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.
Electrical – Reactive Power in Inductor and Capacitor
circuit analysiselectromagnetismpower
Related Solutions
'Both inductor and capacitor absorb power' - NO
They both store energy (the integral of power).
In an AC circuit, a simple one where there's only one capacitor and one inductor, doesn't matter whether series or parallel connection, one will be accumulating energy while the other is discharging energy. One will be taking power, the other will be generating power.
The net change in total energy storage seen at the terminals will be the algebraic sum of the changes at the L and at the C. As one is the opposite sign to the other, the sum is the difference in their magnitudes.
Ther mistake you made was assuming Q meant "reactive power" - it doesn't it's the Q-factor of the coil and quite simply is \$X_L/R\$. Q-factor is something you should look up. Q is sometimes called quality factor. See this wiki page. It's embodied in an RLC resonant tuned circuit for instance by this formula: -
And, for just the inductor on its own is simply \$X_L/R\$
Related Topic
- Electrical – A question about reactive power through a capacitor in LTspice
- Electronic – Can a purely reactive element exhibit resistive properties
- Electrical – Why flow of real power depends on the phase difference of voltage magnitudes
- Power – What Does the Value of Reactive Power Represent Physically?
- Electrical – the difference between a tank circuit in an oscillator and a resonance circuit
- Inductor – What Drives Current if Induced Voltage is Equal and Opposite to Applied Voltage?
Best Answer
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.