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how is that possible drop bieng greater than source voltage??????????

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- Thread starter FizixFreak
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how is that possible drop bieng greater than source voltage??????????

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Bob S

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sophiecentaur

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*If you give a child's swing a very small push each time it returns to you the amplitude of the swing will build up to be much higher than it was from your first push. That's a mechanical analogue.

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*If you give a child's swing a very small push each time it returns to you the amplitude of the swing will build up to be much higher than it was from your first push. That's a mechanical analogue.

so the voltage drops that are greater than the source just cancel that is why we can say that these drops can be greater than the source voltage right ?????????

but it is a bit diffcult for me to understand how the concept of resonance is applied in these circuits

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sophiecentaur

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Left to themselves, a C and L in series have a natural frequency. Charge up the C and leave them to it and you will get an oscillation with a gradually decreasing amplitude as the energy dissipates in the resistance present - just the same as a pendulum or a mass on a spring etc.

The large voltage swing appears across the C two components and can be high - to the source, if applied in parallel with them, it looks like a high impedance and the voltage may not get much higher than the input volts (limited by the source resistance). If you feed them in series, the input looks like a low impedance (low voltage swing) but across the L and C are high voltages - in antiphase with each other and in quadrature with the exciter voltage.

The large voltage swing appears across the C two components and can be high - to the source, if applied in parallel with them, it looks like a high impedance and the voltage may not get much higher than the input volts (limited by the source resistance). If you feed them in series, the input looks like a low impedance (low voltage swing) but across the L and C are high voltages - in antiphase with each other and in quadrature with the exciter voltage.

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hey guyz i was just reading series resonance circuits and a read a confusing statement ''at resonance frequency the voltage drop across capacitor and inductor may be much larger than the source voltage''

how is that possible drop being greater than source voltage??????????

It's NOT possible! Since the source voltage must equal the voltage in the rest of the circuit (capacitor and inductor in series) the voltage of the series combination is never greater than the source voltage.

What they are obviously referring to is that the voltage measured EITHER across the capacitor OR across the Inductor can be greater than the source voltage. This would be because at resonance the impedance of the capacitor and inductor pretty much cancel each other. This makes the total impedance of the series circuit quite low. That leads to a large current flowing. And that large current through EITHER the impedance of the capacitor or inductor gives a high voltage (by ohm's law) But together they cancel leading to the low impedance of the whole circuit. But such a cancellation only happens at one frequency known as "resonance".

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sophiecentaur

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You are right in suggesting that the initial (instantaneous) impedance of the RL combination is at its highest at switch on. As the rate of change of current drops, so does the input impedance.

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how is that possible drop bieng greater than source voltage??????????

Here is a site that explains this: http://www.allaboutcircuits.com/vol_2/chpt_6/3.html

Tesla coils are a phenomenal example of series resonance. Other factors in Tesla coils play a part too like loose coupling, turns ratio, capacitor efficiency, etc., but it is mostly series resonance that produces the high voltages.

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