Suppose a RLC circuit with AC source. Why the net reactive power in the circuit is the difference of reactive power in the inductor and the reactive power in the capacitor ? As both of them store power then why the reactive power in them is not added instead of subtraction.
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As both of them store power then why the reactive power in them is not added instead of subtraction.
Both of them store energy not power. Power is the rate at which energy is stored or transferred. Anyway, moving on...
For a capacitor Q = CV and if you differentiate Q you get: -
\$\dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$ and, of course rate of change of Q is current therefore you can say: -
\$I = C\dfrac{dV}{dt}\$. This means if the applied voltage (V) is a sinewave then the current is a cosine wave: -
Now if you looked at an inductor you would find that \$V = L\dfrac{dI}{dt}\$.
If you integrated both sides to find I then you would see that \$I = \dfrac{\int{V}\cdot dt}{L}\$.
Clearly if V is a sinewave then I must be a negative cosine wave. Here are the two scenarios side by side: -
So, if you have an L and a C in parallel across a sinewave voltage source, then the current in the inductor is exactly opposite in polarity to the current in the capacitor i.e. while one is taking energy from V, the other is delivering energy back to V. This gives rise to the currents being subtracted.
It's also notable that when the impedances are identical (one specific frequency) as indicated on the diagram immediately above, the net current into a parallel LC is zero i.e. they behave together as an infinite impedance. This is called parallel resonance and is extensively used in radio. It's also called power factor correction in electrical engineering; you find a value of capacitance that keeps the PF unity thus keeping the reactive energy "consumed" to a minimum.
answered Jan 23, 2016 at 10:59
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\$\begingroup\$ thanks a lot for your reply. it was very helpful. can you please further tell me that if the circuit is series RLC circuit then what will be the phase difference between inductor voltage and capacitor voltage ? in this case the current in the inductor will be exactly opposite in polarity to the current in the capacitor also ? can you please explain that ? \$\endgroup\$ Commented Jan 27, 2016 at 10:11
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\$\begingroup\$ In a series circuit the current has to be the same so now, the voltage across the inductor will lead the "common" current by 90 degrees and the voltage across the capacitor will lag the "common" current by 90 degrees hence, inductor and capacitor voltages are 180 degrees apart or they are inverse to each other. \$\endgroup\$– Andy akaCommented Jan 27, 2016 at 10:14
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\$\begingroup\$ oh so again the power in capacitor and inductor would be opposite P=VI. great. thanks alot. i really appreciate. :) God bless you. \$\endgroup\$ Commented Jan 27, 2016 at 10:21
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\$\begingroup\$ you have clarified it. i do not have any confusion now. do you think there is any confusion left after your clarification ? \$\endgroup\$ Commented Jan 28, 2016 at 12:21
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\$\begingroup\$ @hurchuchu OK then maybe you can consider pressing the "accept answer" button located below the up/down arrows at the left-top of my answer? \$\endgroup\$– Andy akaCommented Jan 28, 2016 at 12:35
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Current leads voltage in a capacitor. Voltage leads current in an inductor. I was taught this using the CIVIL spelling:
In a C I leads V leads I in an L. (I hope that makes sense.)
The effect is that the voltage or current will be 180° out of phase between the inductor and the capacitor and so in summing them they tend to cancel out rather than add.
As both of them store power then why the reactive power in them is not added instead of subtraction.
Both of them store energy but one will be maximum while the other is minimum. This effect is seen best in the resonant circuits where the energy flows back and forth between the inductive and capacitive elements.
answered Jan 23, 2016 at 10:38