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RLC circuit and rate of increase of current

  1. Mar 16, 2005 #1
    Ok, I am having a problem with a RLC circuit and my teacher was unable to offer a satisfactory explanation. Hope that you'll help.
    Consider a DC battery, a switch, a resistor, a capacitor and an inductor in series. Before t=0 the switch is open and assume that the circuit has achieved steady state. This means that the capacitor voltage is 0 and so is the inductor current. At t=0 the switch is closed. It follows that at t=0+ (i.e. just after t=0) the capacitor voltage will be 0 and so will the inductor current. It also follows that the voltage across the inductor at t=0+ will be V, i.e. the voltage of the source (the capacitor is acting as short-circuit and the inductor as open circuit). Therefore the rate of change of current in the circuit at t=0+ will be some positive value (from v=L di/dt). But the rate at which the capacitor voltage is increasing is ZERO at t=0+ (from i=C dv/dt; i=0). How could the rate of increase of current in the inductor be non-zero, but the rate of increase of voltage in the capacitor be zero? If one is increasing, shouldn't the other as well?
    Please explain in physical terms and not mathematical. I'll be highly grateful.
  2. jcsd
  3. Mar 16, 2005 #2
    I will try to explain as best that I can

    Up till here, you are correct!

    There is a slight misconception here. You are saying that the current i at time t=0 is zero, using the current continuity of the inductor i(0+)=0, i.e. physically, the current cannot have a discontinuity jump at time t=0. But there is no current continuity at the capacitor, i.e. at time t=0, i(0-)=0 AND
    i(0+)=some value. So the rate of increase of capacitor voltage is NOT zero.
    Hence the capacitor voltage is increasing at a positive rate together with the inductor current at time t=0+

    I may be wrong, because it has been one year ago since I touch on this concept. If I am wrong, can someone please advice. Thanks.
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