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I DC RL circuit understanding

  1. Jan 17, 2017 #1
    Hi, I can't figure out what is the physical reason behind the fact that, in DC RL circuit, for example, the series one, the current rises from zero (supposing state zero response), to its maximum value given by ohm's law. I've understood it mathematically but I can't physically explain why, after the current being zero at the very zero instant of time, it starts increasing over time. For current to flow voltage across L must be less than the EMF of the DC power supply but this implies that over time voltage across L, which is back EMF with a minus sign, decreases over time. Again I can't find a physical reason to this fact that voltage across L decreases over time. Thanks in advance to anyone who tries to make me understand this.
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  3. Jan 17, 2017 #2


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    Qualitatively what happens when you switch on the current is that a magnetic field builds up in the coil, but a time-dependent magnetic field implies an EMF, given by the curl of the electric field, and according to Lenz's Law this EMF tries to hinder the current to build up (it also follows of course from Maxwell's equations, particularly Faraday's Law of induction ##\dot{\vec{B}}/c+\vec{\nabla} \times \vec{E}=0##.

    Now you can simplify the task solving the entire set of Maxwell equations in this case by making the quasistationary approximation and lump everything in effective constants of compact circuit elements like resistors, coils, and capacitors. For a coil in this case the self-induction ##L## is the effective quantity describing the above described induction of an EMF due to a time-varying current. For the series of a resistance and a coil you get the differential equation
    $$L \dot{I}+R I = U,$$
    where ##U=\text{const}## is the DC voltage applied to the series circuit. It's easy to see that a particular solution (the stationary final state) is
    The general solution of the homogeneous equation is easily found by
    $$L \dot{I}_h + RI_h =0 \; \Rightarrow \; \frac{\mathrm{d}}{\mathrm{d} t} \ln \left (\frac{I_h}{I_0} \right )=-\frac{R}{L} \; \Rightarrow \; I_h(t)=I_0 \exp \left (-\frac{R}{L} t \right ).$$
    So the general solution is
    $$I(t)=\frac{U}{R} + I_0 \exp \left (-\frac{R}{L} t \right),$$
    and from the initial condition ##I(0)=0## you get ##I_0=-U/R##, so that you finally get
    $$I(t) = \frac{U}{R} \left [1-\exp \left (-\frac{R}{L} t \right) \right ].$$
  4. Jan 17, 2017 #3
    Thanks for the reply, but unfortunately that doesn't answer my question. If I consider the ideal inductor, the self concatenated flux of the magnetic field is givel by: φ=L⋅i , where L is self inductance and its only a geometrical coefficient while i is the current flowing through the inductor. As we turn on the circuit, the inductor feels the instantaneous change in current and react by procucing and induced back EMF according to Faraday and Lenz's law: EMF=-dφ/dt=-L⋅di/dt. Choosing the load references whe change sign to the induced EMF so that now become the potential difference across L. So at zero time voltage across L is the same as the EMF of the DC power supply so by Ohm's law current is zero.
    Now begins the part that Im unable to understand. Via math we derive that i has and positive exponential growth behavior. But what I want to know is from a physical point of view, what causes current to increase? I started wondering this question and I came up with this tought: by Ohm's law, in order to current not be zero the voltage across the inductor must decrease. So what causes this? Again we can derive via math that the voltage across L behaves like a negative exponenxial growth. I want the physical reason not the math explanation. I've already understood it via math but not much physical.
  5. Jan 17, 2017 #4


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  6. Jan 19, 2017 #5


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    Giuseppe Sorrentino:
    First you must realise that the current through an inductor is intimately linked to the magnetic field of that inductor. When a current flows through a conductor it causes a magnetic field at 90° to the current. That magnetic field is coupled back into the conductor, again with a 90° left turn. But then it has turned left twice, so the induced voltage will be facing backwards. That is why a change in the inductor current generates a back emf that opposes the change of current. That is also why the magnetic field cannot come into existence instantly.

    Multiplication of a vector by the operator i = √-1, is equivalent to rotating the vector left by 90°. Multiplication twice by the operator i, is equivalent to multiplying once by i2. We know that i2 = -1, so a vector multiplied by i2 is reversed, or rotated by 180°. That reversal comes up often when magnetic fields induce currents in conductors. It explains why good conductors make good mirrors and why high frequency AC current only flows in the surface of wires.
  7. Jan 20, 2017 #6
    I think I got it. Since at the beginning the inductor fights the change in current, back EMF is induced in the inductance. But this means that current starts to increase according to Faraday's law: EMF=-L⋅di/dt. So current starts to increase until the stationary state is reached where current no longer changes so EMF is zero.
  8. Jan 20, 2017 #7
    I would thank you all for trying help me out.
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