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Im confused with this fact. In books I've seen that when DC current is passed in LCR circuit it behaves as purely resistive circuit.And the impedance in LCR is given by $$Z=\sqrt{R^2+(\omega L-1/\omega C)^2}$$ So if ω (frequency) is taken as 0 $$\frac{1}{\omega C}$$ term becomes infinity then the impedance should also be infinite. Then no current should flow through circuit.

Please help me clear this confusion.

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  • \$\begingroup\$ What is the layout of the LCR circuit? \$\endgroup\$ – Swarles Barkely Feb 20 '16 at 6:29
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    \$\begingroup\$ Means ? @SwarlesBarkely Its a series LCR \$\endgroup\$ – user72436 Feb 20 '16 at 6:32
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    \$\begingroup\$ Your equation gives the steady state frequency response of a series LCR circuit. Steady state means that all the transients have decayed to zero; it does not consider the initial period when the sine wave is first applied and the capacitor and inductor are sorting out their balance condition. 'DC' does mean zero frequency and in this case, when the transient has disappeared, the capacitor has infinite reactance and looks like an open circuit. \$\endgroup\$ – Chu Feb 20 '16 at 8:20
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Yes, a capacitor will always present an open circuit (infinite resistance/reactance) to a dc current.

I domwonder, however, what application are you considering which would involve an RLC network & DC current?

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  • \$\begingroup\$ Is it okay to consider that when DC is passed through LCR circuit V=IR where R is the resistance of resistor and I is the current? \$\endgroup\$ – user72436 Feb 20 '16 at 7:01
  • \$\begingroup\$ DC can pass through an RLC circuit along a parallel path to the capacitor. Once transie ts have settled & 'true' dc current is flowing, the capacitor will present near-infinite impedance, the inductor will present near-zero impedance (only its parasitic resistance remains), and the value of R will remain basically unchanged. So, in a network where L & R are in series, parallel to C, I=V/R. However, if there is an available path through the inductor only, then you'll have to calculate for parallel resistances with the inductor having very low DCR. \$\endgroup\$ – Robherc KV5ROB Feb 20 '16 at 7:07
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First, to answer the topic question, the frequency of DC is indeed 0. As for the rest, we can simulate it to see the transient response (see Flash's answer). The response will vary based on the values of R, L, and C.

enter image description here

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  • \$\begingroup\$ Is it okay to consider that when DC is passed through LCR circuit V=IR where R is the resistance of resistor and I is the current? \$\endgroup\$ – user72436 Feb 20 '16 at 6:57
  • \$\begingroup\$ If by V you mean the voltage across the resistor, then yes it would just be Ohm's Law \$\endgroup\$ – Swarles Barkely Feb 20 '16 at 7:05
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An uncharged capacitor looks like a short circuit, so Instantaneous current will be high at first then taper off as it charges. but in series with an inductor and a resistor, this changes. I think if DC is applied current will oscillate and dampen to zero.

The only time an LCR circuit is purely resistive is at it's resonant frequency. A DC current will not pass, once the oscillations have dampened.

All of this assumes a series LCR circuit. In a parallel circuit, the inductor will pass DC after a short rise time.

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  • \$\begingroup\$ Is it okay to consider that when DC is passed through LCR circuit V=IR where R is the resistance of resistor and I is the current? \$\endgroup\$ – user72436 Feb 20 '16 at 7:04
  • \$\begingroup\$ Yes, IF it is a parallel circuit. In a series circuit, the current will vary as in the graph posted by S. Barkley \$\endgroup\$ – Flash Feb 20 '16 at 7:15
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Your equation gives the steady state frequency response of a series LCR circuit. Steady state means that all the transients have decayed to zero; it does not consider the initial period when the sine wave is first applied and the capacitor and inductor are sorting out their balance condition.

'DC' does mean zero frequency and in this case, when the transient has disappeared, the capacitor has infinite reactance and looks like an open circuit. Also, at DC, the inductor has zero reactance and presents as a short circuit.

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I notice you have multiple answers, and for each one you've asked again 'Is it okay to consider that when DC is passed through LCR circuit V=IR where R is the resitance of the resistor and I is the current.'

The reason you haven't got that question answered is that you haven't supplied a circuit diagram of the RLC circuit you mean, despite being asked for that in comments to your question.

Consider the following 3 RLC circuits

schematic

simulate this circuit – Schematic created using CircuitLab

They all have different assymptotic behaviour at \$\omega\$=0.

Circuit 1, which matches the equation you have put in your question, goes to infinite impedance at \$\omega\$=0, because of the capacitor term you correctly identified. Interestingly, it also goes to infinite impedance at \$\omega\$=\$\infty\$. It goes to impedance R at the LC resonance, when the L and C terms cancel each other out.

Circuit 2 has a different impedance equation, because it's a different topology. That goes to infinity at the LC resonance. At zero and infinity frequency, either the L or C is a short circuit, and then, indeed, the DC current passed through the LCR circuit V=IR where R is the resistance of the resistor and I is the current.

Circuit 3 is an all parallel arrangement, which has a yet different impedance equation. This goes to R at the LC resonance frequency, and to zero at either zero or infinite frequency.

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