Skip to main content
Engineering LibreTexts

8.6: Initial and Steady-State Analysis of RLC Circuits

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    When analyzing resistor-inductor-capacitor circuits, remember that capacitor voltage cannot change instantaneously, thus, initially, capacitors behave as a short circuit. Once the capacitor has been charged and is in a steady-state condition, it behaves like an open. This is opposite of the inductor. As we have seen, initially an inductor behaves like an open, but once steady-state is reached, it behaves like a short. For example, in the circuit of Figure 9.4.1 , initially \(L\) is open and \(C\) is a short, leaving us with \(R_1\) and \(R_2\) in series with the source, \(E\). At steady-state, \(L\) shorts out both \(C\) and \(R_2\), leaving all of \(E\) to drop across \(R_1\). For improved accuracy, replace the inductor with an ideal inductance in series with the corresponding \(R_{coil}\) value. Similarly, practical capacitors can be thought of as an ideal capacitance in parallel with a very large (leakage) resistance.


    Figure 9.4.1 : Basic RLC circuit.

    Example 9.4.1

    Assuming the initial current through the inductor is zero and the capacitor is uncharged in the circuit of Figure 9.4.2 , determine the current through the 2 k\(\Omega\) resistor when power is applied and after the circuit has reached steady-state. Draw each of the equivalent circuits.


    Figure 9.4.2 : Circuit for Example 9.4.1 .

    For the initial-state equivalent we open the inductor and short the capacitor. The new equivalent is shown in Figure 9.4.3 . The shorted capacitor removes everything to its right from the circuit. All that's left is the source and the 2 k\(\Omega\) resistor.


    Figure 9.4.3 : Initial-state equivalent of the circuit of Figure 9.4.2 .

    We can find the current through the 2 k\(\Omega\) resistor using Ohm's law.

    \[I_{2k} = \frac{E}{R} \nonumber \]

    \[I_{2k} = \frac{14 V}{2k \Omega} \nonumber \]

    \[I_{2k} = 7 mA \nonumber \]

    Steady-state is redrawn in Figure 9.4.4 , using a short in place of the inductor, and an open for the capacitor. We are left with a resistance of 2 k\(\Omega\) in series with the parallel combination of 1 k\(\Omega\) and 4 k\(\Omega\), or 2.8 k\(\Omega\) in total.


    Figure 9.4.4 : Steady-state equivalent of the circuit of Figure 9.4.2 .

    \[I_{2k} = \frac{E}{R} \nonumber \]

    \[I_{2k} = \frac{14 V}{2.8k \Omega} \nonumber \]

    \[I_{2k} = 5mA \nonumber \]

    8.6: Initial and Steady-State Analysis of RLC Circuits is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by James M. Fiore.

    • Was this article helpful?