# Rl circ

Discussion in 'Homework Help' started by caique221, Nov 11, 2013.

1. ### caique221 Thread Starter New Member

Nov 10, 2013
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Sorry about the title. I couldn't post, so I read somewhere that a big title could be the cause. I had to make it that short.

I'm having a hard time with a basic AC RL circuit. Since only the voltage drops are given, I don't have a clue where to start from.

1) Source voltage in polar notation;
2) Power factor of the circuit;
3) Phasor diagram with source voltage, voltage drops and current.

The source voltage frequency can be adopted as 60Hz.

I would like to know just what should I do with those voltage drops. I can't think of anything.

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2. ### WBahn Moderator

Mar 31, 2012
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Since all of the components are in series, what you do know about the current in the circuit?

What do you know about the phase angle for the voltage across a resistor relative to the current through it?

What do you know about the phase angle for the voltage across an inductor relative to the current through it?

Dec 12, 2012
17
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you can use the face that the circuit is in series to draw the phasor diagram since there is a common current passing through the elements and
use CIVIL acronym for reactive elements

4. ### caique221 Thread Starter New Member

Nov 10, 2013
7
0

The current is the same along the circuit; the voltage across the resistor is in phase with the current and the voltage across an inductor is 90 degrees ahead of the current going through him. Still, I don't know how to get the magnitude of the current with that.

5. ### caique221 Thread Starter New Member

Nov 10, 2013
7
0

Sorry, but what did you mean by civil acronym for the elements?

6. ### WBahn Moderator

Mar 31, 2012
17,737
4,789
So let's use the current as our reference and say that it is I@0° (magnitude I at an angle of 0°).

What is the phasor for the voltage across the resistor?

What is the phasor for the voltage across the capacitor?

What does KVL require that the voltage across the source be?

7. ### WBahn Moderator

Mar 31, 2012
17,737
4,789
It is one of a handful of acronyms that people have come up with to help them remember things they don't understand.

Or, that they use to help them recall basic conclusions about things they do understand and could figure out but don't want to waste time over and over figuring it out from scratch each time they need it.

So it's either good or bad, depending on why it is being used.

In this case, you think of CIVIL as being two peices: CIV and VIL

CIV - In a C, the I leads the V
VIL - The V leads the I in an L

Another is ELI the ICE man.

ELI - The E leads the I in an inductor
ICE - The I leads the E in a capacitor

Last edited: Nov 13, 2013
8. ### LvW Active Member

Jun 13, 2013
674
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As the forum member #12 has stated in another thread:
"Still learning something almost every day on this site."

9. ### tbinder3 Member

Jun 30, 2013
30
1
I can't view your schematic, but adding voltage drops will give you your total voltage. From there, find total current.

Polar Notation is a complex number, a Magnitude and Vector angle.
Example: 12v<0° The 12v being your magnitude, and 0° being your vector angle.

When solving for total current, you must divide two complex numbers. I=V/Z
Hoping that you know how to solve for impedance.

In my example i'm using the voltage source as a reference, at 0°
Continuing my example: Giving us 12v<0° / 12Ω<30°

Now, when dividing vectors, you divide the magnitudes (12v / 12Ω =1A) AND subtract the angles (0° - 30° =-30°). Giving us I total= 1A<-30°.

Just want to point out that, when subtracting angles, it is common to subtract by a negative number, example: (0° - -30°) resulting in a positive angle.

If you were to add or subtract a vector, you would have to put it in Rectangular Notation first. Rectangular notation has 2 parts as well, Real and "Imaginary" . It will look something like this (-1 + J6) The "-1" being the real part, and "J6" being the so-called "imaginary" part.

When converting a Polar Notation number to Rectangular Notation, I memorized this example: C(magnitude)<θ°(angle) (polar notation)
The Real part= C * cos(θ) or Magnitude * cos(Angle)
Imaginary= C * sin(θ) or Magnitude * sin(Angle)

Example: 1A<-30°
Real = 1 * cos(-30) = .866
Imaginary = 1 * sin(-30) = -.5