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A resistor is capable of absorbing positive power. Why wouldn't this be true for capacitors and inductors?

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    \$\begingroup\$ As Shamtam says. It is true by basic definition of the component. One way to look at it is that energy is the time summation of the vector product of voltage and current. In an ideal inductor or capacitor BY DEFINITION current and voltage are at exactly 90 degrees to each other so their vector product over time is zero. In real world components there are no ideal resistance component - this is the part that behaves "non ideally" as an L or C and ideally as a resistor :-). \$\endgroup\$ – Russell McMahon Sep 13 '12 at 6:06
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An ideal resistor dissipates (converts into heat) electrical power. They are not capable of delivering power. Capacitors and inductors both are capable of absorbing and delivering (positive) power. When power is absorbed by an ideal capacitor, all of it is stored in the form of an electric field. Likewise, all of the power absorbed by an ideal inductor is stored in the form of a magnetic field. These devices can deliver this stored energy, but cannot produce energy.

Real capacitors and inductors, however, are not ideal, and will dissipate some power due to imperfections within the device (leakage within a capacitor, for example). This is why in simulations, capacitors and inductors will sometimes have very complex models to attempt to simulate real-world behavior (such as a leakage within a capacitor, which can be modeled simply with a high-resistance resistor in parallel with the capacitor).

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By convention,Positive Power is the on that flows from source to load.Negative Power is the one that flows from load to source.

Yes,Capacitors and Inductors absorb positive power and store it in the electrostatic and magnetic field respectively.But as soon as power source is disconnected they release back the absorbed power to the circuit.In case there is no circuit available to provide path for power flow to source the energy remains trapped and this is how a capacitor retains charge.For inductor,the energy stored compenstaes(opposes) the change in current when source is disconnected.

In an ideal capacitor or Inductor Ohmic losses are zero.Thus power absorbed =power released and there is no net power dissipation.

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In DC analysis:

  • Capacitor acts as an open-circuit (I=0 Amperes)
  • Inductor acts as a short-circuit (V=0 Volts)

If \$Power = Current \cdot Voltage\$,

therefore:

  • Power in capacitor is \$0 \cdot Voltage = 0 \; Watts\$
  • Power in inductor is \$0 \cdot Current = 0 \; Watts\$
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  • \$\begingroup\$ What about AC?! \$\endgroup\$ – nidhin Jan 19 '15 at 16:46

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