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I understand that ideal inductor behaves as a short circuit at steady state and the ideal capacitor keeps on filling infinitely. I read that the inductor will be short circuit and capacitor will be open circuit at stead state. But the capacitor never blocks the current, it keeps on charging. So why is it an open circuit?

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    \$\begingroup\$ A schematic would be nice but: "It keeps on charging." Only because an ideal current source keeps raising the voltage till infinity. \$\endgroup\$
    – Oldfart
    Jul 28 '18 at 8:34
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    \$\begingroup\$ But it does not as your ideal L will short circuit the ideal DC current source and take all the current. \$\endgroup\$
    – Oldfart
    Jul 28 '18 at 8:45
  • \$\begingroup\$ Draw your circuit, there is a good tool for this on this site. If you have the circuit I think you have Andy's answer is correct the inductor takes all the current and there is 0V across the capacitor. But a drawing would clarify this. \$\endgroup\$ Jul 28 '18 at 9:58
  • \$\begingroup\$ A capacitor isn’t an open circuit, it’s a capacitor. It just behaves like an open circuit under quite specific conditions. And a ramp input current isn’t one of them. \$\endgroup\$
    – Chu
    Jul 28 '18 at 12:59
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Inductor

For a steady state applied current, an inductor behaves like a short circuit because: -

$$V = L\dfrac{di}{dt}$$

If the current is steady state it is not changing hence di/dt = 0 therefore V must be zero and the inductor looks like a short circuit..

However, if the voltage is constant, then di/dt has a non-zero value hence current rises with respect to time.


Capacitor

For a steady state applied voltage, a capacitor behaves like an open circuit because: -

$$I = C\dfrac{dv}{dt}$$

If the voltage is constant then the current has to be zero hence it looks like an open circuit.

However, if the current is constant, then dv/dt has a non-zero value hence the voltage rises with respect to time.


LC Parallel Circuit with current source in steady state

In this scenario the steady state situation is that all the current flows through the inductor and the capacitor never charges.

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Connecting an ideal current source with an LC tank shows this:

enter image description here

Here is the falstad simulation: bit.ly/2KkQsk0

  1. A capacitor that is not "fully charged" appears like a short circuit to a voltage source or current source. Its like an empty cup waiting to be filled.

  2. An inductor initially appears like a open circuit. Because it needs time to allow current to change => dI/dt is finite. If inductor appears short circuit right away means it builds current in 0 time which implies dI/dt = infinity.

When both inductor and capacitor are connected in parallel the steady state outcome is oscillations assuming there are no resistances.

The same can be derived if you try to solve the differential equation.

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