0
Why does the current lead the voltage across a capacitor by 90 degree rather than lag it?
Wiki User
∙ 12y ago
A: Because a capacitor have to have time to charge to the voltage
In a capacitor, the current depends on the voltage difference across it. On AC, this makes it charge, if the voltage is increasing above zero, and discharge if the voltage is reducing towards zero.
Because a capacitor has almost no internal resistance, and most loads that it is connected to have only very small resistances in series with the capacitor, the charging and discharging currents depend pretty much on the rate at which the voltage is changing. At the zero crossing point of the sine-wave, when the voltage is actually zero, the rate of change of voltage is very high (the sine-wave is at its steepest), so the current is also very high. If the voltage is positive-going, the current is positive, and if the voltage is negative-going, the current is negative.
At the peak of the voltage waveform, the rate of change of voltage is zero or very low (the sine-wave is flat, and not really changing its voltage) so the current is zero, too.
Since the maximum positive current occurs when the voltage is passing through zero, going positive, and the maximum negative current happens when the voltage is passing through zero, going negative, the current peaks happen 90 degrees before the voltage peaks, so the current is said to lead the voltage. This is the same as saying the voltage lags the current by 90 degrees.
Wiki User
∙ 13y ago
Wiki User
∙ 11y agoDifficult to explain without a diagram. However, the voltage induced into a conductor is maximum when the rate of change of current is greatest. This voltage opposes the change in current, so acts in the opposite sense to the current. Since the supply voltage is equal and opposite the induced voltage, it acts in the positive sense when the current is increasing and passing through zero -therefore, the supply voltage 'peaks' 90 degrees ahead of the current.
Wiki User
∙ 12y agocapacitor is storing energy in the form of electrostatic which requires charges. Ultimately the flow of current is nothing but the charges' movement. If and only if the charges are accumulated/ stored, there will be a voltage build up inside the capacitor. Hence the current leads the voltage in capacitor.
Wiki User
∙ 12y agoBecause the capacitor resists a change in voltage, while an inductor resists a change in current. This causes current to lead voltage in an AC circuit with a capacitor, while current would lag voltage in an AC circuit with an inductor.
Wiki User
∙ 10y agoBecause you have to put some charging current thru a capacitor before you get any voltage across it. Current times time equals charge, voltage equals capacitance times charge.
Wiki User
∙ 12y agobecause capacitor opposes change involtage and allows the current to flow whereas in inductor current lags because it opposes the change in current
Add your answer:
Earn +20 pts
How can a capacitor reduce the effects of ripple voltage?
The smoothing capacitor converts the full-wave rippled output of the rectifier (which is left over AC signal) into a smooth DC output voltage A smoothing capacitor after either a half-wave or full-wave rectifier will be charged up to the peak of the rectified a.c. Between peaks of the a.c. the stored voltage will drop by a degree dependent on how much current is drawn from it by the load. The larger the value of the capacitor, the less drop there will be, and therefore less ripple when loaded.
How can a capacitor raise the voltage level?
A capacitor alone can not raise the voltage level.Capacitors together with a multivibrator (Or an AC source) and some diodes can do the trick. Works well on higher voltages.Capacitors together with a multivibrator and an inductor would be the preferred method for low voltages. It works well even on voltages as low as 0.8 Volt.Capacitors and diodes are by far the easiest available components.Have a look at the related link down below for more information on how to raise the voltage.
In a pure inductance circuit the current lags the voltage by what degree?
90 degrees
What effect capacitor has on alternating current ac?
In the basic configuration, a capacitor is constructed with two parallel conductor plates with a layer of insulating material in between. When the cap is hooked up to the AC power supply, the voltage (v) across the plates and the charge (q) induced on the plates follow this capacitance expression: C = dq/dv or i = C dv/dt, where C is determined by the properties of the insulating material and the geometry of the cap (in the case of the parallel plates, the separation between the two electrodes (t). For the parallel plates, C can be written as (dielectric constant * plate area / t). Electrically, the change in the charge induced on the plates (dq), is directly related to the change in voltage difference (dv) between the two plates, since C is a constant. Theoretically, no energy is lost by charging and discharging the cap with an AC current. When the cap absorbs electrical energy from the power supply, it stores the energy in the electric field in the insulator. When discharging, the cap gives the stored energy back to the circuit -- hence, no energy loss. In a circuit, we use the cap to prolong/smoothen/resist any voltage change in time or to absorb a sudden energy surge (electrostatic discharge and power-line glitches, for example).
Why capacitor make 90 degree phase shift between voltage and current?
To answer this we assume that the current in a passive component can be written as: i(t) = I*cos(wt +phi), where I is the constant current amplitude for a resistor: V=IR, v(t) = A*R*cos(wt+phi) thus, V = I*R angle(phi) for a capacitor: i(t) = C*(dv/dt) v(t) = V*cos(wt +phi) dv/dt = V*w*sin(wt +phi) therefore: i(t) = wCV*sin(wt + phi) v(t) = V*cos(wt +phi) from this it is clear that the current in a capacitor is 90degrees out of phase (sin->cos 90degrees difference) and that the current amplitude is dependent on capacitance value and frequency (w=2*pi*f). remember because capacitors are not ideal the 90degree phase shift will vary and be dependent on paracitic elements such as parallel/series resistance and series inductance.
Relation between voltage and current if you consider capacitor ckt?
In a capacitor ckt, current will be lead ahead from voltage by an angle 90 degree. Because for a capacitor the relationship between voltage and current is given as v=(jx)i , where v= voltage i= current jx=capacitive reactance
What determine the flow of charge through a conductor?
Charges may appear to flow through a capacitor, although in reality they don't.The degree to which charge appears to flow through a capacitor depends on therate at which the voltage across it changes.-- DC voltage doesn't change, so it doesn't appear to pass through a capacitor at all.-- AC voltage is always changing, and the higher its frequency, the more currentit appears to push through a capacitor.
If the voltage applied across a capacitance is triangular in waveform then the waveform of the current?
Depend the value of capacitor. Capacitance in series act like a high pass filter, while in parallel act like low pass filter. By fourier series, triangular wave is combine of series of the sine or cosine waves. Therefore by certain capacitance, sine wave can preduce by applied a triangular signal through a capacitor. Current is just 90 degree shift from voltage, shape is same.
How can a capacitor reduce the effects of ripple voltage?
The smoothing capacitor converts the full-wave rippled output of the rectifier (which is left over AC signal) into a smooth DC output voltage A smoothing capacitor after either a half-wave or full-wave rectifier will be charged up to the peak of the rectified a.c. Between peaks of the a.c. the stored voltage will drop by a degree dependent on how much current is drawn from it by the load. The larger the value of the capacitor, the less drop there will be, and therefore less ripple when loaded.
Does a resister use current?
It may be better to say that a resistor allows current flow through itself rather than to say that a resistor is a device that will "use" current. It does "resist" current flow, and thus limits it to some degree depending on its resistance. (More resistance means more limiting of current flow.) The resistor "drops voltage" as well limits current. A resistor "feels voltage" from some source, and the voltage it "feels" is said to be the "voltage drop" of the resistor. The voltage drop is the voltage that could be measured across that resistor with a meter.
What is lead and lag angles?
Leading angle means that the current lead voltage by 90 degree,which implies a capacitve load. while,lagging angle mean when the current lag the voltage by 90 degree or when the voltage lead the current by 90 degree.which implies an inductive load.
How can a capacitor raise the voltage level?
A capacitor alone can not raise the voltage level.Capacitors together with a multivibrator (Or an AC source) and some diodes can do the trick. Works well on higher voltages.Capacitors together with a multivibrator and an inductor would be the preferred method for low voltages. It works well even on voltages as low as 0.8 Volt.Capacitors and diodes are by far the easiest available components.Have a look at the related link down below for more information on how to raise the voltage.
In a pure inductance circuit the current lags the voltage by what degree?
90 degrees
Why must the instantaneous current wave be exact 90 degree out-of phase with the applied voltage waveform across an ideal inductor?
Because the voltage induced is proportional to the rate of change of current, and the maximum rate of change of current occurs at the point where the current waveform is 'steepest' -i.e. as it passes through zero. So, as the current passes through zero, the corresponding value of induced voltage is maximum, which means the voltage and current waveforms are displaced by a quarter of the wavelength, or 90 degrees.
What effect capacitor has on alternating current ac?
In the basic configuration, a capacitor is constructed with two parallel conductor plates with a layer of insulating material in between. When the cap is hooked up to the AC power supply, the voltage (v) across the plates and the charge (q) induced on the plates follow this capacitance expression: C = dq/dv or i = C dv/dt, where C is determined by the properties of the insulating material and the geometry of the cap (in the case of the parallel plates, the separation between the two electrodes (t). For the parallel plates, C can be written as (dielectric constant * plate area / t). Electrically, the change in the charge induced on the plates (dq), is directly related to the change in voltage difference (dv) between the two plates, since C is a constant. Theoretically, no energy is lost by charging and discharging the cap with an AC current. When the cap absorbs electrical energy from the power supply, it stores the energy in the electric field in the insulator. When discharging, the cap gives the stored energy back to the circuit -- hence, no energy loss. In a circuit, we use the cap to prolong/smoothen/resist any voltage change in time or to absorb a sudden energy surge (electrostatic discharge and power-line glitches, for example).
Why capacitor make 90 degree phase shift between voltage and current?
To answer this we assume that the current in a passive component can be written as: i(t) = I*cos(wt +phi), where I is the constant current amplitude for a resistor: V=IR, v(t) = A*R*cos(wt+phi) thus, V = I*R angle(phi) for a capacitor: i(t) = C*(dv/dt) v(t) = V*cos(wt +phi) dv/dt = V*w*sin(wt +phi) therefore: i(t) = wCV*sin(wt + phi) v(t) = V*cos(wt +phi) from this it is clear that the current in a capacitor is 90degrees out of phase (sin->cos 90degrees difference) and that the current amplitude is dependent on capacitance value and frequency (w=2*pi*f). remember because capacitors are not ideal the 90degree phase shift will vary and be dependent on paracitic elements such as parallel/series resistance and series inductance.
Voltage current relation in passive circuit?
in passive circuit it depends on the type of load 1. if the load is purely resistive the voltage and current will be in phase 2.if the load is purely inductive the current lags the voltage by 90 dgree 3.if the load is purely capacitive the currents leads the voltage by 90 degree
