voltage across inductor? series RL

Discussion in 'Homework Help' started by electrictek, Jan 23, 2012.

  1. electrictek

    Thread Starter New Member

    Jan 23, 2012
    2
    0
    What is the formula for calculating voltage accross an inductor in series RL circuit? My book - I cannot quite find or pinpoint the formula down. Is it that I have to find reactance and impeadance? I learned that with RC circuits but do not know how to incorporate it within the RL circuit. Thanks for any help :)
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,447
    3,362
    Same thing as with capacitance. X = \omega L
     
  3. electrictek

    Thread Starter New Member

    Jan 23, 2012
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    0
    so I use this forumla for reactance? then take that into a voltage a divider rule- to find voltage across inductor?

    and that is winding * L right? equals X
     
  4. MrChips

    Moderator

    Oct 2, 2009
    12,447
    3,362
    X is the reactance
    \omega is the angular frequency = 2\pi f
    L is the inductance

    To calculate the voltage across the inductor you must use vector arithmetic

    Z = R + jX
    Z = R + j\omega L

    where j = \sqrt {-1}
     
  5. Vahe

    Member

    Mar 3, 2011
    75
    9
    If you have a voltage source in series with an inductor and a resistor, the voltage across the inductor is

    <br />
\bar{V}_L = \frac{j \omega L}{R + j \omega L} \bar{V}_s<br />

    and the voltage across the resistor is

    <br />
\bar{V}_R = \frac{R}{R + j \omega L} \bar{V}_s<br />

    where \bar{V}_s is the voltage source in phasor notation.

    To get the amplitude of the voltages, you must compute the magnitude of the complex quantities. For example, the inductor voltage amplitude is

    <br />
| \bar{V}_L | = \frac{\omega L}{\sqrt{R^2 + (\omega L)^2}} | \bar{V}_s |<br />

    where | \bar{V}_s | is the amplitude of the voltage source.

    Best regards,
    Vahe
     
    Last edited: Jan 24, 2012
  6. dlhylton

    New Member

    Jan 24, 2012
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  7. holnis

    Member

    Nov 25, 2011
    50
    4
    If the RL circuit has a resistor and an inductor connected in series. A constant voltage V is applied:
    The voltage across the resistor is given by:
    [​IMG]
    The voltage across the inductor is given by:
    [​IMG]
    'Kirchhof''s voltage law says that the directed sum of the voltages around a circuit must be zero. This results in the following differential equation:
    [​IMG] ..
     
  8. pavan_maddi

    New Member

    Feb 3, 2016
    3
    0
    Same what i was struggling for is but in my book , The case is in DC RL circuit with time variant and the circuit is an open circuit then how to find the voltage across the inductor when the time constant is 0?
     
  9. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492
    Hello,

    The inductor could be treated like a resistance, with the added factor j*w before it and that works for almost all circuits.
    So with a resistor R in series, we just have something like a voltage divider.

    First, the true voltage divider with two R's R1 and R2, and source voltage Vs:
    vR2=Vs*R2/(R1+R2)

    Now for the series circuit in the frequency domain where we want the voltage across the inductor:
    zL=j*w*L
    vL=Vs*zL/(R1+zL)

    So all we did was replace R2 with zL, and zL is j*w*L. This last equation is then simplified and we have to find the amplitude and phase shift sometimes too.

    Since this depends on time or frequency we have another way of looking at it, in the time domain. Instead of frequency we want the time response here. For this we can do it this way...
    zL=s*L
    vL(s)=Vs(s)*zL/(R1+zL)

    and this is the same as before except we let s=j*w, and also we need to know what kind of source we have Vs(s) as it depends on s also.
    For a simple step response, we find the inverse Laplace Transform and end up with:
    vL(t)=e^(a*t)

    where a=-R1/L

    There are other ways to do it also, i am not sure what way you might find more acceptable.

    The two basic solutions as you see are:
    1. Solution in the frequency domain.
    2. Solution(s) in the time domain.

    In the time domain there could be more than one general solution depending on the values in some circuits.
     
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