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Finding Voltage in Series RL Circuit

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data

    7pqrWXp.png

    2. Relevant equations



    3. The attempt at a solution

    This assigned question is an odd one since it doesn't seem to have anything to do with phasors which is the chapter I am currently studying.

    I have this solution:

    FGM825Y.png

    I don't understand where they got that equation comes from. I have a feeling that it has to do with RMS which the textbook hasn't yet covered.
     
  2. jcsd
  3. Mar 21, 2014 #2

    gneill

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    Do you recall how the phases of the voltages in the resistor and inductor are related?
     
  4. Mar 24, 2014 #3
    I know that that the phase relationship in a RL circuit is that voltage leads current. Is that what you are referring to?
     
  5. Mar 24, 2014 #4

    gneill

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    That fact is intimately related to the question I posed. How does that translate into the relative phase of the potential drops across the R and L at any given time? Think of writing KVL around the loop.
     
  6. Mar 24, 2014 #5
    Here is my attempt at KVL:

    $$0={ -\overrightarrow { V } }_{ AC }+85\overrightarrow { I } +{ V }_{ L }\overrightarrow { I } \\ \\ { \overrightarrow { V } }_{ AC }\quad seems\quad to\quad be\quad impossible\quad to\quad find\quad because\quad the\quad angular\quad frequency\quad was\quad not\quad given.\quad \\The\quad amplitude\quad is\quad 110V.\\ { V }_{ L }=j\omega \cdot ?\cdot \overrightarrow { I }$$

    kdethyj.png

    I regret using a question mark, ?, as the variable for the inductance of the inductor but it is too late to change now.
     
  7. Mar 24, 2014 #6

    gneill

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    You are not given component values for the resistor and inductor, you are only given the voltage across the resistor. So in your diagram the "85 Ω" is not correct. That should be "85 V" for the potential drop across the resistor. As a result your KVL equation is also not right.

    However...

    Let's take a slightly different approach. When you have a series circuit with reactive components (capacitors and/or inductors) drivin by some current I, then there will be a phase angle between the net voltage and the current. While the instantaneous voltage across the resistor will always be directly in phase with the current according to V = IR, the voltage across the reactive component will be out of phase with the current. You can represent the resulting mathematics with a voltage triangle.

    attachment.php?attachmentid=67967&stc=1&d=1395686358.gif

    This is essentially what's known as Phasor addition. The net voltage phasor is the 110 V on the hypotenuse. As you can see, the the Reactive (imaginary) component of the voltage phasor is at right angles to the Resistive (real). The components add just like the sides of a triangle, hence the ##110^2 = V_R^2 + V_L^2## equation.
     

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  8. Mar 26, 2014 #7
    Would this KVL equation be right?

    $$0=-\overrightarrow { { V }_{ AC } } +85+\overrightarrow { { V }_{ L } } \overrightarrow { I }$$

    Also, I think I get it know. I didn't realize resistor voltage was considered real and reactive voltage was considered imaginary.
     
  9. Mar 26, 2014 #8

    gneill

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    It doesn't look right to me. You have a voltage, VL multiplied by a current (which yields power) and added to a voltage and a resistance. All in all a real mixup of units. Something like this would be better:
    $$V_{AC} = I \cdot R + I \cdot jωL$$
    with VAC and I being phasor variables.

    Complex numbers are just a convenient way to represent phasor values, which have both magnitude and direction (phase). It makes the math easier than tying to work with rotating vectors...
     
  10. Mar 26, 2014 #9
    What if the circuit is slightly more complicated like below?

    AYX4cUk.png

    $${ V }_{ L }=\sqrt { { (85+30) }^{ 2 }+{ 110 }^{ 2 } }$$

    Is that right?
     
  11. Mar 26, 2014 #10

    gneill

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    Nope. The units in the terms are not the same. 85 and 30 have the units Ohms (Ω). The 110 has units Volts (V).

    Did you mean for the numbers above the resistors to be the voltage drops across those components?
     
  12. Mar 26, 2014 #11
    Yes, I keep getting that wrong. Not sure why.

    Last try:

    lWAERUN.png

    $${ V }_{ L }=\sqrt { { 110 }^{ 2 }+{ 100 }^{ 2 } } $$
     
  13. Mar 26, 2014 #12

    gneill

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    Sorry, nope again. Components in parallel all have the same potential difference across them. In this case VL = VR = 110 V
     
  14. Mar 26, 2014 #13
    But I thought there's a difference between resistive voltage and reactive voltage? Aren't they orthogonal in that right-angled triangle you drew? How can they be equal? Could you maybe draw a vector representation of VL or VR?

    Edit: Would I be right saying that the magnitudes are equal? i.e. both have a magnitude of 110V but VL has 110V in the imaginary direction and VR has 110V in the real direction?
     
  15. Mar 26, 2014 #14

    gneill

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    In series, yes, there's a difference in the voltages. In parallel there cannot be... they are connected to the same nodes and any given node has the same potential everywhere. In a parallel connection it's the currents that can have different phases.

    In a series connection the components all have identical current (it's a single continuous current path), but may have different voltages with different phase angles. In a parallel connection the components all share the same potential difference, but may have different currents with different phase angles.

    In a series RL or RC circuit the current is the same for all components and the reactive components have a voltage phase angle of +/- 90° with respect to the current.

    In a parallel circuit the voltage is the same for all components and the reactive components have a current phase angle of +/- 90° with respect to the voltage.
     
  16. Apr 4, 2014 #15
    I asked a couple of other people and they said VR and VL are 90 degrees out of phase. Would that be correct? This circuit is not a purely inductive circuit though, right? So they can't be 90 degrees out of phase, can they?
     
  17. Apr 4, 2014 #16

    gneill

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    If you're referring to the series RL circuit then yes, the two voltages will be 90° out of phase with each other. Take a look at the voltage triangle in post #6. If you assume that the reference angle is that of the voltage source, while the angles that the Resistor and Inductor voltages make with the reference are not 90°, a right angle angle exists between those two voltages themselves.

    The phase angle associated with the circuit usually refers to the phase shift of the current with respect to the voltage.
     
  18. Apr 26, 2014 #17
    attachment.php?attachmentid=67967&stc=1&d=1395686358.gif

    How do you know that the inductor's voltage is completely imaginary/reactive? Is it because ideal inductors have completely imaginary/reactive voltages? What about the inductor's current? Is that also completely imaginary/reactive?
     
  19. Apr 26, 2014 #18

    gneill

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    The voltage is specified by a phasor, which means that there is a phase angle associated with its waveform with respect to the reference angle (usually the source voltage waveform provides the reference). Complex values are a handy way to specify a phasor since they incorporate magnitude and phase angle. Just because there is a phase angle does not mean that the current is somehow not real, in the sense of being different from any other current. It's simply shifted in phase with respect to some reference angle.

    In a series circuit if you imagine that there is some current flowing through the components, all of the components "see" the same current with the same phase. The voltage that develops across a given component is given by Ohm's law for impedance, namely V = I*Z. Z for resistance is purely real, so the voltage developed has the same angle as the current. Z for reactive components is purely imaginary so creates a phase shift due to the + or - j associated with the impedance. That's where the 90° phase shift comes in.

    Where "real" and "imaginary" components of a current or voltage waveform become important is when we look at power. Resistors dissipate energy by converting it to heat. Reactive components are components that store and release energy and don't convert any of the power to heat; they absorb then return the energy doing no "real" work. However, even if no energy is lost in reactive components (the "imaginary" part of the power), the generator driving the load still has to produce the current that flows through those components. So it's sort of a "phantom" power requirement for the load that the generator must supply.
     
  20. Apr 27, 2014 #19
    Ah, that clarifies a lot and I feel like I finally understand it. Of course, in a RL Parallel circuit, the voltages would all be the same (with the same phase angle) and the phase angles would be different in the current because the impedance must be completely resistive for resistors and completely reactive for inductors.
     
  21. Apr 27, 2014 #20
    For anyone reading this in the future, this is how I would now solve this problem.

    First, draw a diagram:

    f7tHTTS.png

    With basic series addition:

    psqXbMw.png

    And now we can draw this diagram to solve for V (labelled Vx in the diagram):

    attachment.php?attachmentid=67967&stc=1&d=1395686358.gif
     
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