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Series R L circuit

  1. Jun 25, 2013 #1
    1. In a series R L circuit, we get this equation for the current i : i(t) = E/R[1-e^(-Rt/L)] where R: Resistance, L: inductance, E: emf


    2. My problem is how we get this equation?, why we consider dx/dt + px = c to get the above equation? how the above equation connects with steady response and the transient response of the circuit?
     
  2. jcsd
  3. Jun 25, 2013 #2

    Simon Bridge

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    Usually from a mesh analysis - written in differential form, then you solve the differential equation.
    A good text book should show you this.

    Because that is the form the DE takes.
    This way you get to use someone elses work as a shortcut. You could always just solve it yourself of course.

    That should be clear from the definitions of "steady state" and "transient" response.
    Note: to have a response you have to have something to respond to.

    The circuit behaves a bit like a damped and driven harmonic oscillator ... so there will be a short-lived component and a long-lived one.
     
  4. Jul 6, 2013 #3

    rude man

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    This current is the response to a step input of voltage. No other input will yield this response. This IS the transient response. The steady-state response is i = E/R. In this case V = step input voltage V0 U(t).

    The basic differential equation is V = Ri + L di/dt, based on the simple fact that for an inductorr V = L di/dt and for a resistor V = iR.

    If a sinusoidal voltage V = V0 sin(wt) is applied at t=0 there is a transient as well as a steady-state response. The above diffrerential equation allows solving for both.
     
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