0
Consider the instantaneous DC analysis. Initially, the capacitor has zero resistance. You apply a voltage and current is controlled by other resistive elements alone. As the capacitor charges, its effective resistance rises. This adds to the net resistance in the circuit, reducing current. At full charge, the capacitor has infinite resistance, so there is no current. Remember that the equation for a capacitor is dv/dt = i/c.
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
Why does the ratio of the voltage to current in capacitor and inductor depend on frequence?
The ratio of voltage to current, or the impedance, of reactive elements such as capacitors and inductors depends on the frequency of the applied wave because they store energy, and the amount of energy they store is directly related to the frequency of the applied waveform. When a DC voltage is applied to a capacitor, the current through the capacitor initially will be large, and will decay down to zero as the capacitor charges. Also, the voltage across the capacitor will be small initially and will increase over time to be equal to the applied voltage. This behavior results in a varying impedance when an AC waveform is applied. At a very low frequency, the capacitor will charge up and discharge similarly to if a DC source was switched into the capacitor for a long period of time there would be a large voltage drop, and small current = high impedance). As the frequency increases, the capacitor will appear more like a DC source was initially switched into the capacitor (low voltage drop and high current = low impedance).
Why capacitor behave as open circuit against alternating current?
Capacitors store electrical charge. Imagine we have a capacitor. At time 0 seconds we connect a DC voltage across the capacitor - immediately as the voltage is connected the capacitor is at 0 volts and the maximum current (relative to the circuit resistance) flows. At this extreme the capacitor can be treated as a short circuit, so for high frequency AC volts we should treat a capacitor as being a short circuit. As time passes the current in the circuit will go down and the voltage of the capacitor will go up - this is because as the capacitor gains more charge it gains more voltage, lowering the voltage across any resistance in the circuit consequently lowering the current in the circuit. When the capacitor is virtually full no current will flow at all and the voltage across the capacitor will equal the DC source voltage. At this extreme the capacitor can be treated as an open circuit, so for low frequency AC (allowing the capacitor to fill up before the current alternates) we can treat the capacitor as being an open circuit. Technically, it is not an open/closed circuit when it comes to AC because the capacitance will results in a signal lag or lead. However, if the frequency is low/high enough the lag/lead is often negligable.
What is the advantage of the capacitor-start induction-run motor over the resistance-start induction-run motor?
A capacitor-start system will limit the current drawn by an induction machine by correcting the devices poor power factor during the start up. This limits the current needed to start the device without using any real power. A resistance-start system will limit the starting current by drawing current unto the resistor instead of the induction machine. The real power used by the resistor is turned into heat and lost. A capacitive-start system will use no real power and will not create excess heat, it will be more expensive to initially purchase.
How do you wire a capacitor to a electric motor?
A capacitor resists a change of voltage, proportional to current, and inversely proportional to capacitance. dv/dt = i/c Capacitors can do various things. They can filter out high frequency transients, or power supply ripples, they can block DC while passing AC, they can participate with other components such as resistors and inductors to form filters - the list is endless.
Does current actually flow through a capacitor?
No. Current (or more specifically, charge) flows into one plate, and an opposing current (charge) flows out of the other plate, but the current (except for leakage current) does not flow across the dielectric. The result is that there is a charge differential between the plates.
Why does the ratio of the voltage to current in capacitor and inductor depend on frequence?
The ratio of voltage to current, or the impedance, of reactive elements such as capacitors and inductors depends on the frequency of the applied wave because they store energy, and the amount of energy they store is directly related to the frequency of the applied waveform. When a DC voltage is applied to a capacitor, the current through the capacitor initially will be large, and will decay down to zero as the capacitor charges. Also, the voltage across the capacitor will be small initially and will increase over time to be equal to the applied voltage. This behavior results in a varying impedance when an AC waveform is applied. At a very low frequency, the capacitor will charge up and discharge similarly to if a DC source was switched into the capacitor for a long period of time there would be a large voltage drop, and small current = high impedance). As the frequency increases, the capacitor will appear more like a DC source was initially switched into the capacitor (low voltage drop and high current = low impedance).
WHY does the ratio of the voltage to current in capacitor and inductor depend on frequency?
The ratio of voltage to current, or the impedance, of reactive elements such as capacitors and inductors depends on the frequency of the applied wave because they store energy, and the amount of energy they store is directly related to the frequency of the applied waveform. When a DC voltage is applied to a capacitor, the current through the capacitor initially will be large, and will decay down to zero as the capacitor charges. Also, the voltage across the capacitor will be small initially and will increase over time to be equal to the applied voltage. This behavior results in a varying impedance when an AC waveform is applied. At a very low frequency, the capacitor will charge up and discharge similarly to if a DC source was switched into the capacitor for a long period of time there would be a large voltage drop, and small current = high impedance). As the frequency increases, the capacitor will appear more like a DC source was initially switched into the capacitor (low voltage drop and high current = low impedance).
What to expect if a resistor is placed in a dc circuit with a capacitor?
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"
Why capacitor behave as open circuit against alternating current?
Capacitors store electrical charge. Imagine we have a capacitor. At time 0 seconds we connect a DC voltage across the capacitor - immediately as the voltage is connected the capacitor is at 0 volts and the maximum current (relative to the circuit resistance) flows. At this extreme the capacitor can be treated as a short circuit, so for high frequency AC volts we should treat a capacitor as being a short circuit. As time passes the current in the circuit will go down and the voltage of the capacitor will go up - this is because as the capacitor gains more charge it gains more voltage, lowering the voltage across any resistance in the circuit consequently lowering the current in the circuit. When the capacitor is virtually full no current will flow at all and the voltage across the capacitor will equal the DC source voltage. At this extreme the capacitor can be treated as an open circuit, so for low frequency AC (allowing the capacitor to fill up before the current alternates) we can treat the capacitor as being an open circuit. Technically, it is not an open/closed circuit when it comes to AC because the capacitance will results in a signal lag or lead. However, if the frequency is low/high enough the lag/lead is often negligable.
What are the ways to stop lagging current?
use more impedense load and use suitable capacitor bank with the load.
How do you use a capacitor as a power saver?
You cannot use a capacitor as a 'power saver' or, more accurately, 'energy saver'! A capacitor may improve the power-factor of a load, and this may reduce the value of its load current, but this does not reduce the energy consumed by the load. For a residence, a so-called 'power save' capacitor is nothing more than a rip-off.
What s the role of capacitor in a single phase induction motor?
to start the motor and give more current to motor to run
What is difference among resistor capacitor and inductor?
In an AC circuit through a resistor the voltage and current are said to be in phase. Ie on the oscilloscope in the sine wave as the voltage rises so does the current in tandem. Through an inductor an electromagnetic field is created in the coil which produces a back emf which acts against any change in voltage. This slows down any sharp change in voltage as when a circuit is energised. With an inductor the voltage and current are out of phase. The voltage is said to lead over the current. It is displaced by 90 degrees. It is ahead of the current by 90 degrees on the sine wave. Through a capacitor the current rises until it is fully charged and then no more current flows. An AC current is constantly changing direction at 50 Hz in the UK. Through a capacitor the current leads over the voltage by 90 degrees in the sine wave. A capacitor can balance out the effect of the inductor. The inductor decreases the power factor, the capacitor increases the power factor. With an inductor a circuit draws more current, with a capacitor the current draw decreases. P=IxVxPf. Where Pf is the power factor or power correction factor. Factories will have a big capacitor to correct the power factor for induction motors. This improves the electrical efficiency and lowers the electric bill!
Why does the run capacitor have a lower MFD rating than the start capacitor?
The run capacitor has to absorb the VARs while the motor is running, but the start capacitor has to provide running current to start the motor. The latter is higher, so more microfarads are needed to pass the greater current. Current in a capacitor is 2pi x frequency x capacitance x voltage so, on a 240 v 50 Hz system, 100 mfd would take 2pi x 50 x 100 x 10-6 x 240 amps, and that is multiplied by the voltage 240 to find the VARs.
What is the advantage of the capacitor-start induction-run motor over the resistance-start induction-run motor?
A capacitor-start system will limit the current drawn by an induction machine by correcting the devices poor power factor during the start up. This limits the current needed to start the device without using any real power. A resistance-start system will limit the starting current by drawing current unto the resistor instead of the induction machine. The real power used by the resistor is turned into heat and lost. A capacitive-start system will use no real power and will not create excess heat, it will be more expensive to initially purchase.
How do you wire a capacitor to a electric motor?
A capacitor resists a change of voltage, proportional to current, and inversely proportional to capacitance. dv/dt = i/c Capacitors can do various things. They can filter out high frequency transients, or power supply ripples, they can block DC while passing AC, they can participate with other components such as resistors and inductors to form filters - the list is endless.
Does current actually flow through a capacitor?
No. Current (or more specifically, charge) flows into one plate, and an opposing current (charge) flows out of the other plate, but the current (except for leakage current) does not flow across the dielectric. The result is that there is a charge differential between the plates.
