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.
The PF will increase
What do you mean by a 'parallel delta' circuit -is there such a connection.
depends on the simple circuit. please describe it.
Then the current will stop flowing.
In the last episode of Breaking Bad Walter White dies .
The PF will increase
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
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. == == == ==
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.
The capacitance counter acts the inductivity (decreases it) without impacting the resistivivity, thus increasing the power factor, or resistivity / inductivity ratio.
Current in a purely inductive circuit lags the voltage by 90 degrees. The apparent power in such a circuit will be zero, because the power factor is zero, however, energy will still be transferred, and VARs (Volt-Amps-Reactive) will be non-zero.
Inductors tend to oppose a change in current, so the initial current is low, and rises according to the RC time constant of the circuit to a final value.
Nothing. An open circuit means no current is flowing. When the circuit is closed, current flows, the filament of the bulb is heated by the current and glows, giving off light. But when the circuit is open, nothing happens.
Its breaking """"dawn""""
when a circuit is closed, electricity can move though it.
when a circuit is closed, electricity can move though it.
The series circuit becomes an open circuit because there is no remaining path.