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In a Parallel Resonance circuit (C parallel with RL)

  1. May 10, 2016 #1
    Why does the total supply current increase when the capacitance value is further increased after parallel resonance is reached (where the supply current is minimum) ? I know it has to do with the phase angles but I cannot reason it out.
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. May 10, 2016 #2


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    You are applying a (alternating) voltage to the inductor and this gives a certain current.
    You apply the same voltage to the capacitor and this gives a certain current.
    At resonance these currents are equal - not a coincidence, that's why resonance occurs. But they have opposite phase: one is 90o ahead of the voltage and the other 90o behind the voltage, ie. they are 180o apart, opposite in phase.
    That means, when current is flowing into the capacitor, it is flowing out of the inductor and vice versa. Since the currents are equal in size, no current is supplied by the applied voltage - the circuit has infinite impedance, voltage is applied, but no current flows. EDIT: from the source.
    In fact of course, the resistance means that the currents are not exactly 90o from the voltage and therefore not exactly 180o apart and not exactly balanced, so a small current will flow from the applied voltage and the impedance will be very high, rather than infinite.
    But lets just think of the perfect balance for the moment. IC exactly equals IL. Now increase the capacitance, so IC increases. IL is unaffected, since it still has the same applied voltage (we'll assume a relatively low impedance source.) So all the extra current for the capacitor must come from, and go back to, the source. So the source supplies an extra current. Since it supplied very little at resonance (ideally none, but we know there is a bit) any extra current is a big % increase and appears as a big change in impedance.
    If the circuit has a very low resistance and a high Q, then the resonance source current is very small and the change, as the capacitance increases from resonance, is very marked. A low Q tuned circuit with relatively high resistance (or other losses equivalent to resistance) will have a relatively high source current at resonance, so the increase in current as the capacitance increases away from resonance, is less marked and you get a broader resonance.

    If the capacitance were decreased below the resonant value, you would still get an increase in current. IC would decrease, but IL stays the same and the supply has to provide the deficit. Similarly you can get an increase in current when L changes and C stays the same.
    If you change both L and C so that both currents increase (raise C and lower L) or both currents decrease (raise L and lower C) then the circuit is still resonant.
  4. May 11, 2016 #3


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    At Zero Hz, the inductor is a dead short and at ∞Hz, the Capacitor is a dead short. Add in some realistic component resistances and your 'dead shorts' become low resistances.
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