#52 Transient Analysis #52 — Second-order RLC problem solving

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Build the second-order problem workflow

   

   1) Choose the response variable and circuit form

   2) Compute α and ω₀; classify the poles

   3) If a source is present, find x(∞) first

   4) Write the correct natural or full-response form

   5) Use continuity to obtain x(0⁺)

   6) Use KCL or KVL to obtain x′(0⁺)

   7) Solve two constants; verify initial and final values

   Narration transcript

   For second-order problems, the workflow is short and reliable. First, identify the response variable and the circuit form. Second, compute alpha and omega zero to classify the case. Third, use two initial conditions to determine the constants. If a source is present, first find the final value, then add the transient shape around it.

2. 2. Problem 1: source-free parallel RLC

   

   Given: R=1 Ω, L=0.5 H, C=0.5 F

   $v_{\mathrm{C}}(0)=6 V$

   $i_{\mathrm{L}}(0)=0 A$

   Find: v_C(t), t≥0

   No source → the response is entirely transient

   $v_{\mathrm{C}}(\infty )=0$

   The natural variable in the parallel form is v_C(t)

   Narration transcript

   Problem one is a source-free parallel R L C circuit. We are given R equals one ohm, L equals zero point five henry, and C equals zero point five farad. The initial capacitor voltage is six volts, and the initial inductor current is zero. We want the capacitor voltage for t greater than or equal to zero.

3. 3. Solve Problem 1

   

   α=1/(2RC)=1 s⁻¹

   ω₀=1/√(LC)=2 rad/s

   α<ω₀ → underdamped; ω_d=√3

   $v=e^{-t}[A\cos (\surd 3t)+B\sin (\surd 3t)]$

   $v(0)=6 \to A=6$

   KCL: C v′(0)+v(0)/R+i_L(0)=0

   $v′(0)=-12 V/s \to B=-2\surd 3$

   $v_{\mathrm{C}}=e^{-t}[6\cos (\surd 3t)-2\surd 3\sin (\surd 3t)] V$

   Narration transcript

   This is a natural response problem, so the final value is zero. For the parallel form, alpha equals one over two R C, which gives one. Omega zero equals one over square root of L C, which gives two. Since alpha is smaller than omega zero, the response is underdamped. So we write v of t as e to the minus t times a cosine sine combination with omega d equal to square root of three. The first condition gives A equals six. For the derivative, use K C L at t equals zero plus: C d v over d t plus v over R plus i sub L equals zero. Substituting the initial values gives d v over d t at zero plus equals minus twelve. That leads to B equals minus two root three. So the capacitor voltage is e to the minus t times open bracket six cosine root three t minus two root three sine root three t close bracket.

4. 4. Problem 2: step-driven series RLC

   

   An 8 V source is connected at t=0

   $R=4 \Omega, L=1 H, C=0.25 F$

   v_C(0)=0 V and i(0)=0 A

   Find: v_C(t), t≥0

   A source is present → use the full response

   At DC steady state, C is open

   $v_{\mathrm{C}}(\infty )=8 V$

   Narration transcript

   Problem two is a switched series R L C step response. At t equals zero, the switch closes and an eight volt source drives a series circuit with R equals four ohms, L equals one henry, and C equals zero point two five farad. Initially, the capacitor voltage and loop current are both zero. We want v sub C of t.

5. 5. Solve Problem 2

   

   $\alpha=R/(2L)=2 s⁻¹$

   ω₀=1/√(LC)=2 rad/s

   α=ω₀ → critically damped

   v_C=8+(A+Bt)e^−2t

   $v_{\mathrm{C}}(0)=0 \to A=-8$

   $i(0)=C v_{\mathrm{C}}′(0)=0 \to v_{\mathrm{C}}′(0)=0$

   $B-2A=0 \to B=-16$

   $v_{\mathrm{C}}=8-(8+16t)e^{-2t} V$

   Narration transcript

   Now the source is present, so we first find the final value. At steady state, the capacitor is open, the current becomes zero, and the capacitor voltage reaches the source voltage. So v sub C of infinity is eight volts. For the series circuit, alpha equals R over two L, which gives two. Omega zero also equals two, so the case is critically damped. The correct form is eight plus open bracket A plus B t close bracket e to the minus two t. Using v sub C of zero equals zero gives A equals minus eight. Since i equals C d v over d t and the initial current is zero, the initial slope of the capacitor voltage is also zero. That gives B minus two A equals zero, so B equals minus sixteen. Therefore v sub C of t equals eight minus open bracket eight plus sixteen t close bracket e to the minus two t.

6. 6. Unify both problems with one method

   

   Source-free: x(∞)=0 and the response is entirely transient

   Driven: find x(∞), then attach the transient around it

   α and ω₀ select the pole type

   Continuity preserves the first condition

   KCL/KVL supplies the second condition or derivative

   Two conditions → two unknown constants

   Final check: t=0⁺ and t→∞

   The same workflow works for parallel and series RLC

   Narration transcript

   These two examples show the whole pattern. In a source-free problem, the response is entirely transient. In a driven problem, find the final value first, then attach the correct transient form. In both cases, alpha and omega zero classify the poles, and two initial conditions determine the constants.