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What are the initial conditions in this simple RCL circuit?

  1. Nov 25, 2011 #1

    So, let's assume the switch has been closed for a long time. The capacitor is charged to 12V, and the coil acts like a short-circuit.

    Immediately after opening the switch, is it right to say the Voltage in the coil is 12V? My notes specify something along the lines of: the conditions at time 0+ must be the same as the conditions at time 0- to preserve continuity, because instantaneous variations are impossible. Are my notes wrong? Is the voltage in the coil at 0+ really 12V?
  2. jcsd
  3. Nov 25, 2011 #2


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    No, the capacitor has the 100 K resistor and the inductor effectively in series across it, so the majority of the voltage will appear across the resistor.
  4. Nov 25, 2011 #3
    Really..? Doesn't that contradict the fact that V = L di/dt in the inductor? Isn't di/dt the biggest when the current goes from 0A to whatever it is right after that?
  5. Nov 25, 2011 #4


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    I did a simulation of this and there is a very brief negative going pulse of about 4.9 volts across the inductor when the switch is opened. This pulse is about 1μS wide.

    The 100 K limits the current which is almost unchanged after the switch is opened. It gets 12 volts from the battery before the switch is opened and it gets 12 volts from the capacitor after the switch is opened.

    After the initial pulse across the inductor, the voltage drops close to zero and the current in the inductor is about 120 μA which is due to the capacitor discharging via the 100 K resistor. This decreases with time.
  6. Nov 26, 2011 #5
    Thanks, I was asking the question to find the initial conditions of the differential equation. I'm assuming that VL(0) is either 0 or 12V, which one is it and why?
  7. Nov 26, 2011 #6


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    At equilibrium with the switch closed and the capacitor fully charged, therei s no current through L1 or C1. The voltage across R2 is determined by the values of R1 and R2. The charge of C1 is at the same voltage as across R2. At the instant the switch is opened current through R1 and R2 goes to zero. L1 field is at zero due to zero current and resists trying to discharge C1. The entire voltage (slightly less than battery voltage due to R1 and R2) is across L1. As the discharge current increases the voltage drop across the resistor increases to a peak at maximum discharge current. After this peak the capacitor continues to discharge along with decrease in the inductor field which is now trying to sustain current. As current decreases voltage across the resistor and the inductor drops exponentially to zero as the charge on C1 is fully depleted.

    That is a word description that goes along with the simulation explanation. The circuit mathematics will confirm that behavior as well if you are at the stage where you have the math tools.

    Edit I just noticed that you are looking at the initial conditions of the Dif EQ. So just before the switch is opened di/dt is zero because the capacitor is fully charged. At the instant the switch is closed di/dt is large and drops the complete capacitor voltage.
    Last edited: Nov 26, 2011
  8. Nov 27, 2011 #7
    Thank you for your time guys!
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