ECE 211 · Circuit Analysis · Interactive Notes

Lecture 9: Applying the Universal First-Order Formula

This page presents the universal first-order response and the four-step procedure that produces it: the steady state before the switch, the continuity of the capacitor voltage and the inductor current, the steady state after the switch, and the time constant obtained from the resistance seen at the terminals of the storage element. Three worked examples carry dependent sources, an ideal op amp, and a step source. The material is reconstructed from the Lecture 9 board notes as interactive figures.

Teal indicates linked content. Hovering over (or tabbing to) a boxed term such as VxV_x highlights the corresponding circuit element; the linkage also operates in the reverse direction, from the circuit to the equations.

§1 · One formula, three numbers

Every first-order response has the same shape

A circuit that contains resistors, sources, and exactly one energy-storage element (one capacitor or one inductor) is governed by a first-order differential equation. When the sources are DC and the circuit is switched at t=0t = 0, every voltage and every current in that circuit follows one formula:

f(t)=[f(0+)−f(∞)] e−t/τ+f(∞)τ=ReqC   or   τ=LReqf(t) = [f(0^+) - f(\infty)]\, e^{-t/\tau} + f(\infty) \qquad\quad \tau = R_{eq} C \;\ \text{or}\ \; \tau = \frac{L}{R_{eq}}

Three numbers therefore determine the entire waveform: the value immediately after the switching instant, the value the circuit settles to, and the time constant. The formula is read directly: at t=0t = 0 the exponential equals one and f=f(0+)f = f(0^+); as tt grows the exponential vanishes and f→f(∞)f \to f(\infty).

The response leaves f(0+)f(0^+) and approaches f(∞)f(\infty) along an exponential whose only free parameter is τ\tau. The indigo marks stand at t=τt = \tau and at t=5τt = 5\tau.

f(t) = [ 12 − ( −4 ) ] e−t/0.4 + −4 V

f(τ) = 1.88 V   (63.2 % of the change is complete)     5τ = 2.0 s: f = −3.89 V   (99.3 % complete)

The four-step procedure

  1. t=0−t = 0^-: DC steady state. The circuit before the switch is solved for vCv_C or iLi_L, and for nothing else.
  2. t=0+t = 0^+: apply continuity. Knowing vC(0−)v_C(0^-) or iL(0−)i_L(0^-), the storage element is replaced by a source of that value and f(0+)f(0^+) is computed.
  3. t=∞t = \infty: DC steady state again. The circuit after the switch is solved for f(∞)f(\infty).
  4. t>0t > 0: the time constant. The capacitor or the inductor is removed, the resistance RT=ReqR_T = R_{eq} seen at the terminals where it was removed is found, and τ=RTC\tau = R_T C or τ=L/RT\tau = L / R_T.
Why the resistance is measured at the element terminals

For t>0t > 0 the storage element sees the rest of the circuit as a two-terminal network, and Lecture 4 already reduced any such network to its Thevenin equivalent. The exponential rate is set by that equivalent resistance alone, which is why step 4 is a Thevenin resistance computation. When dependent sources are present, the resistance must be obtained from a test source rather than from series and parallel combinations.

One element only

The formula holds for a single equivalent capacitance or a single equivalent inductance. Several capacitors that combine into one equivalent capacitance still qualify; a circuit holding both a capacitor and an inductor does not, and it is second order.

§2 · What the two elements do

Two constitutive laws and one continuity rule

The capacitor and the inductor are dual to each other:

iC=CdvCdtvL=LdiLdti_C = C \frac{dv_C}{dt} \qquad\qquad v_L = L \frac{di_L}{dt}

Both statements are read twice in this lecture: once in DC steady state, where every derivative is zero, and once across the switching instant, where the stored quantity is required to be continuous.

The elements and their laws

The capacitor current is proportional to the rate of change of its voltage, and the inductor voltage is proportional to the rate of change of its current. The stored energies are 12CvC2\tfrac{1}{2} C v_C^2 and 12LiL2\tfrac{1}{2} L i_L^2.

In DC steady state

Nothing changes in DC steady state, so both derivatives vanish. The capacitor carries no current and therefore behaves as an open circuit; the inductor holds no voltage and therefore behaves as a short circuit. This substitution is what makes steps 1 and 3 ordinary resistive problems.

Across the switching instant

A finite current cannot change the capacitor voltage instantaneously, and a finite voltage cannot change the inductor current instantaneously:

vC(0+)=vC(0−)iL(0+)=iL(0−)v_C(0^+) = v_C(0^-) \qquad\qquad i_L(0^+) = i_L(0^-)

At t=0+t = 0^+ the capacitor is therefore replaced by a voltage source of value vC(0−)v_C(0^-) and the inductor by a current source of value iL(0−)i_L(0^-). The rest of the circuit is then resistive again.

Only these two quantities are continuous

Every other quantity is free to jump at t=0t = 0, and in the examples below several of them do. A capacitor current and an inductor voltage jump routinely, because nothing forbids it.

§3 · Worked example A: an RL circuit with a dependent source

Find vx(t)v_x(t)

A 5 A source and a 2 Ω resistor feed the network through a switch that opens at t=0t = 0. The storage element is the 1/6 H inductor, and the network on the right carries a dependent voltage source 2Vx2V_x controlled by the voltage across the 3 Ω resistor.

The problem

Find vx(t)v_x(t) for t>0t > 0. The four steps are applied in order: the inductor current at 0−0^-, the output at 0+0^+, the output at ∞\infty, and the time constant from the resistance seen by the inductor.

Step 1 · t=0−t = 0^-: DC steady state

The inductor is replaced by a short, so node A is pinned to the reference at 0 V and the node between the 3 Ω and the 1 Ω resistors sits at −Vx-V_x. KCL there (currents out) gives

−Vx−03+−Vx−2Vx1+−Vx−05=0  ⟹  −5315Vx=0  ⟹  Vx(0−)=0\tm{ea.r3}{\frac{-V_x - 0}{3}} + \tm{ea.r1 ea.dep}{\frac{-V_x - 2V_x}{1}} + \tm{ea.r5}{\frac{-V_x - 0}{5}} = 0 \;\Longrightarrow\; -\tfrac{53}{15} V_x = 0 \;\Longrightarrow\; V_x(0^-) = 0

No independent source reaches that network while node A is held at 0 V, so it carries nothing. KCL at node A then delivers the quantity that matters:

−5+0−02+iL+0−03=0  ⟹  iL(0−)=5 A\tm{ea.cs5}{-5} + \tm{ea.r2}{\frac{0 - 0}{2}} + \tm{ea.il}{i_L} + \tm{ea.r3}{\frac{0 - 0}{3}} = 0 \;\Longrightarrow\; \boxed{i_L(0^-) = 5\ \text{A}}

Step 2 · t=0+t = 0^+: continuity

The switch has opened, so the 5 A source and the 2 Ω resistor are gone. The inductor current cannot jump, and it becomes a 5 A source pointing downward:

iL(0+)=iL(0−)=5 Ai_L(0^+) = i_L(0^-) = 5\ \text{A}

Node A now touches only the inductor and the 3 Ω resistor, so those 5 A must arrive through the 3 Ω resistor, that is, against the direction of the VxV_x polarity marks. Ohm's law finishes the step:

vx(0+)=(−5 A)⋅(3 Ω)  ⟹  vx(0+)=−15 Vv_x(0^+) = \tm{ea.i3}{(-5\ \text{A})} \cdot \tm{ea.r3}{(3\,\Omega)} \;\Longrightarrow\; \boxed{v_x(0^+) = -15\ \text{V}}

Step 3 · t=∞t = \infty: DC steady state

With the switch open, not one independent source remains in the circuit. The dependent source is not an input: it can only scale what already exists. A linear network holding resistors and dependent sources alone settles at zero.

vx(∞)=0\boxed{v_x(\infty) = 0}

Step 4 · t>0t > 0: the resistance seen by the inductor

The inductor is removed, and a 1 A test source is connected at its terminals; the dependent source makes this the only reliable route to RTR_T. The test current has one path out of node A, namely the 3 Ω resistor, so

Vx=(3 Ω)⋅(1 A)=3 VVB=Vy−VxV_x = \tm{ea.r3}{(3\,\Omega)} \cdot \tm{ea.i3t}{(1\ \text{A})} = 3\ \text{V} \qquad\quad V_B = V_y - V_x

KCL at node B, with VyV_y the voltage at node A:

−1+Vy−Vx−2Vx1+Vy−Vx−05=0  ⟹  −5+6Vy−16Vx=0\tm{ea.i3t}{-1} + \tm{ea.r1 ea.dep}{\frac{V_y - V_x - 2V_x}{1}} + \tm{ea.r5}{\frac{V_y - V_x - 0}{5}} = 0 \;\Longrightarrow\; -5 + 6V_y - 16V_x = 0

Substituting Vx=3V_x = 3 V gives 6Vy=536 V_y = 53, so Vy=53/6V_y = 53/6 V, and the resistance and the time constant follow:

RT=Vy1 A=536 Ω≈8.83 Ωτ=LRT=1/653/6  ⟹  τ=153 sR_T = \frac{V_y}{1\ \text{A}} = \frac{53}{6}\ \Omega \approx 8.83\ \Omega \qquad\quad \tau = \frac{L}{R_T} = \frac{1/6}{53/6} \;\Longrightarrow\; \boxed{\tau = \tfrac{1}{53}\ \text{s}}

Assemble the response

vx(t)=[ −15⏟f(0+)−0⏟f(∞) ]e−53t+0  ⟹  vx(t)=−15e−53t V,t≥0v_x(t) = [\,\underbrace{-15}_{f(0^+)} - \underbrace{0}_{f(\infty)}\,] e^{-53t} + 0 \;\Longrightarrow\; \boxed{v_x(t) = -15 e^{-53t}\ \text{V},\quad t \ge 0}

The response starts at −15-15 V, which the second step produced, and decays to zero with a time constant of about 19 ms; after 5τ≈945\tau \approx 94 ms the circuit is settled for practical purposes.

Check: the inductor current gives the same answer

For t>0t > 0 the inductor and the 3 Ω resistor are in series at node A, so the inductor current is exactly the current arriving through that resistor: vx=−3iLv_x = -3 i_L. The inductor current itself follows the same universal formula, with iL(0+)=5i_L(0^+) = 5 A and iL(∞)=0i_L(\infty) = 0:

iL(t)=5e−53t A  ⟹  vx(t)=−3⋅5e−53t=−15e−53t V ✓i_L(t) = 5 e^{-53t}\ \text{A} \;\Longrightarrow\; v_x(t) = -3 \cdot 5 e^{-53t} = -15 e^{-53t}\ \text{V}\ \checkmark

Two quantities in the same first-order circuit share one time constant; only the two end values differ. Any conflict between them would indicate an arithmetic error.

§4 · Worked example B: an op amp driving an inductor

Find vL(t)v_L(t)

The 1/3 H inductor sits between a resistive node and the inverting input of an ideal op amp, and the dependent source below the node is controlled by the op-amp output vov_o. The switch closes at t=0t = 0 and shorts out the 4 Ω resistor.

The two ideal op-amp rules

No current enters either input, and negative feedback drives the two input voltages together. The non-inverting input is grounded here, so the inverting input is a virtual ground at 0 V at all times, and the entire inductor current must continue through the 12 Ω feedback resistor.

The problem

Find vL(t)v_L(t), the voltage across the inductor, for t>0t > 0. Because the right terminal of the inductor is the virtual ground, the inductor voltage equals the node voltage at its left terminal, which shortens every equation below.

Step 1 · t=0−t = 0^-: DC steady state

The switch is open, so the 2 Ω and 4 Ω resistors are in series, 6 Ω in total. The inductor is a short, so its left node is also held at 0 V. The feedback resistor carries the whole inductor current, which fixes the output:

vo=−(12 Ω)⋅iL  ⟹  3vo=−36iLv_o = -\tm{eb.r12}{(12\,\Omega)} \cdot \tm{eb.ind}{i_L} \;\Longrightarrow\; \tm{eb.dep}{3 v_o} = -36 i_L

KCL at the shorted node (currents out) then gives one equation in one unknown:

0+66+0+36iL4+iL=0  ⟹  1+10iL=0  ⟹  iL(0−)=−110 A\tm{eb.r2 eb.r4}{\frac{0 + 6}{6}} + \tm{eb.r4b}{\frac{0 + 36 i_L}{4}} + \tm{eb.ind}{i_L} = 0 \;\Longrightarrow\; 1 + 10 i_L = 0 \;\Longrightarrow\; \boxed{i_L(0^-) = -\tfrac{1}{10}\ \text{A}}

Step 2 · t=0+t = 0^+: continuity

The switch closes and shorts the 4 Ω resistor, leaving 2 Ω between the −6-6 V source and the node. The inductor becomes a 0.1 A source pointing left, since iL(0+)=iL(0−)=−1/10i_L(0^+) = i_L(0^-) = -1/10 A. That current still runs through the feedback resistor, so

vo=−12⋅(−110)=1.2 V3vo=3.6 Vv_o = -12 \cdot \left(-\tfrac{1}{10}\right) = 1.2\ \text{V} \qquad\quad \tm{eb.dep}{3 v_o} = 3.6\ \text{V}

KCL at the left inductor terminal, whose node voltage is vLv_L:

vL+62+vL−3.64−110=0  ⟹  3vL+8=0  ⟹  vL(0+)=−83 V\tm{eb.r2}{\frac{v_L + 6}{2}} + \tm{eb.r4b}{\frac{v_L - 3.6}{4}} - \tm{eb.i0}{\frac{1}{10}} = 0 \;\Longrightarrow\; 3 v_L + 8 = 0 \;\Longrightarrow\; \boxed{v_L(0^+) = -\tfrac{8}{3}\ \text{V}}

Step 3 · t=∞t = \infty: DC steady state

The quantity sought is an inductor voltage, and in DC steady state the inductor is a short. No computation is required:

vL(∞)=0\boxed{v_L(\infty) = 0}

Step 4 · t>0t > 0: the resistance seen by the inductor

The inductor is removed, the −6-6 V source is turned off (its terminal becomes 0 V), and a 1 V test source is applied across the terminals with its positive mark on the left. The right terminal remains the virtual ground, so the left node sits at exactly 1 V, and the test current IxI_x leaves the op-amp node through the source. The feedback resistor carries that same current:

vo=(12 Ω)⋅Ix  ⟹  3vo=36Ixv_o = \tm{eb.r12}{(12\,\Omega)} \cdot \tm{eb.ix}{I_x} \;\Longrightarrow\; \tm{eb.dep}{3 v_o} = 36 I_x

KCL at the 1 V node:

1−02+1−36Ix4−Ix=0  ⟹  2+1−36Ix−4Ix=0  ⟹  Ix=340 A\tm{eb.r2}{\frac{1 - 0}{2}} + \tm{eb.r4b}{\frac{1 - 36 I_x}{4}} - \tm{eb.ix}{I_x} = 0 \;\Longrightarrow\; 2 + 1 - 36 I_x - 4 I_x = 0 \;\Longrightarrow\; I_x = \tfrac{3}{40}\ \text{A}
RT=1 VIx=403 Ωτ=LRT=1/340/3  ⟹  τ=140 sR_T = \frac{1\ \text{V}}{I_x} = \frac{40}{3}\ \Omega \qquad\quad \tau = \frac{L}{R_T} = \frac{1/3}{40/3} \;\Longrightarrow\; \boxed{\tau = \tfrac{1}{40}\ \text{s}}

Assemble the response

vL(t)=[−83−0]e−40t+0  ⟹  vL(t)=−83e−40t V,t≥0v_L(t) = \left[-\tfrac{8}{3} - 0\right] e^{-40t} + 0 \;\Longrightarrow\; \boxed{v_L(t) = -\tfrac{8}{3} e^{-40t}\ \text{V},\quad t \ge 0}

The time constant is 25 ms, so the inductor voltage has effectively collapsed after about 125 ms, at which point the inductor is again a short and the circuit is a resistive amplifier.

Check: the Thevenin equivalent reproduces vL(0+)v_L(0^+)

For t>0t > 0 the inductor sees a Thevenin source in series with RT=40/3 ΩR_T = 40/3\ \Omega. The open-circuit voltage is obtained by removing the inductor while the −6-6 V source is active: no current then reaches the feedback resistor, so vo=0v_o = 0, the dependent source holds 0 V, and the left node satisfies

V+62+V−04=0  ⟹  VT=−4 V\frac{V + 6}{2} + \frac{V - 0}{4} = 0 \;\Longrightarrow\; V_T = -4\ \text{V}

The inductor current at 0+0^+ is −1/10-1/10 A, so the port voltage at that instant is

vL(0+)=VT−RT iL(0+)=−4+403⋅110=−4+43=−83 V ✓v_L(0^+) = V_T - R_T\, i_L(0^+) = -4 + \tfrac{40}{3} \cdot \tfrac{1}{10} = -4 + \tfrac{4}{3} = -\tfrac{8}{3}\ \text{V}\ \checkmark

The nodal solution of step 2 and the Thevenin equivalent of step 4 agree, which checks both of them at once.

§5 · Worked example C: an RC circuit under a step source

Find v1(t)v_1(t)

Nothing is switched mechanically in this circuit. The source itself steps:

vs=[ 2+10 u(t) ] Vv_s = [\,2 + 10\,u(t)\,]\ \text{V}

The unit step u(t)u(t) is zero for t<0t < 0 and one for t>0t > 0, so the source holds 2 V before the instant and 12 V after it. A source written this way is equivalent to a switch, and the same four steps apply.

The step source jumps by 10 V at t=0t = 0. The 2 V pedestal sets the initial steady state, and the 12 V level sets the final steady state.

The problem

Find v1(t)v_1(t), the voltage across the 5 Ω resistor, for t>0t > 0. The storage element is the 1/10 F capacitor, and the dependent current source delivers 12V1\tfrac{1}{2} V_1 upward.

Step 1 · t=0−t = 0^-: DC steady state

The source stands at 2 V and the capacitor is an open circuit. No current can flow through the 12 Ω resistor into that open branch, so the top plate of the capacitor sits at the source voltage, 2 V. KCL at V1V_1 (currents out):

V1−210+V1−05−12V1=0  ⟹  V1−2+2V1−5V1=0  ⟹  V1(0−)=−1 V\tm{ec.r10}{\frac{V_1 - 2}{10}} + \tm{ec.r5}{\frac{V_1 - 0}{5}} - \tm{ec.dep}{\frac{1}{2} V_1} = 0 \;\Longrightarrow\; V_1 - 2 + 2V_1 - 5V_1 = 0 \;\Longrightarrow\; V_1(0^-) = -1\ \text{V}

KVL down the capacitor branch supplies the quantity that must be carried across the switching instant:

−2+vC+V1=0  ⟹  vC(0−)=3 V-2 + \tm{ec.c}{v_C} + V_1 = 0 \;\Longrightarrow\; \boxed{v_C(0^-) = 3\ \text{V}}

Step 2 · t=0+t = 0^+: continuity and a supernode

The source jumps to 12 V, but the capacitor voltage cannot jump: the capacitor becomes a 3 V source. That source stands between two non-reference nodes, so the two nodes are enclosed by a supernode and the top node is written as V1+3V_1 + 3. One KCL equation over the surface then suffices:

V1+3−1212+V1−1210+V15−12V1=0\tm{ec.r12}{\frac{V_1 + 3 - 12}{12}} + \tm{ec.r10}{\frac{V_1 - 12}{10}} + \tm{ec.r5}{\frac{V_1}{5}} - \tm{ec.dep}{\frac{1}{2} V_1} = 0

Multiplying by 60 gives 5V1−45+6V1−72+12V1−30V1=05V_1 - 45 + 6V_1 - 72 + 12V_1 - 30V_1 = 0, hence −7V1=117-7 V_1 = 117:

v1(0+)=−1177≈−16.71 V\boxed{v_1(0^+) = -\tfrac{117}{7} \approx -16.71\ \text{V}}

The output jumped from −1-1 V to −16.71-16.71 V at the switching instant. Only vCv_C was required to be continuous.

Step 3 · t=∞t = \infty: DC steady state

The capacitor is again an open circuit, the 12 Ω branch again carries nothing, and the source now stands at 12 V:

V1−1210+V15−12V1=0  ⟹  V1−12+2V1−5V1=0  ⟹  v1(∞)=−6 V\tm{ec.r10}{\frac{V_1 - 12}{10}} + \tm{ec.r5}{\frac{V_1}{5}} - \tm{ec.dep}{\frac{1}{2} V_1} = 0 \;\Longrightarrow\; V_1 - 12 + 2V_1 - 5V_1 = 0 \;\Longrightarrow\; \boxed{v_1(\infty) = -6\ \text{V}}

Step 4 · t>0t > 0: the resistance seen by the capacitor

The capacitor is removed, the 12 V source is turned off (a short), and a 1 A test source is applied at the terminals. The test current has one path out of the top terminal, through the 12 Ω resistor to the grounded left node, so that node stands at 12 Ω⋅1 A=1212\,\Omega \cdot 1\ \text{A} = 12 V. KCL at V1V_1, where the same 1 A leaves upward:

1+V1−010+V15−12V1=0  ⟹  10+V1+2V1−5V1=0  ⟹  V1=5 V\tm{ec.test}{1} + \tm{ec.r10}{\frac{V_1 - 0}{10}} + \tm{ec.r5}{\frac{V_1}{5}} - \tm{ec.dep}{\frac{1}{2} V_1} = 0 \;\Longrightarrow\; 10 + V_1 + 2V_1 - 5V_1 = 0 \;\Longrightarrow\; V_1 = 5\ \text{V}
V=12−5=7 V⟹RT=7 V1 A=7 Ωτ=RTC=7⋅110  ⟹  τ=0.7 sV = 12 - 5 = 7\ \text{V} \quad\Longrightarrow\quad R_T = \frac{7\ \text{V}}{1\ \text{A}} = 7\ \Omega \qquad \tau = R_T C = 7 \cdot \tfrac{1}{10} \;\Longrightarrow\; \boxed{\tau = 0.7\ \text{s}}

Assemble the response

v1(t)=[−1177+6]e−t/0.7−6  ⟹  v1(t)=−757e−10t/7−6 V,t≥0v_1(t) = \left[-\tfrac{117}{7} + 6\right] e^{-t/0.7} - 6 \;\Longrightarrow\; \boxed{v_1(t) = -\tfrac{75}{7} e^{-10t/7} - 6\ \text{V}, \quad t \ge 0}

At t=0t = 0 the two terms give −75/7−42/7=−117/7-75/7 - 42/7 = -117/7 V, and as tt grows the response climbs to −6-6 V. This response rises toward its final value, whereas the first two examples decayed to zero; the formula does not distinguish between the two cases.

Check: the capacitor current at 0+0^+

The capacitor current follows from the constitutive law and from KCL, and the two routes must agree. Differentiating the answer and using vC=(V1+3)−V1v_C = (V_1 + 3) - V_1 is awkward, so KCL is used instead. At t=0+t = 0^+ the top node stands at −117/7+3=−96/7-117/7 + 3 = -96/7 V, and the current arriving from the 12 Ω resistor is

iC(0+)=12−(−96/7)12=180/712=157 Ai_C(0^+) = \frac{12 - (-96/7)}{12} = \frac{180/7}{12} = \frac{15}{7}\ \text{A}

The same current is predicted by iC=C dvC/dti_C = C\, dv_C/dt with vC(t)=18−15e−10t/7v_C(t) = 18 - 15 e^{-10t/7} V (the capacitor voltage runs from 3 V to 18 V through the same time constant):

iC(0+)=110⋅15⋅107=157 A ✓i_C(0^+) = \tfrac{1}{10} \cdot 15 \cdot \tfrac{10}{7} = \tfrac{15}{7}\ \text{A}\ \checkmark

§6 · Your turn

Practice: a capacitor with a switch that closes

The switch in the circuit below closes at t=0t = 0, which places the 4 Ω resistor in parallel with the 12 Ω resistor. The four steps should be carried out on paper before the hints are opened, in order.

Find vC(t)v_C(t) for t≥0t \ge 0, and then the current in the 4 Ω resistor. Given: a 24 V source, a 6 Ω resistor, a 12 Ω resistor, a 0.5 F capacitor, and the switched 4 Ω branch.

Hint 1: The two steady states

Before t=0t = 0 the switch is open and the capacitor is an open circuit, so the 24 V source drives a plain voltage divider through the 6 Ω and 12 Ω resistors. After the switch closes, the same divider is computed with 12∥412 \parallel 4 in the lower leg. Note that the capacitor is connected directly across node A, so vCv_C is the node voltage.

Hint 2: The time constant

No dependent source is present, so no test source is needed. The 24 V source is turned off, which shorts it, and the resistance seen at the capacitor terminals is read off directly. All three resistors then share the same two nodes.

Hint 3: The three numbers

vC(0+)=vC(0−)v_C(0^+) = v_C(0^-) from the first divider, vC(∞)v_C(\infty) from the second divider, and τ=RTC\tau = R_T C from the parallel combination. Substituting them into f(t)=[f(0+)−f(∞)]e−t/τ+f(∞)f(t) = [f(0^+) - f(\infty)]e^{-t/\tau} + f(\infty) completes the problem.

Solution

Step 1. With the switch open, vC(0−)=24⋅126+12=16v_C(0^-) = 24 \cdot \dfrac{12}{6 + 12} = 16 V.

Step 2. Continuity gives vC(0+)=16v_C(0^+) = 16 V.

Step 3. With the switch closed, 12∥4=3 Ω12 \parallel 4 = 3\ \Omega, so vC(∞)=24⋅36+3=8v_C(\infty) = 24 \cdot \dfrac{3}{6 + 3} = 8 V.

Step 4. Turning the source off places all three resistors in parallel at the capacitor terminals:

RT=6∥12∥4=(16+112+14)−1=2 Ωτ=RTC=2⋅0.5=1 sR_T = 6 \parallel 12 \parallel 4 = \left(\tfrac{1}{6} + \tfrac{1}{12} + \tfrac{1}{4}\right)^{-1} = 2\ \Omega \qquad \tau = R_T C = 2 \cdot 0.5 = 1\ \text{s}
vC(t)=[16−8] e−t+8=8+8e−t V,t≥0\boxed{v_C(t) = [16 - 8]\, e^{-t} + 8 = 8 + 8 e^{-t}\ \text{V}, \quad t \ge 0}

The 4 Ω resistor is connected across the same node pair, so its current is i4(t)=vC(t)/4=2+2e−ti_4(t) = v_C(t) / 4 = 2 + 2 e^{-t} A: it starts at 4 A and settles at 2 A.

Check. The constitutive law gives iC(0+)=CdvCdt∣0+=0.5⋅(−8)=−4i_C(0^+) = C \left. \dfrac{dv_C}{dt} \right|_{0^+} = 0.5 \cdot (-8) = -4 A. KCL at node A at t=0+t = 0^+ gives the same value:

16−246+1612+164+iC=0  ⟹  −43+43+4+iC=0  ⟹  iC=−4 A ✓\frac{16 - 24}{6} + \frac{16}{12} + \frac{16}{4} + i_C = 0 \;\Longrightarrow\; -\tfrac{4}{3} + \tfrac{4}{3} + 4 + i_C = 0 \;\Longrightarrow\; i_C = -4\ \text{A}\ \checkmark