#46 Transient Analysis #46 — First-order problem solving

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

1. 1. Universal three-step method

   

   Organize every RC/RL problem with three values

   1) x(0⁺): initial value from continuity

   2) x(∞): final value from DC steady state

   3) τ: equivalent seen by the storage element

   Capacitor voltage cannot jump

   Inductor current cannot jump

   At DC, C is open and L is short

   RC: τ=R_eqC · RL: τ=L/R_eq

   $x(t)=x\infty +(x0-x\infty )e^{(}-t/\tau)$

   Narration transcript

   Welcome to our first order problem solving session. In the previous lessons, we learned how to analyze R C and R L circuits with step inputs. Today we will practice by solving four problems of increasing difficulty. Before we begin, let us recall the universal three step method. Step one: find x of zero, the initial value. This comes from the circuit conditions just before the switch operates. Step two: find x of infinity, the final value. This is the steady state value long after the switch. For a capacitor, that means no current flows, so the capacitor acts as an open circuit. For an inductor, no voltage drop, so it acts as a short circuit. Step three: find tau, the time constant. For R C circuits, tau equals R C. For R L circuits, tau equals L over R. Then the complete solution is: x of t equals x of infinity plus the quantity x of zero minus x of infinity, times e to the negative t over tau.

2. 2. Problem 1: RC with initial voltage

   

   $V_{\mathrm{S}}=20 V, R=5 k\Omega, C=2 \muF, v_{\mathrm{C}}(0⁻)=5 V$

   $v_{\mathrm{C}}(0⁺)=5 V$

   Steady-state current is zero

   $v_{\mathrm{C}}(\infty )=20 V$

   τ=RC=(5 kΩ)(2 μF)

   τ=10 ms

   Substitute into the general formula

   $v_{\mathrm{C}}(t)=20-15e^{(}-100t) V$

   Checks at t=0 and t→∞ pass

   Narration transcript

   Problem one. A simple R C step response. We have a voltage source V S equals twenty volts in series with a switch that closes at t equals zero, a resistor R equals five kilohms, and a capacitor C equals two microfarads. The initial capacitor voltage is v C of zero minus equals five volts. Find v C of t for t greater than or equal to zero. Step one: initial value. By the continuity principle, the capacitor voltage cannot change instantly. So v C of zero plus equals v C of zero minus equals five volts. Step two: final value. As t approaches infinity, the capacitor fully charges to the source voltage. No current flows, so v C of infinity equals twenty volts. Step three: time constant. Tau equals R C equals five times ten to the three, times two times ten to the negative six, which gives tau equals ten milliseconds. Now we plug into the formula: v C of t equals twenty plus five minus twenty times e to the negative t over zero point zero one. This simplifies to v C of t equals twenty minus fifteen e to the negative one hundred t volts. Let us verify: at t equals zero, v C equals twenty minus fifteen equals five volts. Correct. As t goes to infinity, the exponential vanishes and v C equals twenty volts. Also correct.

3. 3. Problem 2: RL with a Thevenin equivalent

   

   V_S=30 V, R₁=6 Ω, R₂=3 Ω, L=12 mH

   Remove the inductor and find Thevenin at its terminals

   $V_{\mathrm{T}}h=30\cdot 3/(6+3)=10 V$

   $R_{\mathrm{T}}h=6∥3=2 \Omega$

   $i_{\mathrm{L}}(0⁺)=0 A$

   $i_{\mathrm{L}}(\infty )=V_{\mathrm{T}}h/R_{\mathrm{T}}h=5 A$

   τ=L/R_Th=6 ms

   Exponential rate is about 166.67 s⁻¹

   i_L(t)=5(1−e⁽−t/6ms)) A

   Narration transcript

   Problem two. An R L circuit that requires Thevenin simplification. We have a thirty volt source, connected to R one equals six ohms in series, which then splits into two parallel branches. The first branch has R two equals three ohms. The second branch has an inductor L equals twelve millihenrys. The switch closes at t equals zero, and the initial inductor current is zero. Find i L of t. Before applying the three step method, we need the Thevenin equivalent seen by the inductor. The Thevenin voltage: V Th equals V S times R two over R one plus R two, equals thirty times three over nine, equals ten volts. The Thevenin resistance: R Th equals R one parallel R two equals six times three over six plus three, equals two ohms. Now the three steps. Step one: i L of zero plus equals zero amperes. Step two: i L of infinity equals V Th over R Th equals ten over two equals five amperes. Step three: tau equals L over R Th equals zero point zero one two over two equals six milliseconds. The solution: i L of t equals five plus zero minus five times e to the negative t over zero point zero zero six. This gives i L of t equals five times one minus e to the negative one hundred sixty seven t amperes. The key lesson here: when the circuit seen by the storage element has multiple resistors and sources, always find the Thevenin equivalent first.

4. 4. Problem 3: sequential switching

   

   $R₁=R₂=20 k\Omega, C=1 \muF, V_{\mathrm{S}}=10 V$

   Region 1: 0≤t<40 ms

   τ₁=20 ms; target 10 V

   v_C=10(1−e⁽−t/20ms))

   v_C(40 ms)=8.6466 V

   Region 2: t≥40 ms with t′=t−40 ms

   R_Th=10 kΩ, τ₂=10 ms, target 5 V

   Continuity gives the 8.6466 V initial value

   v_C=5+3.6466e⁽−t′/10ms) V

   Narration transcript

   Problem three. Sequential switching. This is the most interesting problem today. We have a ten volt source, R one equals twenty kilohms, R two equals twenty kilohms, and a capacitor C equals one microfarad. Switch one closes at t equals zero, connecting the source through R one to the capacitor. The capacitor starts at zero volts. Switch two closes at t equals forty milliseconds, adding R two in parallel with the capacitor. We must solve this in two regions. Region one: zero less than or equal to t less than forty milliseconds. Only R one and C are in the circuit. Tau one equals R one times C equals twenty times ten to the three times one times ten to the negative six, equals twenty milliseconds. v C of zero equals zero. v C of infinity equals ten volts. So v C of t equals ten times one minus e to the negative t over zero point zero two. At t equals forty milliseconds: v C of forty milliseconds equals ten times one minus e to the negative two, equals eight point six five volts. Region two: t greater than or equal to forty milliseconds. Now R two is in parallel with C. The Thevenin resistance becomes R one parallel R two equals ten kilohms. Tau two equals ten kilohms times one microfarad equals ten milliseconds. The Thevenin voltage is V S times R two over R one plus R two, equals ten times twenty over forty, equals five volts. Using t prime equals t minus zero point zero four: v C of t prime equals five plus eight point six five minus five times e to the negative t prime over zero point zero one. This simplifies to: v C equals five plus three point six five times e to the negative one hundred t prime volts. The capacitor was charging toward ten volts, but when R two joins, the new target drops to five volts, and the time constant halves.

5. 5. Problem 4: time to a target

   

   $V_{\mathrm{S}}=10 V, R=2 k\Omega, C=50 \muF, v_{\mathrm{C}}(0)=0$

   τ=RC=100 ms

   $v_{\mathrm{C}}(t)=10(1-e^{(}-10t))$

   Set v_C=8 V

   $0.8=1-e^{(}-10t)$

   $e^{(}-10t)=0.2$

   $-10t=\ln (0.2)=-1.609$

   t=0.16094 s≈161 ms

   $t\approx 1.61\tau$

   Narration transcript

   Problem four. Finding when a signal reaches a specific value. We have V S equals ten volts, R equals two kilohms, C equals fifty microfarads, and v C of zero equals zero. The switch closes at t equals zero. Question: how long does it take for the capacitor to reach eight volts? First, the time constant: tau equals R C equals two thousand times fifty times ten to the negative six, equals one hundred milliseconds. The response is: v C of t equals ten times one minus e to the negative ten t. We set this equal to eight: eight equals ten times one minus e to the negative ten t. Dividing both sides by ten: zero point eight equals one minus e to the negative ten t. Rearranging: e to the negative ten t equals zero point two. Taking the natural logarithm: negative ten t equals the natural log of zero point two, which equals negative one point six zero nine. Therefore t equals zero point one six one seconds, or one hundred sixty one milliseconds. This is approximately one point six one tau. So the capacitor reaches eighty percent of its final value in about one point six time constants. This is a very practical result. For example, in digital circuits, this tells you the minimum time before a signal is reliably read.

6. 6. First-order problem-solving summary

   

   Organize the circuit before using the formula

   Find initial, final, and tau separately

   Q1: RC step with continuity

   Q2: Thevenin first, then RL response

   Q3: a new region at every switch

   The switching value is the next initial value

   Q4: set the target and take a logarithm

   Always check t=0, t→∞, and units

   Next topic: second-order parallel RLC

   Narration transcript

   Let us summarize. We solved four first order circuit problems. Problem one was a basic R C step response. We applied the three step formula directly: initial value five volts, final value twenty volts, tau ten milliseconds. Problem two introduced Thevenin simplification. When the inductor sees multiple components, find V Th and R Th first, then apply the same formula. Problem three demonstrated sequential switching. The circuit changes at t equals forty milliseconds, creating two separate regions, each with its own time constant and final value. The key is to use the voltage at the switching moment as the initial condition for the next region. Problem four showed how to find when a voltage reaches a target value. Set the formula equal to the target and solve for t using the natural logarithm. In the next lesson, we begin second order circuits with the parallel R L C configuration.