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When you close an inductive circuit, since an inductor resists a change in current, the initial reaction of the load is to look like a high resistance. As current builds, the resistance falls. With a theoretical source and inductor, current would eventually reach infinity, that is after infinite time, but practical sources and inductors will reach a plateau current.

When you open an inductive circuit, again, since an inductor resists a change in current, the inductor attempts to maintain that current, but there is no conductivity for that current so, the inductor presents a high voltage spike in the reverse direction it was initially "charged" with. With a theoretical inductor, and theoretical infinite impedance, the voltage spike would be infinite. Again, practical inductors have a maximum voltage spike, but this spike can still be quite high, even thousands of volts, which can damage the circuit, so it is important to maintain a conduction path for the collapsing field, often a diode, or a resistor/capacitor filter.

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Q: What happens when closing and breaking an inductive circuit?
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In a circuit with resistive and inductive load what happens to the power factor as the resistive load increases?

The PF will increase


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The properties of a series alternating-current L-R-C circuit at resonance are:the only opposition to current flow is resistance of the circuitthe current flowing through the circuit is maximumthe voltage across the resistive component of the circuit is equal to the supply voltagethe individual voltages across the inductive and capacitive components of the circuit are equal, but act in the opposite sense to each otherthe voltage appearing across both the inductive and capacitive components of the circuit is zeroif the resistance is low, then the individual voltages appearing across the inductive and capacitive components of the circuit may be significantly higher than the supply voltage


What happens if you overcompensate with 20mF capacitor for the inductive circuit instead of 10mF capacitor which already satisfy the power factor to reach 0.95?

there would be no problem with this == == The inductive circuit has a 'lagging' power factor. If you over-compensate with too much capacitive reactance, you could go over the top (past 1.0) , and end up with a leading power factor that may even be numerically worse than when you started. == == == ==


What happens when inductance is equal capacitance?

Inductance and capacitance are never equal, since they have different units.It's like asking "What happens when temperature is equal to cost ?"It's possible for the inductive and capacitive reactances to be numerically equal,though. That only happens at one frequency, and when it does, your circuit isat resonance.


What happens to an inductive circuit when capacitance is added for power factor correction?

The capacitance counter acts the inductivity (decreases it) without impacting the resistivivity, thus increasing the power factor, or resistivity / inductivity ratio.


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