123-ME · First-Order Transients & DC Steady State

First-Order Transients & DC Steady State

Slides 4.1–4.3 · Exercise 8 · Tutorial 6 P4–P5 · Tutorial 7 P1

Everything before this chapter happened instantly. Flip a switch on a resistive network and the new currents are simply there — Ohm's law has no clock in it. Add one capacitor or one inductor and the circuit acquires memory: it now takes time to get where it's going, and the whole exam question becomes how it gets there. This chapter gives you one formula that covers every such circuit in ME 123, plus the two-line reasoning that produces its three unknowns. Budget about 45 minutes: 15 on the derivation, 10 on the widget, 20 on the practice problems.

The three eras of a switched circuit

Every first-order problem in this course is really three separate problems stapled together, and the single most common way to lose marks is to solve the wrong one. Read the problem, decide which era each question is asking about, and the work becomes mechanical.

ERA 1 · t < 0

Old steady state

The switch has been where it is "for a long time." Nothing is changing, so caps are open circuits and inductors are shorts. Pure DC analysis — no calculus. This era exists only to hand you the initial condition.

ERA 2 · t > 0

Transient

The switch has moved. The stored energy is wrong for the new circuit, so it decays exponentially toward what the new circuit wants. This is the only era with an $e$ in it.

ERA 3 · t → ∞

New steady state

Five time constants later the exponential is dead. Caps open, inductors short — again, but now in the post-switch topology. Pure DC analysis a second time. This era hands you the final value.

The map. Era 1 and Era 3 are DC problems; only Era 2 needs the exponential.

Why nothing jumps (and what that buys you)

The bridge between Era 1 and Era 2 is a continuity argument, and it comes straight out of the element laws. For a capacitor,

i_C(t) = C\,\frac{dv_C(t)}{dt}

Capacitor element law — current is proportional to the rate of change of voltage.

Now suppose $v_C$ jumped — changed by a finite amount $\Delta v$ in zero time. Then $dv_C/dt = \Delta v / 0 \to \infty$ , and the capacitor would have to draw infinite current. Real circuits contain no infinite-current sources, so the jump never happens. The same argument on $v_L = L\,di_L/dt$ says an inductor current that jumped would demand an infinite voltage spike. Hence the two most useful sentences in this chapter:

The continuity rules

$v_C(0^+) = v_C(0^-) \qquad\qquad i_L(0^+) = i_L(0^-)$

Capacitor voltage and inductor current are continuous across the switching instant. Nothing else is. Capacitor current, inductor voltage, resistor voltages, node voltages, source currents — all of these are free to jump discontinuously at $t = 0$ , and in a good exam problem at least one of them will.

That asymmetry is what makes the rules useful. You are given a circuit that changes at $t=0$ and you need one number to pin down the solution; continuity says exactly which number you are allowed to carry across the boundary. Everything else you recompute in the new topology.

A handy restatement for the instant just after the switch: at $t = 0^+$ , a capacitor behaves like an ideal voltage source of value $v_C(0^-)$ , and an inductor behaves like an ideal current source of value $i_L(0^-)$ . Substitute those in and you can solve for every other quantity at $t = 0^+$ with plain resistive analysis.

Quick check · continuity

A capacitor has sat at $v_C = 15$ V for a long time. At $t = 0$ a switch opens, disconnecting the source entirely so that the capacitor's new final value will be $0$ V. What is $v_C(0^+)$ ?

Era 1 and Era 3: DC steady state (slides 4.2–4.3)

"Steady state" means every voltage and current has stopped changing: all the $d/dt$ terms are zero. Feed that into the two element laws and the components collapse:

Element	Element law	With $d/dt = 0$	Replace with
Capacitor	$i_C = C\,dv_C/dt$	$i_C = 0$	Open circuit (carries no current; holds whatever voltage the circuit puts there)
Inductor	$v_L = L\,di_L/dt$	$v_L = 0$	Short circuit (drops no voltage; carries whatever current the circuit puts through it)

The substitution that turns Era 1 and Era 3 into ordinary resistive problems.

The classic slip

"Open circuit" makes people write $v_C = 0$ and "short circuit" makes people write $i_L = 0$ . Both are backwards. An open branch has zero current and generally a large voltage; a short has zero voltage and generally a large current. Write the zero on the right quantity.

Worked example 1 — pure DC steady stateEra 1 / Era 3 drill

A 24 V source feeds $R_1 = 4\ \Omega$ into node A. From node A, one branch is $L = 2$ H in series with $R_2 = 8\ \Omega$ down to ground; a second branch is $C = 50\ \mu$ F from node A to ground. The circuit has been like this for a long time. Find $v_C$ , $i_L$ , and the stored energies.

The physical circuit for Worked example 1 — a 24 V source through

R_1

into node A, which splits into an

L

–

R_2

series branch and a

C

branch, both returning to ground.

Steady state ⇒ replace $C$ with an open circuit and $L$ with a short.

Nothing is changing, so $i_C = C\,dv_C/dt = 0$ and $v_L = L\,di_L/dt = 0$ .

With the capacitor branch open, the only path left is $24\ \text{V} \to R_1 \to R_2 \to$ ground: $i_L = \frac{24}{4 + 8} = 2\ \text{A}$

The capacitor sits across node A to ground, and node A is the top of $R_2$ (the inductor drops nothing): $v_C = v_A = i_L R_2 = 2 \times 8 = 16\ \text{V}$

Cross-check the other way: $v_A = 24 - i_L R_1 = 24 - 2(4) = 16$ V. ✓

$E_C = \tfrac12 C v_C^2 = \tfrac12(50\times10^{-6})(16)^2 = 6.4\ \text{mJ}$ $E_L = \tfrac12 L i_L^2 = \tfrac12(2)(2)^2 = 4\ \text{J}$

i_L = 2

v_C = 16

E_C = 6.4

mJ,

E_L = 4

J. Notice

i_C = 0

and

v_L = 0

— the zeros land on current for the cap and voltage for the inductor.

Era 2: deriving the transient from scratch

Here is the circuit the whole chapter turns on. A DC source, a resistor, a capacitor, and a switch that closes at $t = 0$ with the capacitor uncharged.

Series RC. Switch closes at

t=0

; capacitor starts uncharged,

v(0) = 0

Write KVL around the loop for $t > 0$ , using the capacitor's own element law for the current:

V_s = i(t)R + v(t), \qquad i(t) = C\frac{dv}{dt} \quad\Longrightarrow\quad RC\frac{dv}{dt} + v = V_s

One first-order linear ODE. Every problem in slides 4.1–4.3 reduces to this shape.

Now the method the course uses: guess a form, substitute, and let the algebra tell you the constants. Guess

v(t) = k_1 + k_2 e^{st}

A constant plus a decaying exponential — three unknowns,

k_1

k_2

, and

s

Differentiate the guess. $\dfrac{dv}{dt} = k_2\,s\,e^{st}$ .

Substitute into the ODE.

$RC\left(k_2 s e^{st}\right) + \left(k_1 + k_2 e^{st}\right) = V_s$

Group the constant terms and the $e^{st}$ terms separately.

$\underbrace{\left(k_1 - V_s\right)}_{\text{constant}} \;+\; \underbrace{k_2 e^{st}\left(RCs + 1\right)}_{\text{time-varying}} \;=\; 0$

This is the load-bearing step. The equation must hold at every instant $t$ , and $e^{st}$ is not a constant — so the constant part and the $e^{st}$ part can't cancel each other. Each group must vanish on its own.

Constant group ⇒ $k_1$ . $k_1 - V_s = 0 \Rightarrow \boxed{k_1 = V_s}$ .

$k_1$ is whatever's left when the exponential dies, i.e. $k_1 = v(\infty)$ — the new steady-state value. That is exactly the Era 3 number.

Exponential group ⇒ $s$ . Since $k_2 \ne 0$ and $e^{st} \ne 0$ , we need $RCs + 1 = 0$ :

$s = -\frac{1}{RC} = -\frac{1}{\tau}, \qquad \tau \equiv RC$

$s$ comes only from $R$ , $C$ , and the circuit's shape — never from the source value or the initial condition. Note $s$ is negative, which is what makes the transient decay rather than blow up.

Initial condition ⇒ $k_2$ . Apply $v(0) = 0$ to the now-partly-solved guess $v(t) = V_s + k_2 e^{-t/\tau}$ :

$v(0) = V_s + k_2 e^{0} = V_s + k_2 = 0 \;\Rightarrow\; \boxed{k_2 = -V_s}$

$k_2$ is always applied last, after $k_1$ and $s$ are known. Applying the initial condition before you have $k_1$ is the single most common algebra error here.

Assemble.

$v(t) = V_s - V_s e^{-t/\tau} = V_s\left(1 - e^{-t/\tau}\right), \qquad t \ge 0$

and the current follows from the element law:

$i(t) = C\frac{dv}{dt} = C \cdot \frac{V_s}{\tau}e^{-t/\tau} = \frac{V_s}{R}e^{-t/\tau}$

Sanity check: at $t = 0^+$ the uncharged capacitor looks like a short, so the current should be $V_s/R$ — and it is. At $t \to \infty$ the capacitor is fully charged and the current is zero. Both limits check out. ✓

The general result — memorize this one

Nothing in steps 3–6 used the fact that the source was a voltage source, or that the element was a capacitor. Repeat the derivation on any single-energy-storage circuit and you get the same shape, which is why one formula covers the whole chapter:

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

x

v_C

i_L

. Read it as: final value, plus the leftover error, decaying away. Matching to the derivation,

k_1 = x(\infty)

and

k_2 = x(0^+) - x(\infty)

You need	Where it comes from	Formula
$x(0^+)$	Era 1 — DC analysis before the switch, then continuity	$v_C(0^+) = v_C(0^-)$ ; $i_L(0^+) = i_L(0^-)$
$x(\infty)$	Era 3 — DC analysis after the switch (caps open, inductors short)	whatever the new resistive network gives
$\tau$	The Thévenin resistance seen from the storage element's own terminals, in the post-switch circuit, with independent sources deactivated	$\tau = R_{\text{Th}}C \;$ or $\; \tau = L/R_{\text{Th}}$

Finding τ properly

$\tau = RC$ is shorthand for " $\tau = R_{\text{Th}}C$ ." Pull the capacitor (or inductor) out, kill the independent sources — voltage sources become shorts, current sources become opens — and find the equivalent resistance looking back into the two terminals you just vacated. In a plain series loop that is just $R$ ; in anything with a divider it is not, and using the wrong resistor is the most common way to get a right-shaped answer with a wrong exponent. Units check: $\Omega \cdot \text{F} = \text{s}$ , and $\text{H}/\Omega = \text{s}$ .

Quick check · which R sets τ

A $2\ \mu$ F capacitor sits from node A to ground. A 20 V source in series with $5\ \text{k}\Omega$ feeds node A, and a $20\ \text{k}\Omega$ resistor runs from node A to ground. What is $\tau$ ?

What τ actually looks like

The time constant is the only number that sets the shape of the curve. Slide 4.1 gives the two landmarks worth memorizing: after one $\tau$ a decaying quantity has fallen to $e^{-1} = 36.8\%$ of its starting value (equivalently, a rising one has covered $63.2\%$ of its journey), and after $5\tau$ only $e^{-5} = 0.7\%$ remains — close enough to call it settled.

Both curves share one

\tau

. The tangent at

t=0

hits the asymptote exactly at

t=\tau

— a fast way to read

\tau

off a scope trace.

A useful inversion: if you know two points on the curve you can solve for $\tau$ (or for whatever else is unknown) by taking logs. Given $v(t) = V_0e^{-t/\tau}$ ,

\tau = \frac{t}{\ln\!\big(V_0/v(t)\big)}, \qquad\text{and in particular the half-life is } t_{1/2} = \tau\ln 2 \approx 0.693\,\tau

Worked example 2 — solving for an unknown R (Exercise 8, P4.11)log inversion

A $10\ \mu$ F capacitor is connected at $t=0$ to a voltmeter, which behaves as a resistor $R$ to ground. The meter reads 50 V at $t = 0$ and 25 V at $t = 30$ s. Find $R$ .

The capacitor discharges through

R

— the voltmeter's own input resistance — once the switch closes: a pure decay from

50

V toward

0

No source in the post-switch circuit, so $v(\infty) = 0$ and the general formula collapses to pure decay: $v(t) = 50\,e^{-t/\tau}$ .

The reading halved, so we are exactly at the half-life: $25 = 50e^{-30/\tau} \;\Rightarrow\; e^{-30/\tau} = 0.5 \;\Rightarrow\; \frac{30}{\tau} = \ln 2 = 0.693$ $\tau = \frac{30}{0.693} = 43.3\ \text{s}$

$R = \frac{\tau}{C} = \frac{43.3}{10\times10^{-6}} = 4.33\times10^{6}\ \Omega = 4.33\ \text{M}\Omega$

Sanity check: a voltmeter should have a huge input resistance so it barely disturbs the circuit it measures. Megohms is exactly right; had we gotten kilohms, the arithmetic would be suspect.

R = 4.33\ \text{M}\Omega

(equivalently

\tau = 43.3

s).

Play with τ

Drag the sliders. Watch two things: the shape of the curve never changes — it is always the same exponential — but the $\tau$ marker slides along the time axis, dragging the whole curve with it. Note also that changing $R$ in the RL circuit moves $\tau$ the opposite way from changing $R$ in the RC circuit, because $\tau = RC$ but $\tau = L/R$ .

τ Explorer — first-order response

R = 5.0 kΩ

C = 4.0 µF

τ = 20.0 ms

5τ (settled) = 100 ms

final value = 10.0 V

t = 0.0 ms

value now = 0.00 V

% complete = 0.0 %

The RL case

Swap the capacitor for an inductor and every step of the derivation repeats with the roles of voltage and current exchanged. The continuous quantity is now $i_L$ , the time constant is $\tau = L/R_{\text{Th}}$ , and the general formula is unchanged.

Before

t=0

: inductor is a short, so

i_L(0^-) = V_s/R_1

and node A sits at

0

V. After the switch opens,

i_L

can only circulate through

R_2

, so

\tau = L/R_2

and

i_L \to 0

The RL duality, in one line each

RC: $v_C$ is continuous · cap = open in steady state · $\tau = R_{\text{Th}}C$ · at $t=0^+$ the cap acts as a voltage source $v_C(0^-)$ .

RL: $i_L$ is continuous · inductor = short in steady state · $\tau = L/R_{\text{Th}}$ · at $t=0^+$ the inductor acts as a current source $i_L(0^-)$ .

The RL circuit above also shows off the "everything else can jump" half of the continuity rule. Before the switch opens, the inductor is a short, so node A sits at $0$ V. The instant the switch opens, $i_L$ must still be $V_s/R_1$ — but that current now has to flow up through $R_2$ from ground, which forces $v_A = -i_L R_2$ immediately. If $R_2 \gg R_1$ that is a very large negative spike out of nowhere. This is exactly why a relay coil needs a flyback diode.

Quick check · the t = 0⁺ substitution

An inductor has been carrying $3$ A in steady state. At $t = 0$ the circuit around it changes. For the purpose of computing every other voltage and current at the instant $t = 0^+$ , the inductor is best replaced by:

Worked example 3 — a current-source charge (Exercise 8, P4.9)Norton form

A 1 mA DC current source, a $10\ \text{k}\Omega$ resistor, and a $1\ \mu$ F capacitor are all in parallel. A switch closes at $t=0$ ; the capacitor starts uncharged. Find $v(t)$ .

Norton form: current source in parallel with

R

and

C

. Closing the switch is equivalent to suddenly applying the Thévenin-equivalent 10 V source through

10\ \text{k}\Omega

$x(0^+)$ : given uncharged, $v(0^-) = 0$ , so by continuity $v(0^+) = 0$ .

$x(\infty)$ : in steady state the capacitor is open, so all 1 mA flows through the $10\ \text{k}\Omega$ : $v(\infty) = I_s R = (1\times10^{-3})(10\times10^{3}) = 10\ \text{V}$

This is just the Norton-to-Thévenin conversion: the source pair is equivalent to a 10 V source behind $10\ \text{k}\Omega$ , which is the circuit we already derived.

$\tau$ : deactivate the current source (open circuit), leaving only the $10\ \text{k}\Omega$ across the capacitor: $\tau = R_{\text{Th}}C = (10^{4})(10^{-6}) = 10\ \text{ms} \;\Rightarrow\; 1/\tau = 100\ \text{s}^{-1}$

Assemble: $v(t) = 10 + (0 - 10)e^{-100t} = 10\left(1 - e^{-100t}\right)\ \text{V},\quad t \ge 0$

Cross-check by the $k_1/k_2/s$ route: KCL gives $10^{-3} = v/10^4 + 10^{-6}\,dv/dt$ , i.e. $10^{-2}\,dv/dt + v = 10$ . Substituting $v = k_1 + k_2e^{st}$ gives $k_1 = 10$ and $10^{-2}s + 1 = 0 \Rightarrow s = -100$ ; then $v(0)=0 \Rightarrow k_2 = -10$ . Same answer. ✓

v(t) = 10\left(1 - e^{-100t}\right)

V. It reaches 63.2% (6.32 V) at

t = 10

ms and is settled by

t = 50

ms.

The recipe

Step	Do this	Watch out for
1	Redraw the circuit for $t < 0$ . Caps open, inductors short. Solve for $v_C(0^-)$ / $i_L(0^-)$ .	"Has been closed a long time" is the phrase that licenses this. If the problem doesn't say it, don't assume it.
2	Carry only $v_C$ or $i_L$ across $t=0$ by continuity.	Don't carry currents through capacitors or voltages across inductors — those jump.
3	Redraw the circuit for $t > 0$ in its new configuration.	Branches that got disconnected are gone entirely; don't leave them in.
4	In the new circuit, solve the DC steady state for $v_C(\infty)$ / $i_L(\infty)$ .	If the source got disconnected, the final value is $0$ — but only if there's no other source.
5	Find $R_{\text{Th}}$ at the storage element's terminals in the new circuit; get $\tau$ .	Deactivate independent sources first. Use $\tau = R_{\text{Th}}C$ or $L/R_{\text{Th}}$ .
6	Write $x(t) = x(\infty) + [x(0^+)-x(\infty)]e^{-t/\tau}$ , valid for $t \ge 0$ .	State the range of validity. The expression is wrong for $t < 0$ .
7	Get any other quantity from $x(t)$ by the element law or Ohm's law.	$i_C = C\,dv_C/dt$ , $v_L = L\,di_L/dt$ . Differentiating brings down a factor of $-1/\tau$ — keep the minus sign.
8	Check the two endpoints: does your expression give $x(0^+)$ at $t=0$ and $x(\infty)$ as $t\to\infty$ ?	Ten seconds, catches most sign and constant errors.

Practice

Three problems, all original, all cross-solved two independent ways before shipping. Work them before you open the solutions.

Problem 1 — RC discharge after a switch throwRC · ~8 min

A 24 V source in series with $R_1 = 4\ \text{k}\Omega$ has been connected to a $C = 5\ \mu$ F capacitor for a long time (switch at position A). At $t = 0$ the switch throws to position B, disconnecting the source and connecting the capacitor across $R_2 = 6\ \text{k}\Omega$ .

Before

t=0

the switch sits at A, charging

C

through

R_1

; at

t=0

it throws to B, disconnecting the source and letting

C

discharge through

R_2

(a) Find $v_C(t)$ for $t \ge 0$ . (b) At what time does $v_C$ fall to 6 V? (c) How much energy does $R_2$ dissipate in total?

(a) Initial condition (Era 1). With the switch at A the circuit has settled, so the capacitor is an open circuit. No current flows, so no voltage drops across $R_1$ , and the full source appears across the capacitor: $v_C(0^-) = 24$ V. By continuity $v_C(0^+) = 24$ V.

Final value and τ (Era 3). After the throw, the only elements left in the loop are $C$ and $R_2$ — no source — so $v_C(\infty) = 0$ , and $\tau = R_2C = (6\times10^{3})(5\times10^{-6}) = 0.03$ s $= 30$ ms. Note $R_1$ plays no part after $t=0$ ; it isn't in the circuit any more.

Route 1 — general formula. $v_C(t) = 0 + (24 - 0)e^{-t/0.03} = 24e^{-33.33t}\ \text{V}$

Route 2 — $k_1/k_2/s$ from the ODE. KVL around the $R_2C$ loop for $t > 0$ : $v_C + R_2 i = 0$ with $i = C\,dv_C/dt$ taken in the passive direction, giving $R_2C\,\dfrac{dv_C}{dt} + v_C = 0$ , i.e. $0.03\,\dot v + v = 0$ . Substituting $v = k_1 + k_2e^{st}$ : the constant group gives $k_1 = 0$ (there is no forcing term), the exponential group gives $0.03s + 1 = 0 \Rightarrow s = -33.33\ \text{s}^{-1}$ , and $v(0) = 24 \Rightarrow k_2 = 24$ . So $v_C(t) = 24e^{-33.33t}$ V. Both routes agree. ✓

(b) $6 = 24e^{-t/0.03} \Rightarrow e^{-t/0.03} = 0.25 \Rightarrow t = 0.03\ln 4 = 0.03(1.386) = 41.6$ ms. Cross-check: $6$ V is a quarter of $24$ V, i.e. two halvings, so $t = 2t_{1/2} = 2(0.03)(0.693) = 41.6$ ms. ✓

(c) Energy conservation: with no source connected, everything stored in the capacitor ends up in $R_2$ : $E = \tfrac12 Cv_C(0)^2 = \tfrac12(5\times10^{-6})(24)^2 = 1.44\ \text{mJ}$ Cross-check by integrating the power: $\int_0^\infty \frac{v_C^2}{R_2}dt = \frac{576}{6000}\int_0^\infty e^{-2t/0.03}dt = 0.096\left(\frac{0.03}{2}\right) = 1.44\ \text{mJ}$ ✓

(a)

v_C(t) = 24e^{-33.33t}

t \ge 0

· (b)

t = 41.6

ms · (c)

E = 1.44

Problem 2 — RL decay and the voltage spikeRL · ~10 min

A 12 V source in series with $R_1 = 2\ \Omega$ feeds node A. From node A, an inductor $L = 0.4$ H runs to ground, and in parallel with it a resistor $R_2 = 6\ \Omega$ also runs from node A to ground. The switch has been closed for a long time. At $t=0$ the switch opens, removing the source and $R_1$ entirely.

Before

t=0

L

is a short, so

i_L(0^-) = 12/2 = 6

A and node A sits at

0

V. After the switch opens,

i_L

can only circulate through

R_2

, so

\tau = L/R_2

and

i_L \to 0

(a) Find $i_L(0^+)$ . (b) Find $i_L(t)$ for $t \ge 0$ , with $i_L$ defined positive flowing from node A down through $L$ . (c) Find the node-A voltage $v_A(0^+)$ and comment on it. (d) How much energy was stored in $L$ at $t=0$ ?

(a) Era 1. Long-settled DC: the inductor is a short circuit, so node A is tied to ground, $v_A(0^-) = 0$ . With $0$ V across $R_2$ , no current goes that way, so the source current all goes through $L$ : $i_L(0^-) = \frac{12}{R_1} = \frac{12}{2} = 6\ \text{A}$ By continuity, $i_L(0^+) = 6$ A.

(b) Route 1 — general formula. After the switch opens, $L$ and $R_2$ form a closed loop with no source, so $i_L(\infty) = 0$ , and $\tau = L/R_2 = 0.4/6 = 0.0667$ s, i.e. $1/\tau = 15\ \text{s}^{-1}$ : $i_L(t) = 0 + (6-0)e^{-15t} = 6e^{-15t}\ \text{A}$

Route 2 — $k_1/k_2/s$ from the ODE. KVL around the $L$ – $R_2$ loop: $L\,\dfrac{di}{dt} + R_2 i = 0$ , i.e. $0.4\,\dot i + 6i = 0$ . Substituting $i = k_1 + k_2e^{st}$ : the constant group gives $6k_1 = 0 \Rightarrow k_1 = 0$ ; the exponential group gives $0.4s + 6 = 0 \Rightarrow s = -15\ \text{s}^{-1}$ ; and $i(0) = 6 \Rightarrow k_2 = 6$ . So $i_L(t) = 6e^{-15t}$ A. Both routes agree. ✓

(c) At $t=0^+$ the inductor still forces $6$ A downward out of node A. That current can only be resupplied through $R_2$ , flowing up from ground into node A. KCL at node A: $i_L + v_A/R_2 = 0$ , so $v_A(0^+) = -i_L(0^+)R_2 = -(6)(6) = -36\ \text{V}$ Cross-check via the element law: $v_L = L\,di_L/dt = 0.4\big(6\big)(-15)e^{-15t} = -36e^{-15t}$ V, which at $t=0$ is $-36$ V — and $v_L$ is $v_A$ here since the inductor's bottom end is grounded. ✓

Comment: node A jumped from $0$ V to $-36$ V instantaneously, a swing three times the source voltage. Inductor current is continuous; nothing forbids the voltage from jumping, and with a larger $R_2$ the spike would be larger still. This is the mechanism behind ignition coils and behind why switching an inductive load without a flyback diode destroys transistors.

(d) $E = \tfrac12 Li_L(0)^2 = \tfrac12(0.4)(6)^2 = 7.2$ J. Cross-check by integrating the power in $R_2$ : $\int_0^\infty i^2R_2\,dt = 36(6)\int_0^\infty e^{-30t}dt = 216/30 = 7.2$ J. ✓

(a)

6

A · (b)

i_L(t) = 6e^{-15t}

t \ge 0

· (c)

v_A(0^+) = -36

V (a legal jump) · (d)

7.2

Problem 3 — charging with a divider (τ is not what it looks like)RC · ~10 min

A 20 V source in series with $R_1 = 5\ \text{k}\Omega$ feeds node A. A resistor $R_2 = 20\ \text{k}\Omega$ runs from node A to ground, and a capacitor $C = 2\ \mu$ F also runs from node A to ground. The capacitor is initially uncharged; the switch closes at $t = 0$ .

R_1

and

R_2

form a divider once

C

is open (Era 3); before that,

\tau

needs

R_1 \parallel R_2

— the point of this problem.

(a) Find $v_C(t)$ for $t \ge 0$ . (b) Find the capacitor current $i_C(t)$ . (c) How long until $v_C$ reaches 90% of its final value?

(a) Route 1 — the three ingredients.

$x(0^+)$ : uncharged, so $v_C(0^+) = v_C(0^-) = 0$ .

$x(\infty)$ : capacitor open, so $R_1$ and $R_2$ form a plain voltage divider: $v_C(\infty) = 20\cdot\frac{R_2}{R_1+R_2} = 20\cdot\frac{20}{25} = 16\ \text{V}$

$\tau$ : deactivate the 20 V source (short it). Looking back from the capacitor's terminals, $R_1$ and $R_2$ are now in parallel: $R_{\text{Th}} = \frac{(5)(20)}{5+20} = 4\ \text{k}\Omega, \qquad \tau = R_{\text{Th}}C = (4\times10^{3})(2\times10^{-6}) = 8\ \text{ms}, \quad 1/\tau = 125\ \text{s}^{-1}$ This is the trap: neither $5\ \text{k}\Omega$ alone nor $20\ \text{k}\Omega$ alone is the right resistance.

$v_C(t) = 16 + (0-16)e^{-125t} = 16\left(1 - e^{-125t}\right)\ \text{V}$

Route 2 — $k_1/k_2/s$ straight from KCL. At node A (with $v \equiv v_C$ ): $\frac{v-20}{5000} + \frac{v}{20000} + 2\times10^{-6}\frac{dv}{dt} = 0$ Multiply through by $20000$ : $\;4(v-20) + v + 0.04\,\dot v = 0$ , i.e. $0.04\,\dot v + 5v = 80$ . Substitute $v = k_1 + k_2e^{st}$ :

constant group: $5k_1 = 80 \Rightarrow k_1 = 16$ ✓ (matches the divider);
exponential group: $0.04s + 5 = 0 \Rightarrow s = -125\ \text{s}^{-1}$ ✓ (matches $-1/R_{\text{Th}}C$ );
initial condition: $v(0) = k_1 + k_2 = 0 \Rightarrow k_2 = -16$ ✓.

$v_C(t) = 16 - 16e^{-125t} = 16\left(1-e^{-125t}\right)\ \text{V}.$ Both routes agree. ✓

(b) $i_C = C\,\dfrac{dv_C}{dt} = (2\times10^{-6})(16)(125)e^{-125t} = 4\times10^{-3}e^{-125t}$ A $= 4e^{-125t}$ mA.

Cross-check at $t=0^+$ independently: an uncharged capacitor is a momentary short, so node A is pinned to $0$ V. Then $R_2$ carries no current, and everything the source pushes through $R_1$ goes into the capacitor: $i_C(0^+) = 20/5000 = 4$ mA. ✓ And at $t\to\infty$ the formula gives $0$ , as it must for a fully charged capacitor. ✓

(c) $0.9 = 1 - e^{-t/\tau} \Rightarrow e^{-t/\tau} = 0.1 \Rightarrow t = \tau\ln 10 = 8\ \text{ms}\times 2.303 = 18.4$ ms. (Compare: fully settled at $5\tau = 40$ ms.)

(a)

v_C(t) = 16\left(1-e^{-125t}\right)

V · (b)

i_C(t) = 4e^{-125t}

mA · (c)

t = 18.4

Error checklist before you hand in

① Is the exponent negative? A positive $s$ means you dropped a sign — first-order passive circuits never grow. ② Did you use $R_{\text{Th}}$ at the element's terminals, not just the nearest resistor? ③ Did you use the post-switch circuit for $\tau$ and $x(\infty)$ , and the pre-switch circuit only for $x(0^-)$ ? ④ Does $t=0$ in your expression reproduce $x(0^+)$ ? ⑤ Did you state "for $t \ge 0$ "? ⑥ If you differentiated to get $i_C$ or $v_L$ , did the factor $-1/\tau$ come along?