ECE 211 · Lecture 8: Deriving the First-Order Response

§1 · Chapter 7 opens

One energy storage element changes everything

Every circuit considered so far has been purely resistive, and every response has therefore been instantaneous: a change at the source appeared at the output in the same instant. The introduction of a single energy storage element, either a capacitor or an inductor, replaces the algebraic problem with a differential one. Because one storage element contributes one derivative, the governing equation is of first order and requires exactly one initial condition.

1 energy storage device RL or RC 1st-order differential equation 1 initial condition

The response of such a circuit is the sum of two parts, and the vocabulary for them is worth fixing at the outset.

The two parts of the response

The transient response diminishes in time and carries the memory of the initial condition. The steady-state response is what remains after the transient has decayed; for the DC sources treated in this chapter, it is a constant. Energy is stored in the element during one part of the interval and supplied back to the circuit during another, which is precisely what the transient describes.

The resistive part is always reduced first

Whatever the surrounding network contains, it is linear and resistive, so it may be replaced by its equivalent as seen from the terminals of the storage element. A capacitor is most conveniently driven by a Thevenin equivalent, because the capacitor voltage is the natural variable and a voltage source states it directly. An inductor is most conveniently driven by a Norton equivalent, because the inductor current is the natural variable. Only two circuits therefore need to be solved in the whole chapter.

RC: the capacitor sees a Thevenin equivalent

RL: the inductor sees a Norton equivalent

§2 · Switches and the unit step function

The switching instant defines $t = 0$

A switched circuit has two topologies: one that has been in place long enough before the switching instant for all transients to have died out, and one that applies afterwards. The instant of the change is taken as $t = 0$ by convention, and the two symbols below distinguish the two possible actions.

A switch that closes at $t = 0$ is drawn open, and a switch that opens at $t = 0$ is drawn closed; in both cases the symbol shows the state that holds before the switching instant.

The unit step function

The same change may be written into the source itself rather than drawn as a mechanism. The unit step function is defined by

u(t) = \begin{cases} 0, & t < 0 \\ 1, & t > 0 \end{cases} \qquad\qquad u(-t) = \begin{cases} 1, & t < 0 \\ 0, & t > 0 \end{cases}

and its value at $t = 0$ itself is left undefined, which costs nothing because the circuit variables are evaluated at $0^-$ and $0^+$ rather than at $0$ . A source labeled $2u(t)$ V is therefore off before the switching instant and delivers 2 V afterwards, while $2u(-t)$ V does the opposite. Reversing the argument reverses time.

Any single-step waveform can be written as a constant plus a scaled step. The two sliders below set the constant $A$ and the step amplitude $B$ , and the tabs select the direction of the argument; the four cases from the board notes are available as presets.

5u(t) −5u(−t) 3 − 10u(t) 3 − 10u(−t)

A, the constant (V) 3

B, the step amplitude (V) −10

v_s(t) = 3 − 10u(−t) V ⟹ t < 0: −7 V t > 0: 3 V

With the argument $t$ , the step is off before the switching instant and on afterwards. The constant $A$ therefore sets the value that holds for $t < 0$ , and $A + B$ sets the value that holds for $t > 0$ . This is the natural form when a source is applied at $t = 0$ .

With the argument $-t$ , the step is on before the switching instant and off afterwards. The constant $A$ now sets the value that holds for $t > 0$ , so the final value is read directly from $A$ and the initial value from $A + B$ . This is the natural form when a source is removed at $t = 0$ , which is the case in the RC example of §7.

A step source is a switch and two ordinary sources

A source expression such as $3 - 10u(-t)$ V is not exotic hardware. It is realized by two constant sources and one single-pole double-throw switch, as the board notes show. The value before the switching instant is $3 - 10 = -7$ V and the value afterwards is 3 V, which fixes the polarity of the lower source.

Before $t = 0$ : the blade rests on the lower throw

The 7 V source is connected with its positive terminal at the reference node, so the port sees $v_s = -7$ V. This matches $3 - 10u(-t)$ evaluated for $t < 0$ , where $u(-t) = 1$ .

After $t = 0$ : the blade moves to the upper throw

The 3 V source is connected instead, so the port sees $v_s = 3$ V. This matches $t > 0$ , where $u(-t) = 0$ and only the constant term survives. Any source written as a constant plus a step admits the same two-source construction.

§3 · The governing differential equation

Two circuits, two equations, one structure

The reduced circuits are solved once and for all. The element laws $i_C = C\,dv_C/dt$ and $v_L = L\,di_L/dt$ are substituted into a single Kirchhoff equation, and the result is arranged with the unknown and its derivative on the left.

RC: one loop, so KVL is written

RL: one node pair, so KCL is written

KVL is applied around the single loop, and the resistor current is the capacitor current:

\tm{rc.vt}{-V_T} + \tm{rc.rt rc.ic}{R_T \cdot C \frac{dv_C}{dt}} + \tm{rc.vc}{v_C} = 0

Division by $R_TC$ puts the equation in standard form:

\frac{dv_C}{dt} + \frac{1}{R_TC}\,\tm{rc.vc}{v_C} = \frac{1}{R_TC}\,\tm{rc.vt}{V_T}

The coefficient of the unknown is the reciprocal of $\tau = R_TC$ , a quantity with units of seconds, since one ohm-farad is one second.

KCL is applied at the upper node, and the resistor voltage is the inductor voltage:

\tm{rl.in}{-I_N} + \tm{rl.rn rl.vl}{\frac{L}{R_N} \frac{di_L}{dt}} + \tm{rl.il}{i_L} = 0

Multiplication by $R_N/L$ puts the equation in standard form:

\frac{di_L}{dt} + \frac{R_N}{L}\,\tm{rl.il}{i_L} = \frac{R_N}{L}\,\tm{rl.in}{I_N}

The coefficient of the unknown is again the reciprocal of a time constant, here $\tau = L/R_N$ , since one henry per ohm is one second.

Why the initial condition belongs to the storage element

The capacitor voltage and the inductor current cannot change instantaneously, because a step in either would require infinite current or infinite voltage respectively. These two variables are therefore continuous across the switching instant: $v_C(0^-) = v_C(0^+)$ and $i_L(0^-) = i_L(0^+)$ . Every other variable in the circuit is free to jump, and in general it does.

Superposition applies to the two parts, not to convenience

The response is written as a transient part plus a steady-state part, and both parts must satisfy the differential equation individually before they may be added. The derivation of §4 enforces exactly this by matching the two groups of terms separately.

§4 · Solving by an assumed form

Guess the shape, then let the equation fix the constants

The equation is linear with constant coefficients and a constant forcing term, so the shape of the answer is known in advance: a decaying exponential for the transient part and a constant for the steady-state part. The unknown constants are then determined by substitution. The derivation is carried out for the inductor; the capacitor result follows by the same algebra.

Assume the form

f(t) = \underbrace{K_1e^{st}}_{\text{transient}} + \underbrace{K_2}_{\text{steady state}}

Three unknowns are introduced, namely $K_1$ , $K_2$ , and $s$ . Two of them are fixed by the differential equation itself and the third by the initial condition.

Substitute into the equation

\frac{d}{dt}\left[K_1e^{st} + K_2\right] + \frac{R_N}{L}\left[K_1e^{st} + K_2\right] = \frac{R_N}{L}I_N

Carrying out the differentiation and grouping the exponential terms against the constant terms gives

\underbrace{sK_1e^{st} + \frac{R_N}{L}K_1e^{st}}_{\text{must sum to zero}} + \underbrace{\frac{R_N}{L}K_2}_{\phantom{x}} = \underbrace{\frac{R_N}{L}I_N}_{\text{must be equal}}

The transient terms must sum to zero

An exponential is not a constant, so no constant on the right can cancel it. The exponential group must therefore vanish on its own:

sK_1e^{st} + \frac{R_N}{L}K_1e^{st} = 0 \;\Longrightarrow\; s + \frac{R_N}{L} = 0 \;\Longrightarrow\; \boxed{s = -\frac{R_N}{L} = -\frac{1}{\tau}}

The root $s$ is a property of the circuit alone. It does not depend on the sources or on the initial condition, and its reciprocal magnitude is the time constant.

The constant terms must be equal

What remains is an equality between constants:

\frac{R_N}{L}K_2 = \frac{R_N}{L}I_N \;\Longrightarrow\; \boxed{K_2 = I_N}

The steady-state part is simply the value the inductor current settles to, which is the entire Norton source current once the inductor behaves as a short circuit.

Apply the initial condition

Only now is the initial condition used, and it is applied to the complete solution rather than to either part alone. With $i_L(0^+) = I_0$ carried across the switching instant by continuity:

i_L(0^+) = K_1e^{0} + I_N = K_1 + I_N = I_0 \;\Longrightarrow\; \boxed{K_1 = I_0 - I_N}

Assemble the two parts

i_L(t) = \left[I_0 - I_N\right]e^{-t/\tau} + I_N, \qquad \tau = \frac{L}{R_N}

The dual result for the capacitor is obtained by the same steps, with the Thevenin voltage in place of the Norton current:

v_C(t) = \left[V_0 - V_T\right]e^{-t/\tau} + V_T, \qquad \tau = R_TC

Both have the identical structure of an initial value, a final value, and one exponential that carries the response from one to the other.

§5 · The general first-order response

One formula covers every variable in the circuit

The two results above are stated for the storage-element variable, but the restriction is unnecessary. Every voltage and every current in a first-order circuit obeys the same differential equation with the same root $s = -1/\tau$ , because the root is a property of the circuit rather than of the variable chosen. Any response $f(t)$ may therefore be written as

\boxed{\;f(t) = \left[f(0^+) - f(\infty)\right]e^{-t/\tau} + f(\infty)\;} \qquad \tau = R_TC \;\text{ or }\; \frac{L}{R_T}

Three numbers are all that is required: the value just after switching, the value long afterwards, and the time constant. The plot below is driven by exactly those three.

The solid curve is the response, the dashed horizontal line is $f(\infty)$ , and the short dashed line from $f(0^+)$ is the tangent at the origin. The tangent reaches the final value at exactly $t = \tau$ , which is the standard graphical construction for the time constant.

f(0⁺) 8

f(∞) 3

τ (s) 0.10 s

f(t) = [8 − 3] e^{−t / 0.10 s} + 3 = 5 e^−t/τ + 3

at t = τ: f = 4.84 (63.2 % of the way) at t = 5τ: f = 3.03 settled within 0.7 % by 0.50 s

Reading the time constant off a measurement

After one time constant the response has covered $1 - e^{-1} = 63.2\%$ of the total change, and after five time constants it has covered $1 - e^{-5} = 99.3\%$ . The second figure is the practical definition of settled, and it explains the common engineering rule that a first-order circuit is finished after roughly $5\tau$ .

§6 · The five-step process

From a switched circuit to $f(t)$

The formula of §5 needs three numbers, and each is obtained from a resistive circuit. No differential equation is ever solved again after this point; the following procedure replaces it.

Draw the $t = 0^-$ circuit and solve it at DC steady state. Because the circuit has been in this state indefinitely, the capacitor is an open circuit and the inductor is a short circuit. Find $v_C$ or $i_L$ at $0^-$ .
Carry that one value across the switching instant. Continuity gives $v_C(0^+) = v_C(0^-)$ or $i_L(0^+) = i_L(0^-)$ . Draw the $t = 0^+$ circuit with the capacitor replaced by a voltage source of that value, or the inductor replaced by a current source of that value, and solve it for the requested variable $f(0^+)$ . (1)
Draw the $t = \infty$ circuit and solve it at DC steady state. The capacitor is again an open circuit and the inductor again a short circuit. Solve for $f(\infty)$ . (2)
Find the time constant from the $t > 0$ circuit. Remove the storage element and mark its terminals $a$ and $b$ ; turn off the independent sources; compute $R_T$ looking into $a$ - $b$ ; then $\tau = R_TC$ or $\tau = L/R_T$ . (3)
Assemble. $f(t) = [f(0^+) - f(\infty)]e^{-t/\tau} + f(\infty)$ .

At DC steady state no variable is changing, so $i_C = C\,dv_C/dt = 0$ and $v_L = L\,di_L/dt = 0$ : the capacitor is an open circuit and the inductor a short circuit. These two replacements are used in step 1 and again in step 3.

Two habits that cause most errors

The DC steady-state replacements apply only in steps 1 and 3, never in step 2: at $t = 0^+$ the circuit is not at steady state, and the storage element is represented by a source holding its carried-over value. Independent sources are turned off in step 4 only; dependent sources are never turned off, exactly as in the Thevenin procedure of Lecture 4.

§7 · Worked example A: an RC circuit

Find $i_x(t)$

The current source is written $2u(-t)$ mA, so it supplies 2 mA before the switching instant and nothing afterwards. The requested variable is $i_x$ , the current in the 2 kΩ resistor, which is not the capacitor variable and is therefore free to jump at $t = 0$ .

The problem

Find $i_x(t)$ for $t > 0$ . The single storage element is the 1 μF capacitor, so the circuit is first order. The five-step process of §6 is applied without modification.

Step 1: the $t = 0^-$ circuit at DC steady state

The source supplies 2 mA and the capacitor is an open circuit, so no current is diverted into it and the same $i_x$ flows through the 2 kΩ and the lower 1 kΩ. KCL at the right node then forces $i_x + 2$ mA through the upper 1 kΩ, and KVL around the top loop gives

\tm{xa.r1t xa.itop}{-1000\,[i_x + 0.002]} \; \tm{xa.r1b xa.ix}{-\; 1000\,[i_x]} \; \tm{xa.r2k xa.ix}{-\; 2000\,[i_x]} = 0 \;\Longrightarrow\; -4000\,i_x = 2

\boxed{i_x(0^-) = -0.5\ \text{mA}}

The whole 2 mA therefore returns through the 4 kΩ resistor, and KVL around the left loop delivers the quantity that actually matters:

\tm{xa.r4k}{-4000\,(0.002)} + \tm{xa.r2k xa.ix}{2000\,i_x} + \tm{xa.vc}{v_C} = 0 \;\Longrightarrow\; \boxed{v_C(0^-) = 9\ \text{V}}

Step 2: the $t = 0^+$ circuit

Continuity of the capacitor voltage carries 9 V across the switching instant, so the capacitor is replaced by a 9 V source. The current source is now off and therefore an open circuit, which leaves the upper 1 kΩ and the lower 1 kΩ in series between the left node and the capacitor node. Taking the bottom rail as the reference and the left node as $v_L = 9 + 2000\,i_x$ , KCL at the left node gives

\tm{xa.r4k}{\frac{9 + 2000\,i_x}{4000}} + \tm{xa.r2k xa.ix}{i_x} + \tm{xa.r1t xa.r1b}{\frac{9 + 2000\,i_x - 9}{2000}} = 0 \;\Longrightarrow\; 0.00225 + 2.5\,i_x = 0

\boxed{i_x(0^+) = -0.9\ \text{mA}}

The jump from $-0.5$ mA to $-0.9$ mA is entirely legitimate: only $v_C$ is required to be continuous.

Step 3: the $t = \infty$ circuit at DC steady state

The source remains off for all $t > 0$ , so no independent source is left in the circuit at all. With the capacitor again an open circuit, every branch current must vanish:

\boxed{i_x(\infty) = 0}

Step 4: the time constant

The capacitor is removed and the resistance looking into its terminals is computed with the current source turned off, that is, replaced by an open circuit. From the capacitor node, the 2 kΩ and the series pair 1 kΩ + 1 kΩ run in parallel back to the left node, and the 4 kΩ continues from there to the reference:

\tm{xa.rt}{R_T} = (1\,\text{k} + 1\,\text{k}) \parallel 2\,\text{k} + 4\,\text{k} = 1\,\text{k} + 4\,\text{k} = 5\ \text{k}\Omega

\tm{xa.tau}{\tau} = R_TC = 5000 \cdot (1 \times 10^{-6}) \;\Longrightarrow\; \boxed{\tau = 5\ \text{ms}}

Step 5: assemble

i_x(t) = \left[i_x(0^+) - i_x(\infty)\right]e^{-t/\tau} + i_x(\infty) = \left[-0.9 - 0\right]e^{-t/0.005} + 0

\boxed{i_x(t) = -0.9\,e^{-200t}\ \text{mA}, \qquad t \ge 0}

Because the final value is zero, the response is a pure transient: all the energy stored on the capacitor before the switching instant is delivered to the resistors afterwards.

The waveform of $i_x$ . The step down at $t = 0$ from $-0.5$ mA to $-0.9$ mA is the discontinuity permitted to a resistor current, and the dashed vertical line marks one time constant.

Independent check by node voltages

The result may be confirmed without the KVL bookkeeping of step 2. For $t > 0$ the only source is the capacitor itself, so every variable is proportional to $v_C$ . Writing node equations at $t = 0^+$ with $v_C = 9$ V, $v_R$ at the right node, and $v_L$ at the left node:

\text{@}R:\quad \frac{v_R - v_L}{1000} + \frac{v_R - 9}{1000} = 0 \;\Longrightarrow\; v_R = \frac{v_L + 9}{2}

\text{@}L:\quad \frac{v_L}{4000} + \frac{v_L - 9}{2000} + \frac{v_L - v_R}{1000} = 0 \;\Longrightarrow\; 5\,v_L = 36 \;\Longrightarrow\; v_L = 7.2\ \text{V}

Then $i_x = (v_L - 9)/2000 = -1.8/2000 = -0.9$ mA, in agreement. ✓ The same relation $i_x = -v_C/10000$ holds at every instant, and since $v_C(t) = 9e^{-200t}$ V, it returns $i_x(t) = -0.9e^{-200t}$ mA once more. ✓

§8 · Worked example B: an RL circuit

Find $v_x(t)$

The switch is closed before the switching instant and opens at $t = 0$ , which is the opposite action to Example A. While it is closed it short-circuits the 4 Ω resistor, so the requested voltage $v_x$ is held at zero until the switch releases it.

The problem

Find $v_x(t)$ , the voltage across the 4 Ω resistor. The single storage element is the ½ H inductor, so $f(t) = v_x(t)$ is a first-order response and the process of §6 applies.

Step 1: the $t = 0^-$ circuit at DC steady state

The inductor is a short circuit at DC, and the closed switch places another short across the 4 Ω resistor. Both the 3 Ω and the 4 Ω are therefore short-circuited, and the entire source current is carried by the inductor branch:

\boxed{i_L(0^-) = 2\ \text{A}} \qquad\qquad v_x(0^-) = 0\ \text{V}

Step 2: the $t = 0^+$ circuit

Continuity of the inductor current carries 2 A across the switching instant, so the inductor is replaced by a 2 A source. The switch is now open, so that 2 A has only the 4 Ω resistor to flow through:

\tm{xb.vx}{v_x(0^+)} = \tm{xb.il}{2\ \text{A}} \cdot \tm{xb.r4}{4\ \Omega} \;\Longrightarrow\; \boxed{v_x(0^+) = 8\ \text{V}}

KCL at the left node leaves nothing for the 3 Ω resistor: the 2 A delivered by the source is exactly the 2 A taken by the inductor branch, so that resistor carries no part of this current.

Step 3: the $t = \infty$ circuit at DC steady state

The inductor is a short circuit once more, which places the 3 Ω and the 4 Ω in parallel across the 2 A source. KCL at the upper node gives

\tm{xb.cs}{-2} + \tm{xb.r3 xb.vx}{\frac{v_x}{3}} + \tm{xb.r4 xb.vx}{\frac{v_x}{4}} = 0 \;\Longrightarrow\; \frac{7}{12}v_x = 2

\boxed{v_x(\infty) = \tfrac{24}{7}\ \text{V} \approx 3.43\ \text{V}}

Step 4: the time constant

The inductor is removed and its terminals are marked $a$ and $b$ . The 2 A source is an independent source and is therefore turned off, that is, replaced by an open circuit, which detaches the leftmost branch entirely. What remains between $a$ and $b$ is the 3 Ω in series with the 4 Ω:

\tm{xb.rt}{R_T} = 3 + 4 = 7\ \Omega \qquad\qquad \tm{xb.tau}{\tau} = \frac{L}{R_T} = \frac{1/2}{7} \;\Longrightarrow\; \boxed{\tau = \tfrac{1}{14}\ \text{s}}

The two resistors are in series here, not in parallel, because turning off the current source opens its branch and leaves only one path from $a$ to $b$ .

Step 5: assemble

v_x(t) = \left[v_x(0^+) - v_x(\infty)\right]e^{-t/\tau} + v_x(\infty) = \left[8 - \tfrac{24}{7}\right]e^{-14t} + \tfrac{24}{7}

\boxed{v_x(t) = \tfrac{32}{7}\,e^{-14t} + \tfrac{24}{7}\ \text{V}, \qquad t \ge 0}

Here the final value is not zero, so the response is a genuine sum of a transient part and a steady-state part, decaying from 8 V toward $24/7 \approx 3.43$ V.

The waveform of $v_x$ . The voltage is held at zero by the closed switch, jumps to 8 V the instant the switch opens, and decays toward the steady-state value $24/7$ V with the time constant $\tau = 1/14$ s.

Independent check of the final value and the jump

The value at $t = \infty$ may be checked by current division rather than by KCL. With the inductor shorted, the 2 A source divides between the 3 Ω and the 4 Ω resistors, and the current in the 4 Ω branch is $2 \cdot 3/(3 + 4) = 6/7$ A, so $v_x = (6/7)(4) = 24/7$ V. ✓

The value at $t = 0^+$ is checked by noting that the inductor current cannot change, so the 3 Ω resistor and the 2 A source must together carry zero net current into the upper node: the source delivers 2 A, the inductor removes 2 A, and the 3 Ω resistor is left with none. The full 2 A therefore reaches the 4 Ω resistor and $v_x(0^+) = 8$ V. ✓ Substituting $t = 0$ into the final expression returns $32/7 + 24/7 = 56/7 = 8$ V, as required. ✓

§9 · Your turn

Practice: find $v_C(t)$

The switch in the circuit below is open for a long time and closes at $t = 0$ . The problem should be attempted on paper before the hints are opened, in order. All three numbers required by the general formula come out as whole values, which makes the answer easy to check.

Find $v_C(t)$ for $t \ge 0$ , and then the current delivered through the 2 kΩ branch once the switch has closed.

Hint 1: the initial condition

With the switch open, the 2 kΩ branch carries nothing and the capacitor is an open circuit at DC steady state. The circuit reduces to a single loop through the 3 kΩ and the 6 kΩ resistors, so a voltage divider gives the voltage at node A, which is $v_C(0^-)$ .

Hint 2: the final value

With the switch closed and the capacitor again an open circuit, the 6 kΩ and 2 kΩ resistors appear in parallel between node A and the reference. Combine them, then apply the voltage divider a second time against the 3 kΩ.

Hint 3: the time constant

Remove the capacitor and turn off the 12 V source, which means replacing it by a short circuit. Three resistors then run in parallel from node A to the reference. Multiply the result by $C = 100\ \mu\text{F}$ .

Solution

Step 1. With the switch open, $v_C(0^-) = 12 \cdot \dfrac{6}{3 + 6} = 8$ V, and continuity gives $\boxed{v_C(0^+) = 8\ \text{V}}$ .

Step 3. With the switch closed, $6\,\text{k} \parallel 2\,\text{k} = 1.5\ \text{k}\Omega$ , so

v_C(\infty) = 12 \cdot \frac{1.5}{3 + 1.5} = \boxed{4\ \text{V}}

Step 4. With the 12 V source shorted, all three resistors are in parallel:

R_T = 3\,\text{k} \parallel 6\,\text{k} \parallel 2\,\text{k} = 1\ \text{k}\Omega \qquad \tau = R_TC = 1000 \cdot (100 \times 10^{-6}) = \boxed{0.1\ \text{s}}

Step 5. Assembling the three numbers:

\boxed{v_C(t) = \left[8 - 4\right]e^{-10t} + 4 = 4e^{-10t} + 4\ \text{V}, \qquad t \ge 0}

The current in the 2 kΩ branch follows directly from Ohm's law, since that branch is connected across the capacitor once the switch is closed:

i_{2\text{k}}(t) = \frac{v_C(t)}{2000} = 2e^{-10t} + 2\ \text{mA}

Check by the differential equation. KCL at node A with the switch closed reads $\dfrac{12 - v_C}{3000} = \dfrac{v_C}{6000} + \dfrac{v_C}{2000} + (100 \times 10^{-6})\dfrac{dv_C}{dt}$ . Multiplying by 6000 and collecting terms gives

\frac{dv_C}{dt} + 10\,v_C = 40

whose time constant is $1/10 = 0.1$ s and whose steady-state value is $40/10 = 4$ V, both matching the results above. ✓ At $t = 0$ the expression returns $4 + 4 = 8$ V, which matches the initial condition. ✓

One energy storage element changes everything

The resistive part is always reduced first

The switching instant defines t=0t = 0t=0

The unit step function

A step source is a switch and two ordinary sources

Before t=0t = 0t=0: the blade rests on the lower throw

After t=0t = 0t=0: the blade moves to the upper throw

Two circuits, two equations, one structure

Guess the shape, then let the equation fix the constants

Assume the form

Substitute into the equation

The transient terms must sum to zero

The constant terms must be equal

Apply the initial condition

Assemble the two parts

One formula covers every variable in the circuit

From a switched circuit to f(t)f(t)f(t)

Find ix(t)i_x(t)ix​(t)

The problem

Step 1: the t=0−t = 0^-t=0− circuit at DC steady state

Step 2: the t=0+t = 0^+t=0+ circuit

Step 3: the t=∞t = \inftyt=∞ circuit at DC steady state

Step 4: the time constant

Step 5: assemble

Find vx(t)v_x(t)vx​(t)

The problem

Step 1: the t=0−t = 0^-t=0− circuit at DC steady state

Step 2: the t=0+t = 0^+t=0+ circuit

Step 3: the t=∞t = \inftyt=∞ circuit at DC steady state

Step 4: the time constant

Step 5: assemble

Practice: find vC(t)v_C(t)vC​(t)

The switching instant defines $t = 0$

Before $t = 0$ : the blade rests on the lower throw

After $t = 0$ : the blade moves to the upper throw

From a switched circuit to $f(t)$

Find $i_x(t)$

Step 1: the $t = 0^-$ circuit at DC steady state

Step 2: the $t = 0^+$ circuit

Step 3: the $t = \infty$ circuit at DC steady state

Find $v_x(t)$

Step 1: the $t = 0^-$ circuit at DC steady state

Step 2: the $t = 0^+$ circuit

Step 3: the $t = \infty$ circuit at DC steady state

Practice: find $v_C(t)$