Why current leads in capacitor

In summary: =\frac{j}{\omega}=\frac{-1}{\omega}=\frac{j}{\omega^3}=\frac{-j^3}{\omega^3}=\frac{-10}{\omega^3}=\frac{1}{10}=\frac{1}{10^3}=\frac{1}{10^6}=\frac{1}{10^9}=\frac{1}{100}
  • #1
mohdfasieh
26
0
HEllo ,

Can anybody answer my question;


" i know that VOltage leads in Inductor by 90 as compared to current.But i want to know WHY?"

Why voltage leads in INDUCTOR




" I know CURRENT leads in CAPACITOR as compare to VOLTAGE but i want to know why ?"

Why current leads in capacitor
 
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  • #2
mohdfasieh said:
HEllo ,

Can anybody answer my question;


" i know that VOltage leads in Inductor by 90 as compared to current.But i want to know WHY?"

Why voltage leads in INDUCTOR

V = L di/dt is the basic V ~ I equation for an inductor. As you put current into an inductor, the change in current generates a voltage. A DC current into an inductor generates no net DC voltage (assuming that the parasitic DC resistance of the inductor windings is negligible).


" I know CURRENT leads in CAPACITOR as compare to VOLTAGE but i want to know why ?"

Why current leads in capacitor

I = C dV/dt is the basic V ~ I equation for a capacitor. As you push more charge onto the plates of a capacitor with the current, that changes the voltage between the plates.

These simplistic (and very useful day-to-day) equations come from Maxwell's equations in E&M. If you'd like to understand their origins better, I'd suggest that you study Maxwell's equations more, and how the simplified versions of those turn into Faraday's Law and Ampere's Law, etc. Just google some of those terms and have some fun... :biggrin:
 
  • #3
In addition to what berkeman wrote:
If the current in an inductor is [tex]i(t) = I sin(\omega.t)[/tex], the voltage will be:
[tex]v(t) = L.\frac{di}{dt} = \omega.L.I cos(\omega.t)[/tex].
Since a cosine wave is a sine wave with a lead of [tex]90^o[/tex] you have your voltage lead.
 
  • #4
This is my reasoning for this affect.

In an inductor the voltage will change 'instantly' however, when the voltage changes it creats a changing magnetic field. This field will inhibit the current from flowing freely. This delay in current is by 90 degrees as shown above.

In a capacitor the current will flow instantly, however the charge will not build up on capacitor's plates instantly. This delay in building the charge is what causes the 90 degree delay in voltage, as shown above.
 
  • #5
A quick mathematical "proof" (which also derives the impedences of these elements) is to apply Laplace transforms to the relations expressed by the capacitor current and inductor voltage, and then substitute s = jw. Multiplying by j causes a forward rotation of 90 of the phasor, dividing by j causes a reverse rotation of the phasor by 90 degrees.

E.g.--

[tex]i(t)=C\frac{dv(t)}{dt}[/tex]

Apply Laplace, use the derivative transform, and take initial conditions to be zero:

[tex]L[i(t)]=I(s)=sC V(s)[/tex]

Set [tex]s=j\omega[/tex]:

[tex]I(j\omega)=j\omega C V(j\omega)[/tex]

Since [tex]j[/tex] is mulplying the voltage phasor the current phasor leads it by [tex]90^o[/tex]

The impedence is *by definition*:

[tex]Z_{C}(j\omega)=\frac{V(j\omega)}{I(j\omega)}=\frac{1}{j\omega C}=\frac{-j}{\omega C}[/tex]
 
Last edited:

1. Why do current leads occur in capacitors?

The presence of current leads in capacitors is due to the flow of electric current between the plates of the capacitor. This flow of current is caused by the difference in charge between the plates, which creates an electric field that allows for the flow of electrons.

2. Can current leads be harmful to capacitors?

In some cases, current leads can be harmful to capacitors. When the flow of current is too high, it can cause the capacitor to overheat and potentially damage its components. This is why it is important to select capacitors with appropriate current ratings for the intended application.

3. How can current leads be minimized in capacitors?

To minimize current leads in capacitors, it is important to select capacitors with low equivalent series resistance (ESR). ESR is the measure of the internal resistance of a capacitor and plays a significant role in reducing current leads. Additionally, using capacitors with higher capacitance values can also help to reduce current leads.

4. Are current leads the same as leakage current in capacitors?

No, current leads and leakage current are two different phenomena in capacitors. Current leads refer to the flow of current between the plates of the capacitor, while leakage current is the small amount of current that flows through the dielectric material of the capacitor. Leakage current can occur even when there is no voltage applied to the capacitor.

5. How do current leads affect the performance of capacitors?

Current leads can have both positive and negative effects on the performance of capacitors. On one hand, they can cause the capacitor to heat up and potentially damage its components. On the other hand, they can also improve the overall performance of the capacitor by reducing the effects of parasitic elements such as inductance and resistance.

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