circuit is additive, while in a parallel circuit, the voltage across each component is the same.

4. What is the role of impedance in AC circuits?

Impedance is the total opposition to the flow of current in an AC circuit, similar to resistance in a DC circuit. It is a combination of resistance, capacitance, and inductance and is measured in ohms. Impedance affects the flow of current and the distribution of voltage in a circuit, and it can be calculated using Ohm's law.

5. How does an AC circuit differ from a DC circuit in terms of power?

In an AC circuit, the power is constantly changing due to the alternating current. The power is calculated using the root mean square (RMS) values of voltage and current. In a DC circuit, the power remains constant as the current flows in one direction. The power in a DC circuit is calculated using the product of voltage and current.

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