First-Order Circuits

Write KVL round the loop, using the capacitor's element law from Set 17:

A first-order linear differential equation with constant coefficients — and this is the only kind a single-storage-element circuit can produce.

Separate the variables:

Identify the constants physically. As \(t \to \infty\) the exponential vanishes, so \(V_s = v_C(\infty)\); and at \(t = 0\), \(v_C(0^+) = V_s + A\). Hence

Nothing about the derivation was specific to this circuit. Any first-order circuit produces the same equation with different constants, so this formula covers every problem in the set.

Apply it here. The capacitor starts uncharged and ends at the source voltage:

The 90% point:

Note that \(t_{90} = 2.303\tau\) whatever the circuit — the fraction reached depends only on \(t/\tau\), never on the individual values.

The two-term reading. The solution splits naturally into the parts named in Set 11, Problem 18:

The forced term is what the sources impose; the natural term is the circuit's own decaying response, and its shape is fixed by \(\tau\) alone regardless of what drives the circuit.

The time constant is set by the resistance the capacitor sees, not by any single resistor in the circuit. Deactivate the independent source — short the 10 V battery — and look back from the capacitor's terminals:

Hence

Why this is Thévenin's theorem doing the work. Set 9 reduced any linear network at a terminal pair to \(V_{Th}\) behind \(R_{Th}\). Do that at the capacitor's terminals and the circuit becomes a single loop — exactly Problem 1's circuit — with

So every first-order RC circuit, however complicated, is Problem 1 after one Thévenin reduction.

The common error is to use the series resistor alone, giving \(\tau = RC\) instead of \(\tfrac23RC\) — a 50% error. The shunt \(2R\) provides a second discharge path and speeds the circuit up.

A useful check: adding any resistor in parallel with the capacitor always reduces \(R_{Th}\) and hence shortens \(\tau\). Adding one in series with the storage element lengthens it. If a change makes the circuit slower when intuition says faster, the wrong resistance has been used.

Initial value. For \(t < 0\) the circuit is in DC steady state, so the capacitor is an open circuit (Set 17) and the source divides across the two resistors:

By continuity, \(v(0^+) = 4\ \text{V}\) as well.

After switching the source and the 10 kΩ are disconnected. The capacitor discharges through the 2 kΩ alone:

Substituting into the general formula, with a final value of zero:

Note which resistor appears in each calculation. The 10 kΩ sets the initial value but plays no part in the time constant, because the switch removed it. Using \(R_{Th} = 12\ \text{k}\Omega\) would be a common and serious error:

This is the natural response — the circuit's behaviour with no sources at all, driven only by stored energy. Its shape is a pure decaying exponential, and the time constant is the only thing that distinguishes one such circuit from another.

Initial value, with the capacitor open in DC steady state:

The source-free response through the remaining 3 kΩ:

The one-third point. The initial amplitude cancels, because the fraction remaining depends only on \(t/\tau\):

The universal table, which applies to every first-order circuit and is worth committing to memory:

"Five time constants" is the usual engineering definition of steady state, at 99.3%. Useful landmarks: half-life is \(\tau\ln2 = 0.693\tau\), and the 10%–90% rise takes \(\tau\ln9 = 2.20\tau\) — Problem 19.

Why the initial value cancels. Any question of the form "how long to reach a given fraction?" has an answer depending on \(\tau\) alone:

Here \(\ln 3 = 1.0986\), so any first-order circuit falls to a third in 1.1 time constants — whether it started at 15 V or 15 kV.

The discharge current:

Note the initial current is \(4/5 = 0.8\ \text{A}\) — the capacitor voltage is continuous, so the resistor immediately sees the full 4 V.

Charge is the integral of current, and the interval is exactly \(1\tau\) to \(4\tau\):

Check by charge directly, which avoids the integral entirely. Since \(q = Cv\):

Identical, and it must be: \(CV_0 = 20\ \mu\text{C}\) and \((V_0/R)\tau = (V_0/R)(RC) = CV_0\) are the same quantity.

The total charge available is \(CV_0 = 20\ \mu\text{C}\), so this interval accounts for 35% of it. The first time constant alone carries 63%, and by \(4\tau\) only 1.8% remains.

A note on the current's discontinuity. Before the switch closes the current is zero; immediately after, it is 0.8 A. Nothing forbids this — only the capacitor voltage must be continuous (Set 17, Problem 13), and it is.

Initial value. The capacitor was charged to the 1 V source and carries that voltage across the switching instant:

Final value. With the new source connected and the capacitor open in DC steady state, the 3 V divides across the 1 Ω and 2 Ω:

Time constant. Deactivating the 3 V source puts the two resistors in parallel at the capacitor's terminals:

The response:

Reading the result. The capacitor is charging from 1 V towards 2 V, and at \(0.75\tau\) it has covered 53% of the 1 V gap. Checking against Problem 4's table: at \(\tau\) it would have covered 63.2%, so 53% at \(0.75\tau\) is consistent.

Why the initial value is not zero here — and why that matters. Many textbook problems start from rest, making the response \(x(\infty)(1-e^{-t/\tau})\). This one does not, and using that simplified form would give \(2(1-e^{-0.75}) = 1.06\ \text{V}\), badly wrong. The general formula handles both cases; the simplified one is a special case worth not memorising separately.

Convert the charge to a voltage, since voltage is the state variable:

Final value and time constant:

The capacitor voltage:

The current, from the element law:

Check the initial current independently. At \(t = 0^+\) the capacitor holds 20 V, so the resistor sees the difference:

This is always worth doing. The current at \(0^+\) can be read straight off the circuit by replacing the capacitor with a voltage source equal to \(v_C(0^+)\) — which is the substitution theorem of Set 14, Problem 10.

The current could have been written directly. The general formula applies to any quantity, so with \(i(0^+) = 40\) and \(i(\infty) = 0\):

— no differentiation needed. Every quantity in a first-order circuit shares the same \(\tau\); only the two endpoint values differ.

A source-free discharge:

At exactly one time constant:

36.8% of the initial value — the defining property of \(\tau\), matching Problem 4's table exactly.

The energy dissipated is the difference in stored energy, since the capacitor is the only source:

Out of a total stored energy of 50 µJ. So 86.5% of the energy is gone after one time constant, even though 36.8% of the voltage remains.

Why energy decays twice as fast. Energy goes as \(v^2\), so

an exponential with time constant \(\tau/2\). The same is true of power and of any squared quantity — a distinction worth watching, since "the circuit has a time constant of 10 ms" refers to the voltage, not the energy.

Practical note. This is the standard discharge circuit fitted across power-supply capacitors as a bleeder resistor. Sizing it is a trade: a small \(R\) discharges quickly but wastes power continuously in normal operation, while a large one may leave a dangerous charge for minutes after the supply is switched off.

The charging current, after reducing the bridge by delta–wye transformation (Set 2):

The energy delivered is the integral of the source's power:

Now show the resistance is irrelevant. The source holds \(V\) constant while the total charge \(Q = CV\) passes through it, so

No \(R\) appears. A different bridge, or a different reduction, would change the time constant but not the energy.

The accounting, which is Set 17, Problem 17 arriving as a transient calculation:

Exactly half is lost, whatever the resistance and however the bridge is arranged.

Confirming the dissipation by integration, as a check on the claim:

The \(R\) cancels, as it must.

The current at \(0^+\), found by treating the capacitor as a 6 V source at that instant:

Final value and time constant:

The energy absorbed:

Note the exponent: squaring \(e^{-t/8}\) gives \(e^{-t/4}\), so the power decays with time constant \(\tau/2 = 4\ \text{s}\) — the point made in Problem 8.

Check by energy balance, which is independent of the integration. The capacitor charges from 6 V to 10 V, so the charge delivered is

Two independent routes agreeing — always worth doing when an integral is involved.

Note the efficiency here is 80%, not 50%. Problem 9's half-and-half result assumed charging from zero. Starting at 6 V means less of the journey is spent at a large voltage difference across the resistor, and less is wasted. Charging in stages exploits exactly this — Set 17, Challenge C2.

The three numbers. The state variable is now the current (Set 16), and in DC steady state the inductor is a short:

The response:

The inductor voltage:

Check at \(t = 0^+\). The inductor voltage starts at 24 V — the entire source voltage:

which must be so: the current is still zero, so the resistor drops nothing and KVL puts all 24 V across the inductor. An inductor with zero initial current behaves momentarily as an open circuit.

The two elements' switching behaviour is worth tabulating, since it supplies the initial values in every problem:

Each behaves at the first instant as the opposite of what it becomes in steady state — provided it starts from rest. With a non-zero initial condition, replace it instead by a source of the appropriate value, as in Problem 7.

Initial value. For \(t < 0\) the inductor is a short in DC steady state, which also shorts out the 16 Ω. The 4 Ω and 12 Ω are then in parallel:

Current division gives the inductor's share — the opposite resistance on top, as in Set 10:

After the switch opens the source is gone and the inductor drives the remaining loop:

The natural response:

Note the role reversal of the 16 Ω. While the switch was closed it was shorted out and contributed nothing; once open, it becomes the discharge path and sets the time constant. The 2 Ω, conversely, mattered before and not after:

Why this matters practically. Removing the 16 Ω would leave the inductor current nowhere to go, forcing a very large \(L\,di/dt\) across the opening switch — the inductive kick of Set 16, Problem 19. The 16 Ω is doing the job of a freewheeling path.

Before switching the inductor is a short and the 6 Ω branch is disconnected:

After switching, the inductor drives the 3 Ω and 6 Ω in parallel:

The inductor voltage, and from it the other two quantities:

Assembling all three for all time:

Which quantities jump. Only \(i\) is continuous, at 2 A. The others step at \(t = 0\):

Check the jump in \(v_o\): at \(0^+\) the inductor still carries 2 A, which now divides between the 3 Ω and 6 Ω, giving \(2 \times 2 = 4\ \text{V}\) across the parallel pair — matching the formula.

The negative sign on \(i_o\) says the 6 Ω current flows opposite to its assumed reference. That is physically right: the inductor is now the source, driving current backwards through a branch that previously carried none.

Method 1 — a test source. Dependent sources must not be deactivated, so \(R_{Th}\) requires the test-source technique of Set 9. Apply \(v_o = 1\ \text{V}\) at the inductor's terminals:

Substituting the second into the first:

Method 2 — write the differential equation directly, as a check that does not use Thévenin at all:

The coefficient \(\tfrac23\) is \(1/\tau\), agreeing with Method 1. The two routes are the same calculation organised differently.

Then \(i_x\) follows from the inductor voltage across the 2 Ω:

The point of interest is \(R_{Th} = \tfrac13\ \Omega\), far smaller than any resistor in the circuit. The dependent source is supplying current in support of the test source, which lowers the effective resistance — the same effect Set 9, Problem 13 found. Had the dependence been reversed in sign, \(R_{Th}\) could have been negative, and the "decaying" exponential would have grown instead.

Three intervals, each a separate first-order problem linked by continuity of \(i_L\).

a\(t \le 0\): both switches open, so \(i(0^-) = i(0^+) = 0\).

b\(0 \le t \le 4\): only \(S_1\) closed, giving 4 Ω and 6 Ω in series:

Note \(i(\infty) = 4\ \text{A}\) is the value this interval is heading towards, not one it reaches — the second switch intervenes first.

c\(t \ge 4\): the new initial value comes from continuity:

Eight time constants have passed, so the first transient is essentially complete.

The new final value, by KCL at the junction node \(P\) with the inductor shorted:

The current falls — adding the 10 V source through 2 Ω loads the node down, since 10 V is below the 16.4 V the node would otherwise reach.

The new time constant:

Shift the exponential by the 4 s delay — the single most-missed step in these problems:

Writing \(e^{-1.467t}\) instead of \(e^{-1.467(t-4)}\) would make the response start decaying from \(t=0\) and give nonsense at \(t=4\).

Evaluating:

The two differential equations, side by side:

Identical in form. Everything proved for one holds for the other under the dictionary of Set 17, Problem 20.

The correspondence:

Why \(R\) appears in opposite places. The time constant is always \(\tau = (\text{storage})/(\text{dissipation rate})\). For a capacitor the resistance limits the current that charges it, so more \(R\) means slower. For an inductor the resistance dissipates its stored energy, so more \(R\) means faster:

This is not an inconsistency but the duality: \(R \leftrightarrow G\), so \(\tau = RC\) becomes \(\tau = GL = L/R\).

Converting Problem 3. There a capacitor charged to 4 V discharged through 2 kΩ with \(\tau = 80\ \text{ms}\), giving \(v = 4e^{-12.5t}\ \text{V}\). The dual circuit has an inductor carrying 4 A decaying through a conductance, and by inspection:

For example \(L = 160\ \text{H}\) with \(R = 2\ \text{k}\Omega\), or more practically \(L = 80\ \text{mH}\) with \(R = 1\ \Omega\). No re-derivation was required.

A practical asymmetry. The two are equivalent mathematically but not physically. RC time constants of seconds are trivial to obtain; an RL circuit with \(\tau = 1\ \text{s}\) would need, say, \(L = 1\ \text{H}\) with \(R = 1\ \Omega\) — a large, heavy coil whose own winding resistance would probably exceed 1 Ω and spoil the design. Long time constants are the capacitor's territory, which is why timing circuits are almost always RC.

Take the ratio of two samples, which eliminates the unknown amplitude:

Back out the initial value:

Note that \(V_0\) was never measured — it is inferred, which is useful when the start of the transient was missed or obscured by switching noise.

Three practical methods:

Why the tangent method works. Differentiating the decay at the origin:

so a line of that slope from \(V_0\) reaches zero after exactly \(\tau\). Equivalently: if the initial rate continued unchanged, the transient would finish in one time constant. That is the cleanest physical definition of \(\tau\).

The best method for noisy data is none of these: plot \(\ln v\) against \(t\), which should be a straight line of slope \(-1/\tau\), and fit it. This uses every sample rather than two, and a visible curvature in the log plot is immediate evidence that the circuit is not first order — which is Problem 18.

The test. A circuit's order equals the number of independent energy-storage elements — that is, the number remaining after all possible series and parallel combinations have been made. Equivalently, it is the number of independent initial conditions the circuit requires.

Applying it:

The delta case is the subtle one: three elements, but the loop constraint \(v_1+v_2+v_3 = 0\) removes one degree of freedom. Counting elements is not always counting order.

The general rule for such constraints. A capacitor loop (a closed path of capacitors and voltage sources) removes one order; an inductor cutset (a set of inductors and current sources whose removal splits the circuit) likewise removes one:

These are exactly the degenerate configurations of Set 17, Problem 13, where an impulsive current was needed — the circuit has fewer free initial conditions than it appears to.

Why order matters. Only a first-order circuit has the single-exponential response of this set:

The diagnostic. A first-order response can never overshoot its final value or oscillate — it moves monotonically from \(x(0^+)\) to \(x(\infty)\). Observing any overshoot in a circuit believed to be first order means a storage element has been overlooked, very often a stray capacitance or a lead inductance.

The rise time. For a step response \(1-e^{-t/\tau}\), solve for the 10% and 90% crossings:

The bandwidth relation. An \(RC\) low-pass has its \(-3\ \text{dB}\) point at \(f_c = 1/2\pi\tau\), so

The standard rule of thumb: a 350 MHz oscilloscope has a 1 ns rise time. Speed in the time domain and bandwidth in the frequency domain are the same specification, and Sets 28 and 30 will make the connection general.

The pulse response. During a pulse of width \(T\) the output rises towards the final value; when the pulse ends it decays back:

The two limiting behaviours are worth naming, since they connect to Set 15, Problem 19:

Taking the output across the resistor instead gives the complementary pair: a differentiator for \(T \gg \tau\), and faithful transmission for \(T \ll \tau\).

The design consequence. To pass a pulse train of period \(T\) without appreciable distortion requires \(\tau \lesssim T/10\). Every stray capacitance in a digital circuit forms such an \(RC\) with the driving resistance, and it is this that sets the maximum clock rate — not the transistors' switching speed.

The procedure, with its dependencies:

Only step 5 is new to this set. Everything else is Part 1 applied three times over.

The single genuinely new idea is that a circuit now has a state — one number carrying information across the switching instant. Before Set 16 a circuit's response depended only on its present sources; now it depends on its history, compressed into one variable.

What a second storage element changes. The differential equation becomes second order, and its solution can no longer be written down by inspection:

The second initial condition is the awkward part. Where this set needed only \(v_C(0^+)\), a second-order circuit needs \(dv_C/dt\) at \(0^+\) as well — which must be obtained from the other element's initial condition through the element law:

So both continuity results are needed together, and finding initial conditions becomes most of the work.

What does not change. Kirchhoff's laws, linearity, superposition, Thévenin, and the division into natural and forced response all survive unaltered. The natural response is still the circuit's own decaying behaviour and the forced response still what the sources impose — there are simply two natural modes instead of one.