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Can anyone elaborate for me: is a capacitor an ohmic material or not? As we know that when the voltage and current graph is linear the material is said to be ohmic. Now consider when DC is applied to a capacitor offer infinite resistance and thus it obeys ohms law, but when AC applied across it, the graph of Voltage vs Current does not pass through the origin. So, am I right? When AC applied across capacitor it isn't an ohmic material.

M. Enns
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user129291
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3 Answers3

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when the Voltage and Current graph is linear the material is said to be ohmic

Yes, if there is a linear relationship between these parameters, it is called Ohmic. But a capacitor doesn't behave like this - it changes it's "resistance" with time.

For example for DC current:

  • When you first turn on the circuit, charge flows to one plate, and this "pushes" charge on the other plate away (it induces a negative charge on the opposite plate). It looks from the outside as if there is no capacitor there but just a straight wire.
  • After some time, charge starts to build up on the plate, which prevents more charge from arriving, so now it looks like a "resistance" from the outside.
  • After a long time, the capacitor is "full" and no more charge flows to the plate. It looks like an open circuit - as a hole in the circuit - from the outside, which would correspond to infinite resistance, if there was an Ohmic relationship.

See this answer to a similar question, which explains these basic features of a capacitor as well.

Community
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Steeven
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As stated by Steeven, the behaviour of a capacitor is more complex. I'll try to give a slightly different answer. In the DC case there are two options, the equilibrium and non-equilibrium case. In the non-equilibrium case, the standard example is the charging or discharging via a resistor. As in the link provided by Steeven, this takes place with an exponential law, so it is non-linear. In the equilibrium case, no current flows. You might consider this as linear as the current is a "linear" function of the voltage, namely $I=0\ \Omega^{-1} \times V$, but this is not very useful as you cannot invert it, i.e. expressing the voltage as function of current does not make sense.

In the AC case there are two ways of looking at it, time resolved and on average. Time resolved there is a phase shift between voltage and current, so if one of them is zero, the other is not. On average, however, the behaviour is linear. I.e., looking at the root-mean-square values $I_\mathrm{RMS}\propto V_\mathrm{RMS}$. This is why (as Steeven pointed out) one use the term impedance here rather than resistance.

Eventually, all of this is wrong in the real world case as there is a leakage current and many more effects taking place. Hence the equivalent circuit model of a capacitor.

mikuszefski
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But isnt V(across capacitor)=Xc I(Current through capacitor)?

Xc= 1/(2(pi)fC) which is constant at a fixed frequency and capacitance. Ohm's law is being obey at a point in time.

Delisle Warner
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