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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.

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0An inductive filter is in fact an LC circuit in which an inductor is connected in series with the capacitor. This arrangement is also known as a low-pass filter. http://www.answers.com/topic/inductive-filter

A capacitor in a circuit reduces noise because a capacitor resists a change in voltage. In combination with resistive and inductive components, even parasitic ones, a capacitor in parallel with a signal or voltage line represents a low pass filter.

because the current in a capacitive circuit leads voltage and current in an inductive circuit lags voltage. Capacitive reactance cancels out inductive reactance which is caused by motors and transformers.

What you are really looking for is total impedance not just inductive reactance. It will be different depending on how the 3 components are connected and which nodes you connect to.Consult your textbook.

Inductors and capacitors are called reactive elements in electric circuits.these reactive elements also offer resistance in the circuit termed as reactancefor inductor it is wL (-j)for capacitor i is 1/wC (j)where L,C and w are inductance , capacitance and frequency of the AC source respectivelywhen clubbed with resistance the the resultant of the resistance and reactance gives us the impedance of a circuitif the impeadence(R=0) of the circuit is of inductor only then these are called as purely inductive circuitsif the impedence of the circuit is dominated by inductor ( wL > 1/wC ) even though the circuit has resistance and capacitor then these circuits are called inductive circuits

A small capacitor can be part of an integrated circuit.

There's no effect since the capacitor was already faulty i.e it was like not in the circuit. Install a healthy capacitor because it will improve the power factor of the fluorescent lamp circuit thus reducing energy wasted.

A mnemonic that was taught in electrical school many years ago is ELI the ICEman. In an inductive circuit, the current through the inductor lags the voltage (E) (L) inductor (I) current.In a capacitive circuit, the voltage through the capacitor lags the current (I) (C)capacitor (E) voltage.See related links below.

A capacitive circuit will more readily pass an AC current, while an inductive circuit will pass a DC current.

What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.

Inductive. Voltage (E) leads current (I) in an inductive (L) circuit and current (I) leads voltage (E) in a capacitive (C) circuit. (ELI the ICEman)

An ideal inductor only has inductance. And ideal resistor only has resistance. And an ideal capacitor only has capacitance. In real life, however, all 3 have some amount of the characteristics of the others. So, in an inductive or capacitive circuit you should only have apparent power in theory, but in an actual circuit you will have resistance from the inductor or capacitor and from the conductors that connect them. This resistance is where the true power is dissipated.

an inductoran electromagnetthe inductive part of a resonant circuitthe inductive part of a ripple filter circuit in a power supplyan RF blocking deviceetc.

Any circuit using a capacitor will not work if the cap is short-circuited.

You want a power factor of 1 or 100%, which is a purely resistive circuit. If you have a motor or some other inductive load in a circuit the total voltage and total current in the circuit will not be in phase (phase shift), your power factor will be less than 1. By adding a capacitor (180 degrees out of phase with inductive load) to the circuit that has a capacitive reactance equal to the inductive reactance of the motor, you can cancel the phase shift and have an ideal power factor (no wasted power). Anything above .9 would be good.

when we replace the resistor with a capacitor ,the current will flow until the capacitor charge when capacitor will fully charged there is no current through the circuit because now capacitor will act like an open circuit. for more info plz E-mailt me at "zaib.zafar@yahoo.com"

A parallel capacitor increase the capacity of the portion of the circuit in which it is installed.

Current lags voltage in an inductive circuit. The angle by which it lags depends on the frequency of the AC, and on the relative size of the inductance compared to the resistance in the circuit.

In general the length of the leads contributes only a negligible amount to the capacitance of a capacitor. However at high enough frequencies excessive lead length can contribute an undesirable amount of parasitic inductive reactance, causing problems in circuit operation.

the capacitor is store the charge and used as back up

Yes, you can connect a polarized capacitor to a direct current. Make sure you get your polarization is correct.A non polarized capacitor can be connected in a DC circuit as well. "Non Polarized" just means it does not matter which side of the capacitor is positive. If you attempt to connect a polarized capacitor in a DC circuit backwards, you will know when the capacitor explodes.

An open circuit, by definition, has no continuity, therefore there is no current flow. A failed capacitor in an open circuit would have absolutely no effect.

inductivecapacitiveresistive

Yes. For a condition called 'series resonance', if the resistance of the circuit is low compared with the inductive reactance and capacitive reactance, then the voltage drop across the capacitor can be VERY much higher than the supply voltage.

In general, no. You need to use the correct capacitor as designed for the circuit.

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