# ESR of capacitor.

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#### rainman1

##### New Member
Does tan(lamda) equal [ (1/jwC) / ESR ], where ESR is the ohmic resistance of the capacitor?

According to datasheet, tan(lamda) = 0.25 @ 120Hz,
And it turns out that R = 1 / (2pi*120Hz*10uF*0.25) = 530ohm
And it seems too large to me.

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#### crutschow

##### Well-Known Member
It should be Tan(δ) = ESR/(1/jwC).

Thus the ESR = 33.2Ω at 120Hz

#### picasm

##### Member
According the the web page below, tan δ is the "dissipation factor" and in order to calculate the ESR you need to multiply it by the reactance value in ohms
reactance = 1/2πfC Capacitor ESR, Dissipation Factor, Loss Tangent and Q :: Radio-Electronics.Com

The esr for an average 10uF capacitor should be less than 2 ohms. (ESR is usually measured at 100Khz)

The tan δ is related to esr: lower=better

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#### Papabravo

##### Well-Known Member
ESR is just the real part of the complex impedance WHICH IS A FUNCTION OF FREQUENCY. As such, no single value will do. To actually see it you should try to get your hands on a network analyzer. If you can't afford one maybe a friend will let you borrow one -- LOL.

Ponder the following. There is a frequency at which the complex impedance of a capacitor has no imaginary part and it looks like a pure resistance.

Even stranger. There is a frequency at which the capacitor behaves like an inductor. It's complex impedance has a positive imaginary part.

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