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Why current leads in capacitor

  1. Aug 16, 2005 #1
    HEllo ,

    Can any body answer my question;

    " i know that VOltage leads in Inductor by 90 as compared to current.But i want to know WHY?"

    Why voltage leads in INDUCTOR

    " I know CURRENT leads in CAPACITOR as compare to VOLTAGE but i want to know why ?"

    Why current leads in capacitor
  2. jcsd
  3. Aug 17, 2005 #2


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    Staff: Mentor

    V = L di/dt is the basic V ~ I equation for an inductor. As you put current into an inductor, the change in current generates a voltage. A DC current into an inductor generates no net DC voltage (assuming that the parasitic DC resistance of the inductor windings is negligible).

    I = C dV/dt is the basic V ~ I equation for a capacitor. As you push more charge onto the plates of a capacitor with the current, that changes the voltage between the plates.

    These simplistic (and very useful day-to-day) equations come from Maxwell's equations in E&M. If you'd like to understand their origins better, I'd suggest that you study Maxwell's equations more, and how the simplified versions of those turn into Faraday's Law and Ampere's Law, etc. Just google some of those terms and have some fun... :biggrin:
  4. Aug 17, 2005 #3


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    In addition to what berkeman wrote:
    If the current in an inductor is [tex]i(t) = I sin(\omega.t)[/tex], the voltage will be:
    [tex]v(t) = L.\frac{di}{dt} = \omega.L.I cos(\omega.t)[/tex].
    Since a cosine wave is a sine wave with a lead of [tex]90^o[/tex] you have your voltage lead.
  5. Sep 12, 2005 #4
    This is my reasoning for this affect.

    In an inductor the voltage will change 'instantly' however, when the voltage changes it creats a changing magnetic field. This field will inhibit the current from flowing freely. This delay in current is by 90 degrees as shown above.

    In a capacitor the current will flow instantly, however the charge will not build up on capacitor's plates instantly. This delay in building the charge is what causes the 90 degree delay in voltage, as shown above.
  6. Oct 20, 2005 #5
    A quick mathematical "proof" (which also derives the impedences of these elements) is to apply Laplace transforms to the relations expressed by the capacitor current and inductor voltage, and then substitute s = jw. Multiplying by j causes a forward rotation of 90 of the phasor, dividing by j causes a reverse rotation of the phasor by 90 degrees.



    Apply Laplace, use the derivative transform, and take initial conditions to be zero:

    [tex]L[i(t)]=I(s)=sC V(s)[/tex]

    Set [tex]s=j\omega[/tex]:

    [tex]I(j\omega)=j\omega C V(j\omega)[/tex]

    Since [tex]j[/tex] is mulplying the voltage phasor the current phasor leads it by [tex]90^o[/tex]

    The impedence is *by definition*:

    [tex]Z_{C}(j\omega)=\frac{V(j\omega)}{I(j\omega)}=\frac{1}{j\omega C}=\frac{-j}{\omega C}[/tex]
    Last edited: Oct 20, 2005
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