ECE 211 · Lecture 7: Capacitors, Inductors & Switching

§1 · The capacitor

Two elements give a circuit a memory

Chapter 6 introduces the two energy storage devices, and they differ from every element studied so far in one decisive respect: their terminal relations involve a derivative rather than a proportionality. A resistor dissipates whatever is applied to it at the instant it is applied, so its response carries no history. A capacitor stores energy in an electric field and an inductor stores energy in a magnetic field, so each responds to how fast the applied quantity is changing, and the circuit acquires a memory of its own past.

Capacitor · C · farads [F] Inductor · L · henrys [H] both store energy; neither dissipates it

The capacitor is labeled under the passive sign convention: the current $i_C$ enters the terminal marked positive for $v_C$ .

The capacitor current is proportional to the rate of change of the capacitor voltage:

\tm{cp.ic}{i_C} = \tm{cp.c}{C} \, \frac{d \tm{cp.vc}{v_C}}{dt}

Three consequences follow directly from that one relation, and the whole of this lecture rests on them.

Read the derivative three ways

At DC steady state the capacitor is an open circuit. Every voltage is constant, so $dv_C/dt = 0$ and therefore $i_C = 0$ , which is precisely the behavior of an open circuit. The capacitor voltage cannot change instantaneously, that is, $v_C(0^-) = v_C(0^+)$ , because a step in $v_C$ would require an infinite derivative and therefore an infinite current. The capacitor current is free to jump, since nothing in the relation constrains $i_C$ itself.

Left: the capacitor voltage is carried continuously through the switching instant, so its value just before the switch operates is also its value just after. Right: the capacitor current is under no such constraint, and a switching instant typically steps it to a new value.

§2 · The inductor

A coil that resists a change of current

An inductor is a coil of conducting wire that stores energy in the magnetic field established by its own current. Its inductance is fixed by geometry and construction rather than by any electrical quantity:

L = \frac{N^2 \, \mu \, \tm{co.area}{A}}{\tm{co.len}{\ell}}

where $N$ is the number of turns, $\mu$ is the permeability of the core, $A$ is the cross-sectional area, and $\ell$ is the length of the winding.

Inductance grows with the square of the turn count, which is why practical inductors are wound rather than merely bent.

The inductor voltage is proportional to the rate of change of the inductor current, the dual of the capacitor relation:

\tm{ic.vl}{v_L} = \tm{ic.l}{L} \, \frac{d \tm{ic.il}{i_L}}{dt}

The same passive sign convention applies: the current $i_L$ enters the terminal marked positive for $v_L$ .

The dual reading

At DC steady state the inductor is a short circuit. Every current is constant, so $di_L/dt = 0$ and therefore $v_L = 0$ , which is the behavior of a short circuit. The inductor current cannot change instantaneously, that is, $i_L(0^-) = i_L(0^+)$ . The inductor voltage is free to jump.

The inductor sketch is the capacitor sketch with the two quantities exchanged. Exactly one state variable per element is continuous: $v_C$ for a capacitor and $i_L$ for an inductor.

§3 · Series and parallel combinations

Inductors follow the resistor rules; capacitors invert them

The combination rules are read off the terminal relations rather than memorized separately. Inductors in series carry a common current, so their voltages add and the inductances add, exactly as resistances do:

Three inductors in series, driven through one port. One loop implies one current $i$ , so a single derivative $di/dt$ appears in every term.

-\tm{ls.port}{v} + \tm{ls.l1}{L_1 \frac{di}{dt}} + \tm{ls.l2}{L_2 \frac{di}{dt}} + \tm{ls.l3}{L_3 \frac{di}{dt}} = 0 \;\;\Longrightarrow\;\; v = [L_1 + L_2 + L_3] \frac{di}{dt} = L_{eq} \frac{di}{dt}

The same argument run on parallel inductors, where the voltage is common and the currents add, produces the reciprocal rule. The capacitor relation carries the derivative on the other side, so its two rules are exchanged:

Element	In series	In parallel
Resistors	values add	reciprocals add
Inductors	values add	reciprocals add
Capacitors	reciprocals add	values add

Written out, the inductor rules are the resistor rules and the capacitor rules are their inverse:

L_{eq} = L_1 + L_2 + L_3 \quad \text{(series)} \qquad \frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} \quad \text{(parallel)}

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \quad \text{(series)} \qquad C_{eq} = C_1 + C_2 + C_3 \quad \text{(parallel)}

Capacitances add in parallel and their reciprocals add in series. The sliders drive both networks with the same three values.

C₁ (µF) 10

C₂ (µF) 20

C₃ (µF) 30

C_parallel = 10 + 20 + 30 = 60.0 µF C_series = 1 / [1/10 + 1/20 + 1/30] = 5.45 µF

With the same three numbers read as inductances in µH: L_series = 60.0 µH L_parallel = 5.45 µH

The series combination is always the smallest

A series combination of capacitors is smaller than the smallest member, and a parallel combination of inductors is smaller than the smallest member. Whenever a computed equivalent violates that bound, the wrong rule has been applied.

§4 · Three snapshots of a switched circuit

Read the circuit at $0^-$ , at $0^+$ , and at $\infty$

A switched circuit that contains one storage element is answered by three separate static analyses rather than by one dynamic one. The recipe is fixed:

At $t = 0^-$ the circuit has been in its old configuration long enough to reach DC steady state. Every capacitor is replaced by an open circuit and every inductor by a short circuit, and the resulting resistive circuit is solved. Only two results are carried forward: $v_C(0^-)$ and $i_L(0^-)$ .
At $t = 0^+$ the switch has operated but the state variables have not moved. The capacitor is replaced by a voltage source of value $v_C(0^-)$ and the inductor by a current source of value $i_L(0^-)$ , and the new resistive circuit is solved for whatever quantity was requested. Everything other than $v_C$ and $i_L$ is free to jump here.
At $t \to \infty$ the new configuration has reached DC steady state, so the capacitors are open circuits and the inductors are short circuits again.

Between the second and the third snapshot the circuit travels along a single exponential, whose time constant is $\tau = R_{th} C$ or $\tau = L / R_{th}$ with $R_{th}$ the resistance seen by the storage element for $t > 0$ :

x(t) = x(\infty) + [\, x(0^+) - x(\infty) \,] \, e^{-t/\tau}

The two endpoints are fixed by the second and third snapshots; the time constant sets only how quickly the curve travels between them.

x(0⁺) 6

x(∞) -4

τ (s) 1

x(t) = -4.0 + [6.0 − -4.0] · e^−t/1.0

x(τ) = -0.32 (63.2 % traveled) x(3τ) = -3.50 (95.0 %) x(5τ) = -3.93 (99.3 %)

The third snapshot is not always needed

The first two snapshots are demanded by continuity and are therefore never optional. The third is required only when the response itself is wanted; a question that asks only for the value immediately after switching is answered at $t = 0^+$ .

§5 · Worked example A: two amplifiers pin two nodes

Find $v_o(0^-)$ , $v_o(0^+)$ , and $v_o(\infty)$

Two ideal operational amplifiers appear in this circuit, and their only role is to hold two node voltages rigid. Neither amplifier is analyzed as an amplifier; each contributes two facts, and those two facts make the rest of the circuit a routine node problem.

What an ideal amplifier contributes

No current enters either input terminal, and negative feedback holds the two input terminals at the same voltage. For the amplifier on the left, the noninverting terminal is grounded, so its inverting terminal, the node marked 0 V, is held at zero. For the amplifier at the bottom, the noninverting terminal is tied to +10 V, so its inverting terminal, the node marked 10 V, is held at ten. Because no current enters either terminal, KCL written at those two nodes contains only the resistor branches.

The problem

The switch closes at $t = 0$ . The output $v_o$ is required at the three snapshots, and the ½ H inductor is the only storage element.

Snapshot 1a: $t = 0^-$ , the inductor is a short

At DC steady state the inductor is replaced by a short circuit. KCL is written at the node held at zero, where only the 25 Ω and the 50 Ω branches meet:

\tm{xa.r25}{\frac{0 - 5}{25}} + \tm{xa.r50}{\frac{0 - V_x}{50}} = 0 \;\;\Longrightarrow\;\; -10 - V_x = 0 \;\;\Longrightarrow\;\; \boxed{V_x = -10\ \text{V}}

The switch is still open, so the upper 5 Ω carries nothing.

Snapshot 1b: the inductor current

With the inductor shorted, the 10 Ω branch is connected directly between the node held at $10$ V and the node at $-10$ V:

\tm{xa.il}{i_L(0^-)} = \frac{10 - (-10)}{\tm{xa.r10}{10}} = \boxed{2\ \text{A}}

This is the one value that must survive the switching instant.

Snapshot 1c: the output before switching

KCL at the node held at $10$ V involves the 10 Ω branch and the 5 Ω feedback branch only:

\tm{xa.r10}{\frac{10 - (-10)}{10}} + \tm{xa.r5b}{\frac{10 - v_o}{5}} = 0 \;\;\Longrightarrow\;\; 20 + 20 - 2 v_o = 0 \;\;\Longrightarrow\;\; \boxed{v_o(0^-) = 20\ \text{V}}

Snapshot 2a: $t = 0^+$ , continuity replaces the inductor

The switch closes, and the inductor current cannot follow it: $i_L(0^+) = i_L(0^-) = 2\ \text{A}$ . For this instant the inductor branch is therefore replaced by a 2 A current source directed from the node at $10$ V toward the node at $-10$ V. The node held at zero is untouched by the switch, so $V_x$ is still $-10$ V.

Snapshot 2b: the output just after switching

Three branches now leave the node held at $10$ V: the feedback 5 Ω, the newly connected 5 Ω, and the 2 A of the inductor branch.

\tm{xa.r5b}{\frac{10 - v_o}{5}} + \tm{xa.r5a}{\frac{10 - (-10)}{5}} + \tm{xa.ind}{2} = 0 \;\;\Longrightarrow\;\; 10 - v_o + 20 + 10 = 0 \;\;\Longrightarrow\;\; \boxed{v_o(0^+) = 40\ \text{V}}

Snapshot 3: $t \to \infty$ , the inductor is a short again

At the new DC steady state the inductor is once more a short circuit, so the 10 Ω branch again spans the two rigid nodes:

\tm{xa.r5a}{\frac{10 - (-10)}{5}} + \tm{xa.r10}{\frac{10 - (-10)}{10}} + \tm{xa.r5b}{\frac{10 - v_o}{5}} = 0 \;\;\Longrightarrow\;\; 40 + 20 + 20 - 2 v_o = 0 \;\;\Longrightarrow\;\; \boxed{v_o(\infty) = 40\ \text{V}}

Why the last two answers agree

The equality $v_o(0^+) = v_o(\infty)$ is not a coincidence and not an arithmetic slip. Both ends of the inductor branch are pinned by the amplifiers: the left node sits at $-10$ V and the right node at $10$ V, before and after the switch operates. The branch therefore always carries $20/10 = 2$ A, the 10 Ω always drops the full $20$ V, and the inductor itself always has zero volts across it. With $v_L = 0$ at every instant, $di_L/dt = 0$ and the inductor current never moves. The circuit reaches its final state at the switching instant, and no exponential follows.

§6 · Worked example B: the differential equation for $i_L$

Assemble a first-order equation from one KVL

The differential equation relating $i_L$ to $i_s$ is required for $t > 0$ . The procedure is the branch-current method with one addition: the inductor contributes $v_L = L \, di_L/dt$ instead of a resistance, and the resulting equation is differential rather than algebraic.

The problem

The switch on the right opens at $t = 0$ ; before that instant it shorts the right side of the circuit. The 2 A source is constant, and the storage element is the ½ H inductor.

Snapshot 1: $t = 0^-$

The inductor is a short at DC steady state, and the closed switch is a short as well. Both resistors are therefore short circuited, they carry no voltage and no current, and the entire source current is forced through the inductor:

\tm{xb.il}{i_L(0^-)} = \tm{xb.is}{2\ \text{A}}

Step 1: label every branch with one unknown

For $t > 0$ the switch is open, so the circuit has two backyards and one current source, which leaves one unknown. Naming the inductor current $i_L$ fixes everything else: KCL at the upper left node assigns $i_L - 2$ upward through the 3 Ω, and KCL at the upper right node assigns $i_L$ downward through the 4 Ω.

Step 2: KVL around the middle backyard

\tm{xb.r3 xb.i32}{3 \, (i_L - 2)} + \tm{xb.vl}{v_L} + \tm{xb.r4 xb.i4}{4 \, i_L} = 0

The inductor voltage is then written as a derivative, $v_L = \frac{1}{2} \, di_L/dt$ , and the equation is arranged with the derivative alone in the leading term:

3 i_L - 6 + \tfrac{1}{2} \frac{d i_L}{dt} + 4 i_L = 0 \;\;\Longrightarrow\;\; \boxed{\frac{d i_L}{dt} + 14 \, i_L = 12}

This is a first-order differential equation with constant coefficients. Its coefficients carry all the information needed for the response: the time constant is the reciprocal of $14$ , and the final value is $12/14$ .

Snapshot 3: $t \to \infty$

At the new DC steady state the derivative vanishes, so the differential equation collapses to an algebraic one:

14 \, i_L = 12 \;\;\Longrightarrow\;\; \boxed{i_L(\infty) = \tfrac{6}{7}\ \text{A}}

The same value is obtained independently by redrawing the circuit with the inductor shorted, which places the 3 Ω and the 4 Ω in parallel across the source, and applying the current divider: $i_L = 2 \cdot \tfrac{3}{3 + 4} = \tfrac{6}{7}$ A. Agreement between the two routes confirms the coefficients.

The complete response

The three snapshots are now available: $i_L(0^+) = i_L(0^-) = 2$ A and $i_L(\infty) = 6/7$ A. The time constant is read from the equation, and it agrees with $L / R_{th}$ where the resistance seen by the inductor for $t > 0$ is $3 + 4 = 7\ \Omega$ :

\tau = \frac{1}{14}\ \text{s} = \frac{L}{R_{th}} = \frac{1/2}{7} \qquad i_L(t) = \tfrac{6}{7} + \left[ 2 - \tfrac{6}{7} \right] e^{-14 t} = \boxed{\tfrac{6}{7} + \tfrac{8}{7} e^{-14 t}\ \text{A}}

§7 · Worked example C: the differential equation for $v_C$

The same assembly with the capacitor relation

The capacitor is handled in the mirror image of the inductor: its current is named $i_C$ , every other branch current is expressed in terms of $i_C$ and $v_C$ , and the substitution $i_C = C \, dv_C/dt$ is postponed until the KVL equation has been written.

The problem

The switch closes at $t = 0$ , connecting the 5 Ω branch. The differential equation for $v_C$ is required for $t > 0$ , so every figure below shows the switch closed.

Step 1: express every branch current with $i_C$ and $v_C$

The capacitor carries $i_C$ downward by definition, and the 5 Ω carries $v_C/5$ downward by Ohm's law, since the capacitor voltage appears directly across it. KCL at the capacitor node then forces $i_C + v_C/5$ through the 3 Ω, and KCL at the source node adds the source current to give $i_C + v_C/5 + 2$ upward through the 2 Ω.

Step 2: KVL around the middle backyard

\tm{xc.r2 xc.i2}{2 \left[ i_C + \tfrac{v_C}{5} + 2 \right]} + \tm{xc.r3 xc.i3}{3 \left[ i_C + \tfrac{v_C}{5} \right]} + \tm{xc.vc}{v_C} = 0

Each resistor contributes its own current multiplied by its own resistance, with the sign set by the polarity marks, and the capacitor contributes its voltage directly.

Step 3: substitute the capacitor relation and clear the fractions

\tm{xc.ic}{i_C} = \tm{xc.cap}{C} \frac{d v_C}{dt} = \tfrac{1}{10} \, v_C'

2 \left[ \tfrac{1}{10} v_C' + \tfrac{v_C}{5} + 2 \right] + 3 \left[ \tfrac{1}{10} v_C' + \tfrac{v_C}{5} \right] + v_C = 0

Multiplying through by $10$ gathers the terms into integers:

2 v_C' + 4 v_C + 40 + 3 v_C' + 6 v_C + 10 v_C = 0 \;\;\Longrightarrow\;\; 5 v_C' + 20 v_C = -40 \;\;\Longrightarrow\;\; \boxed{v_C' + 4 v_C = -8}

Snapshot 3 and the time constant

Setting the derivative to zero gives the final value, and the coefficient of $v_C$ gives the reciprocal of the time constant:

4 v_C = -8 \;\;\Longrightarrow\;\; \boxed{v_C(\infty) = -2\ \text{V}} \qquad \tau = \tfrac{1}{4}\ \text{s}

Both results are confirmed independently. With the capacitor opened, the 2 Ω and 3 Ω carry the same current as the 5 Ω, and node analysis returns $v_C = -2$ V. The resistance seen by the capacitor with the current source opened is $(2 + 3) \parallel 5 = 2.5\ \Omega$ , so $\tau = R_{th} C = 2.5 \cdot \tfrac{1}{10} = \tfrac{1}{4}$ s, in agreement.

The initial condition and the complete response

The board stops at the differential equation, but the first snapshot completes the problem. Before $t = 0$ the switch is open, so the 5 Ω branch is absent and the capacitor is an open circuit at DC steady state. No current then flows in the 3 Ω, the whole source current of 2 A is forced upward through the 2 Ω, and the capacitor node follows the source node:

v_C(0^-) = -2 \cdot 2 = -4\ \text{V}

Continuity carries that value across the switching instant, so $v_C(0^+) = -4$ V and

v_C(t) = -2 + [\, -4 - (-2) \,] e^{-4t} = \boxed{-2 - 2 e^{-4t}\ \text{V}}

A check at $t = 0$ returns $-4$ V and a check at large $t$ returns $-2$ V, both as required. Substituting the expression into $v_C' + 4 v_C = -8$ gives $8 e^{-4t} - 8 - 8 e^{-4t} = -8$ , an identity. ✓

§8 · Worked example D: a current that jumps

Find $i_x(0^-)$ , $i_x(0^+)$ , and $i_x(\infty)$

The quantity requested here is a current in a plain wire, not a state variable, so it is free to jump at the switching instant and it does. The example is the clearest demonstration on the page of why the second snapshot cannot be skipped.

The problem

The switch closes at $t = 0$ , and $i_x$ is the current it carries. The storage element is the ½ F capacitor.

Snapshot 1a: $t = 0^-$ , the capacitor is an open circuit

The switch is open, so $i_x(0^-) = 0$ without any calculation. The capacitor is an open circuit at DC steady state, so the 12 V source and the 4 Ω carry no current either, and the 2 A of the source has only one available return path: upward through the 10 Ω.

Snapshot 1b: the capacitor voltage

KVL is taken around the left loop, through the source, the capacitor, and the 10 Ω. The resistor carries 2 A upward, so its lower terminal is the positive one:

\tm{xd.vs}{-12} + \tm{xd.vc}{v_C} - \tm{xd.r10}{10 \cdot 2} = 0 \;\;\Longrightarrow\;\; \boxed{v_C(0^-) = 32\ \text{V}}

The 4 Ω contributes nothing because it carries no current.

Snapshot 2a: $t = 0^+$ , continuity fixes the capacitor

The switch closes, and the capacitor voltage cannot follow it: $v_C(0^+) = v_C(0^-) = 32$ V. For this instant the capacitor behaves as a 32 V source. The closed switch also short circuits the 10 Ω, which therefore carries nothing while the switch is closed.

Snapshot 2b: the two equations

KVL around the left loop, now closed through the switch, gives the capacitor current:

\tm{xd.vs}{-12} + \tm{xd.vc}{32} + \tm{xd.r4 xd.ic}{4 \, i_C} = 0 \;\;\Longrightarrow\;\; \boxed{i_C = -5\ \text{A}}

KCL at the upper node then gives the requested current, since the capacitor branch delivers $i_C$ into the node while the 2 A source and the switch draw current out of it:

\tm{xd.ix}{i_x} = \tm{xd.ic}{i_C} - \tm{xd.cs2}{2} = -5 - 2 \;\;\Longrightarrow\;\; \boxed{i_x(0^+) = -7\ \text{A}}

Snapshot 3: $t \to \infty$

At the new DC steady state the capacitor is an open circuit again, so the left branch carries nothing. The only current left at the node is the source current, and the closed switch is its only path:

\boxed{i_x(\infty) = -2\ \text{A}}

The capacitor voltage settles at $v_C(\infty) = 12$ V, since with no current in the 4 Ω the capacitor sees the source directly.

The response

The resistance seen by the capacitor for $t > 0$ is the 4 Ω alone, because the 12 V source is turned off into a short and the closed switch returns the loop, so $\tau = R_{th} C = 4 \cdot \tfrac{1}{2} = 2$ s:

i_x(t) = -2 + [\, -7 - (-2) \,] e^{-t/2} = \boxed{-2 - 5 e^{-t/2}\ \text{A}} \qquad (t > 0)

A current of zero jumps to $-7$ A and then relaxes to $-2$ A. Neither of the two extreme values could have been obtained from the other.

The sketch of $i_x$ : zero while the switch is open, a jump to $-7$ A at the switching instant, and an exponential approach to the final value of $-2$ A with a time constant of two seconds.

§9 · Your turn

Practice: a switched resistor across a charged capacitor

The switch closes at $t = 0$ . Find $i_x$ at the three snapshots, then the differential equation for $v_C$ , the time constant, and the complete response. The problem should be attempted on paper before the hints are opened, in order.

A 4 A source drives a 10 Ω resistor and a ½ F capacitor; the 15 Ω branch joins them when the switch closes.

Hint 1: the first snapshot

At $t = 0^-$ the switch is open and the capacitor is an open circuit, so the entire source current is forced through the 10 Ω. Both $v_C(0^-)$ and $i_x(0^-)$ follow from that single observation.

Hint 2: what survives the switching instant

The capacitor voltage is continuous, so $v_C(0^+) = v_C(0^-)$ . The 10 Ω resistor is connected directly across the capacitor, so $i_x = v_C/10$ at every instant. Asking whether $i_x$ can jump therefore answers itself.

Hint 3: the equation and the time constant

KCL is written at the top node with the capacitor relation substituted: $C \, v_C' + v_C/10 + v_C/15 = 4$ for $t > 0$ . The resistance seen by the capacitor is $10 \parallel 15$ , with the current source turned off into an open circuit.

Solution

Snapshot 1. With the switch open and the capacitor open, all 4 A passes through the 10 Ω:

v_C(0^-) = 4 \cdot 10 = 40\ \text{V} \qquad i_x(0^-) = 4\ \text{A}

Snapshot 2. Continuity gives $v_C(0^+) = 40$ V, and the 10 Ω sits directly across the capacitor, so

i_x(0^+) = \frac{40}{10} = 4\ \text{A}

This current does not jump. It is tied to the continuous capacitor voltage by Ohm's law, which is exactly what distinguishes it from the current of Example D. The 15 Ω branch takes the jump instead: it goes from 0 to $40/15 = 8/3$ A, and the capacitor current jumps from 0 to $4 - 4 - 8/3 = -8/3$ A.

Snapshot 3. With the capacitor open again, the source drives $10 \parallel 15 = 6\ \Omega$ :

v_C(\infty) = 4 \cdot 6 = 24\ \text{V} \qquad i_x(\infty) = \frac{24}{10} = 2.4\ \text{A}

The equation and the response. KCL at the top node for $t > 0$ gives

\tfrac{1}{2} v_C' + \tfrac{v_C}{10} + \tfrac{v_C}{15} = 4 \;\;\Longrightarrow\;\; \boxed{v_C' + \tfrac{1}{3} v_C = 8}

so $\tau = 3$ s, which agrees with $R_{th} C = 6 \cdot \tfrac{1}{2}$ , and the final value $8 / (1/3) = 24$ V agrees with the third snapshot. Therefore

v_C(t) = 24 + 16 \, e^{-t/3}\ \text{V} \qquad i_x(t) = \frac{v_C(t)}{10} = 2.4 + 1.6 \, e^{-t/3}\ \text{A}

both valid for $t > 0$ . A check at $t = 0$ returns $40$ V and $4$ A, and a check at large $t$ returns $24$ V and $2.4$ A. ✓

Two elements give a circuit a memory

A coil that resists a change of current

Inductors follow the resistor rules; capacitors invert them

Read the circuit at 0−0^-0−, at 0+0^+0+, and at ∞\infty∞

Find vo(0−)v_o(0^-)vo​(0−), vo(0+)v_o(0^+)vo​(0+), and vo(∞)v_o(\infty)vo​(∞)

The problem

Snapshot 1a: t=0−t = 0^-t=0−, the inductor is a short

Snapshot 1b: the inductor current

Snapshot 1c: the output before switching

Snapshot 2a: t=0+t = 0^+t=0+, continuity replaces the inductor

Snapshot 2b: the output just after switching

Snapshot 3: t→∞t \to \inftyt→∞, the inductor is a short again

Why the last two answers agree

Assemble a first-order equation from one KVL

The problem

Snapshot 1: t=0−t = 0^-t=0−

Step 1: label every branch with one unknown

Step 2: KVL around the middle backyard

Snapshot 3: t→∞t \to \inftyt→∞

The complete response

The same assembly with the capacitor relation

The problem

Step 1: express every branch current with iCi_CiC​ and vCv_CvC​

Step 2: KVL around the middle backyard

Step 3: substitute the capacitor relation and clear the fractions

Snapshot 3 and the time constant

Find ix(0−)i_x(0^-)ix​(0−), ix(0+)i_x(0^+)ix​(0+), and ix(∞)i_x(\infty)ix​(∞)

The problem

Snapshot 1a: t=0−t = 0^-t=0−, the capacitor is an open circuit

Snapshot 1b: the capacitor voltage

Snapshot 2a: t=0+t = 0^+t=0+, continuity fixes the capacitor

Snapshot 2b: the two equations

Snapshot 3: t→∞t \to \inftyt→∞

The response

Practice: a switched resistor across a charged capacitor

Read the circuit at $0^-$ , at $0^+$ , and at $\infty$

Find $v_o(0^-)$ , $v_o(0^+)$ , and $v_o(\infty)$

Snapshot 1a: $t = 0^-$ , the inductor is a short

Snapshot 2a: $t = 0^+$ , continuity replaces the inductor

Snapshot 3: $t \to \infty$ , the inductor is a short again

Snapshot 1: $t = 0^-$

Snapshot 3: $t \to \infty$

Step 1: express every branch current with $i_C$ and $v_C$

Find $i_x(0^-)$ , $i_x(0^+)$ , and $i_x(\infty)$

Snapshot 1a: $t = 0^-$ , the capacitor is an open circuit

Snapshot 2a: $t = 0^+$ , continuity fixes the capacitor

Snapshot 3: $t \to \infty$