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Question about the RLC circuit

  1. Jan 20, 2015 #1
    In the direct current circuit, when you have a switch and you turn it on, you get the coil to generate a current to the opposite direction but the magnitude isn't big enough to offset the main one. But in the case of alternating currents, the coil seems to generate enough current up to the magnitude of the main one (because when you calculate the overall emf, its zero). Why is it different? Perhaps because there is always a change in the alternating current? But how do you know that the overall emf is zero? Also, what happens when you increase the magnitude of the voltage so that the total emf goes over zero? And what happens when you lower it?
  2. jcsd
  3. Jan 20, 2015 #2


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    You ask a lot of questions. I will try to answer - but I am not always certain of what you are really asking.
    1. An inductance (coil) generates EMF proportional to the change in current [tex]E = -L\frac{\mathrm{dI}}{\mathrm{dt} }[/tex] In a DC circuit this means that the inductance tries to hinder the current from changing.
    2. In an AC circuit the current is trying to change all the rime and the inductance tries to resist. And the faster the current tries to change, the more the inductance resists. Therefore the reactance of an inductance is [tex]Z_{L}=\omega L[/tex]
    Your statements about EMF being zero I do not understand.
  4. Jan 20, 2015 #3
    Sorry for not stating my questions properly, I was asking about the magnitude of the emf, how can it equal the voltage in the AC circuit but cannot in the DC circuit? And because in the textbooks you derive the current of the AC circuit by equaling the voltage from the battery and the emf induced from a coil; What happens when this is not true? -- Can you make it so that one is bigger than the other?
    I'm sure you did a nice explanation but I'm still a bit confused :confused:
  5. Jan 20, 2015 #4
  6. Jan 20, 2015 #5


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    Well, in a DC circuit, as I said earlier, the inductance tries to hinder the current from changing. It cannot, however, do it fully, since if the current did change, the back EMF would be zero. Thus, the current will increase until it is limited by the resistive losses in the circuit.

    I suggest you draw a circuit with a battery, an inductance and a resistor in it.Start with a zero current and solve the differential equation, it will show you what happens.
  7. Jan 21, 2015 #6


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    Let's solve your problem of switching on a DC circuit with a coil and a resistor in series. One word of warning an EMF is NOT a voltage although it has some similarities. A voltage is an electric field's potential difference, but the electric field has in general not a potential due to Faraday's Law,
    ##\vec{\nabla} \times \vec{E}=-\partial_t \vec{B},##
    from which in the quasistationary limit Eq. (2) in posting #2 is derives. Kirchhoff's Law tells you that the equation for the current reads
    $$L \dot{I}+R I=U,$$
    where $$U=\text{const}$$ is the voltage of the battery. We assume that at the beginning ##I(0)=0##. Then you turn the switch, and the equation holds.

    It's pretty easy to solve. First there's obviously a particular solution, ##I=I_{\infty}=\text{const}##. Plugging this ansatz into the equation you get
    To find the general solution, make the ansatz
    $$I(t)=I_{\infty} + i(t).$$
    Plugging this in gives the homogeneous equation
    $$L \dot{i}+R I=0$$
    with the general solution
    $$i(t)=C \exp(-R t/L),$$
    and the initial condition demancs that the integration constant is ##C=-I_{\infty}=-U/R##. So the solution reads
    $$I(t)=\frac{U}{R} \left [1-\exp \left (-\frac{R}{L} t \right ) \right ].$$
    So the current reaches its DC value asymptotically with a "relaxation time", ##\tau=L/R##.
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