Second-Order Circuits

The word irreducible matters. Two capacitors that can be combined into one by the rules of Section 6.4 give a first-order circuit, not a second-order one. What is required is a capacitor and an inductor, or two storage elements that cannot be merged.

A second-order equation needs two initial conditions, not one. This is the single greatest source of difficulty in the chapter, and it is worth confronting before any differential equation appears.

Throughout this chapter, unless stated otherwise, \(v\) denotes capacitor voltage and \(i\) denotes inductor current — the two continuous variables. The quantities normally required are

The two continuity conditions remain the foundation, exactly as in Chapter 7:

Take care with polarity: apply the passive sign convention consistently to the capacitor voltage and the inductor current, and do not change the reference direction halfway through a problem.

The derivative conditions are the new work, and they come from the element relations of Chapter 6 read backwards:

In words: to find the initial rate of change of capacitor voltage, find the initial capacitor current; to find the initial rate of change of inductor current, find the initial inductor voltage. Both of those are obtained by analysing the circuit at \(t = 0^{+}\) with the now-known values of \(v(0^{+})\) and \(i(0^{+})\) treated as a fixed voltage source and a fixed current source respectively.

Final values are easier: at \(t = \infty\) the circuit is in DC steady state again, so the capacitor becomes an open circuit and the inductor a short circuit, exactly as in Section 6.8.

Consider a series loop of \(R\), \(L\) and \(C\) with no source, in which the capacitor is initially charged to \(V_{0}\) and the inductor carries an initial current \(I_{0}\).

Derivation. Applying KVL around the loop, with the capacitor voltage written as an integral of the current:

The integral is awkward, so differentiate the whole equation with respect to time and divide by \(L\):

A second-order homogeneous differential equation with constant coefficients. Its solution is assumed exponential, \(i = A e^{st}\), and substituting gives

Neither \(A\) nor \(e^{st}\) can vanish if a solution is to exist, so the bracket must be zero. This is the characteristic equation:

Solving by the quadratic formula gives the two natural frequencies:

Everything now depends on the sign of the quantity under the square root, \(\alpha^{2}-\omega_{0}^{2}\). Three cases arise, and they are qualitatively different.

Now put \(R\), \(L\) and \(C\) in parallel. By duality — the theme of Section 8.9 — everything follows the same pattern with current and voltage exchanged.

Derivation. Applying KCL at the top node, with the inductor current written as an integral:

Differentiating and dividing by \(C\):

The characteristic equation is \(s^{2} + s/RC + 1/LC = 0\), with the same structure as before but a different \(\alpha\):

Note the crucial difference. In the series circuit \(\alpha = R/2L\) rises with \(R\); in the parallel circuit \(\alpha = 1/2RC\) falls as \(R\) rises. A large resistance damps a series circuit heavily but a parallel circuit hardly at all — which makes sense, since a large parallel resistance draws little current and therefore dissipates little. The resonant frequency \(\omega_{0}\) is unchanged.

The three response forms are identical in shape, with \(v\) in place of \(i\):

The second initial condition now comes from KCL at \(t = 0\):

Apply a DC source suddenly and the equation acquires a constant on the right-hand side:

The solution is the sum of two parts, exactly as in Chapter 7:

The steady-state or forced part is simply the final value found from DC analysis, and the natural part takes whichever of the three forms the damping dictates. So:

The constants must be evaluated from the complete response, not from the natural part alone. This is where most errors occur: \(A_{1}\) and \(A_{2}\) are found by imposing \(v(0)\) and \(dv(0)/dt\) on the whole expression including \(V_{s}\).

By duality once more, applying a step of current \(I_{s}\) to a parallel RLC circuit gives

where \(i\) is the inductor current. The complete response is

with \(\alpha = 1/2RC\) and \(\omega_{0} = 1/\sqrt{LC}\) as in Section 8.5. The final value is \(I_{s}\), since in DC steady state the inductor becomes a short circuit and takes the entire source current, leaving nothing for \(R\) or \(C\).

The dual pairing is worth writing out explicitly, because it saves memorising a second set of results:

Many second-order circuits are neither purely series nor purely parallel, and no ready-made formula for \(\alpha\) applies. The method then reverts to first principles, but the structure of the solution is unchanged.

1. Choose a convenient variable — usually a capacitor voltage or inductor current — and apply Kirchhoff's laws to obtain a second-order differential equation in it.

2. Read \(\alpha\) and \(\omega_{0}\) off by comparing with the standard form \(\ddot{x} + 2\alpha\dot{x} + \omega_{0}^{2}x = f\).

3. Classify, write the response form, and evaluate the constants exactly as before.

A useful shortcut for source-free circuits: if the network reduces, by series and parallel combination alone, to a single loop or a single node pair, then the standard formulas apply using the equivalent \(R\), \(L\) and \(C\). Only when it does not reduce is the full derivation needed.

Duality has been used informally since Chapter 6. It can be made systematic, and doing so turns every solved problem into two.

Two circuits are duals if the mesh equations of one have the same form as the node equations of the other. Solve one and the other's solution follows by substituting the paired quantities.

Constructing a dual. For a planar network:

1. Place a node at the centre of each mesh, and one external node outside the network.

2. Draw a line between each pair of nodes, crossing exactly one element, and place the dual of that element on the line.

3. Assign source polarities so that a clockwise mesh current corresponds to a current source directed into the node.

The result explains everything observed in this chapter. The series RLC circuit and the parallel RLC circuit are duals, which is why \(\alpha = R/2L\) becomes \(\alpha = 1/2RC = G/2C\) under the substitutions \(R\to G\) and \(L\to C\). Not a coincidence but a theorem.

The classical ignition circuit is a second-order transient exploited deliberately. An ignition coil of inductance \(L\) and winding resistance \(R\) is connected across the 12 V battery through the contact breaker points, and current builds up according to the RL step response of Chapter 7. When the points open, that current must find another path — and the only one available is the small capacitor deliberately placed across the points.

The inductor and capacitor now form a series RLC circuit with almost no resistance, so the response is heavily underdamped. All the energy stored in the magnetic field transfers into the capacitor's electric field, and since \(\tfrac{1}{2}Li^{2} = \tfrac{1}{2}Cv^{2}\) at the moment of complete transfer, the peak voltage is

The quantity \(\sqrt{L/C}\) has units of ohms and is called the characteristic impedance of the LC pair. It reappears throughout Part 2.

• A second-order circuit has two irreducible storage elements and needs two initial conditions.

• Initial values come from continuity; initial derivatives come from \(dv(0)/dt = i_{C}(0)/C\) and \(di(0)/dt = v_{L}(0)/L\); final values from DC steady state.

• The characteristic equation is \(s^{2}+2\alpha s+\omega_{0}^{2}=0\), with roots \(s_{1,2} = -\alpha \pm \sqrt{\alpha^{2}-\omega_{0}^{2}}\).

• For a series circuit \(\alpha = R/2L\); for a parallel circuit \(\alpha = 1/2RC\). In both, \(\omega_{0} = 1/\sqrt{LC}\).

• Three cases: overdamped (\(\alpha>\omega_{0}\), two real roots), critically damped (\(\alpha=\omega_{0}\), repeated root, needs the \(t\) factor), underdamped (\(\alpha<\omega_{0}\), complex pair, oscillates at \(\omega_{d}\)).

• Critical damping reaches the final value fastest without overshoot.

• The damping ratio \(\zeta = \alpha/\omega_{0}\) determines the shape of the response; \(\omega_{0}\) determines its speed.

• A step response is the final value plus the natural response, with the constants evaluated from the complete expression.

• Series and parallel RLC circuits are duals, as are voltage and current, \(R\) and \(G\), \(L\) and \(C\), KVL and KCL.

• Using \(\alpha = R/2L\) for a parallel circuit. The parallel value is \(1/2RC\), and the two behave oppositely as \(R\) changes. Identify the topology before reaching for a formula.

• Writing \(A_{1}e^{-\alpha t}+A_{2}e^{-\alpha t}\) for the critically damped case. That collapses to a single constant and cannot satisfy two initial conditions. The correct form is \(\left(A_{1}+A_{2}t\right)e^{-\alpha t}\).

• Evaluating the constants from the natural response alone in a step problem. The initial conditions apply to the complete response, final value included.

• Forgetting the product rule when differentiating. Both \(\left(A_{1}+A_{2}t\right)e^{-\alpha t}\) and \(e^{-\alpha t}\cos\omega_{d}t\) are products; differentiating only one factor loses a term.

• Confusing \(\omega_{0}\) with \(\omega_{d}\). The circuit oscillates at \(\omega_{d}\), which is always less than \(\omega_{0}\). They coincide only when \(R=0\).

• Assuming two storage elements always give a second-order circuit. Two capacitors that combine into one give a first-order circuit. The elements must be irreducible.

• Applying the derivative condition before the continuity condition. You need \(v(0^{+})\) and \(i(0^{+})\) first; the derivatives are computed from the circuit at \(t=0^{+}\) using those values.

• Treating \(dx(0)/dt = 0\) as meaning the variable is constant. It means the response is momentarily at a turning point, nothing more.

• Mixing radians and degrees in the phase angle. When writing \(Ae^{-\alpha t}\cos(\omega_{d}t+\phi)\), the argument \(\omega_{d}t\) is in radians; state the units of \(\phi\) explicitly.

• Expecting critical damping to be achievable in practice. It requires an exact equality between component values that have tolerances. Real designs aim slightly to one side, usually a little underdamped.

Work these before opening the answers. Always check \(t=0\) and \(t\to\infty\) against your assembled expression.

1. P8.1 A series RLC circuit has \(R = 10~\Omega\), \(L = 0.5~\mathrm{H}\) and \(C = 2~\mathrm{mF}\). Find \(\alpha\), \(\omega_{0}\), \(\zeta\), and classify the response.

   Show answer

   \[\alpha = \frac{10}{2(0.5)} = 10, \qquad \omega_{0} = \frac{1}{\sqrt{(0.5)(0.002)}} = \frac{1}{\sqrt{0.001}} = 31.62\]

   \[\zeta = \frac{10}{31.62} = 0.3162 \;\Longrightarrow\; \text{underdamped}, \qquad \omega_{d} = \sqrt{1000-100} = 30~\mathrm{rad/s}\]

2. P8.2 A source-free series RLC circuit has \(R = 25~\Omega\), \(L = 1~\mathrm{H}\), \(C = 10~\mathrm{mF}\), with \(i(0) = 2~\mathrm{A}\) and \(v(0) = 0\). Find \(i(t)\).

   Show answer

   \[\alpha = 12.5, \quad \omega_{0} = 10 \;\Longrightarrow\; \text{overdamped}, \quad s_{1,2} = -12.5 \pm 7.5 = -5, \; -20\]

   \[\frac{di(0)}{dt} = -\frac{(25)(2)+0}{1} = -50\]

   From \(A_1+A_2 = 2\) and \(-5A_1-20A_2 = -50\): \(A_1 = -2/3\), \(A_2 = 8/3\).

   \[i(t) = -\tfrac{2}{3}e^{-5t} + \tfrac{8}{3}e^{-20t}~\mathrm{A}\]

3. P8.3 What value of \(R\) gives critical damping in a series circuit with \(L = 4~\mathrm{H}\) and \(C = 1~\mathrm{mF}\)?

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   Setting \(\alpha = \omega_{0}\) gives \(R/2L = 1/\sqrt{LC}\), so

   \[R = 2\sqrt{\frac{L}{C}} = 2\sqrt{\frac{4}{0.001}} = 2\sqrt{4000} = 126.5~\Omega\]

4. P8.4 A parallel RLC circuit has \(R = 10~\Omega\) and \(C = 5~\mathrm{mF}\). What \(L\) makes it critically damped?

   Show answer

   Critical damping requires \(L = 4R^{2}C\):

   \[L = 4(100)(0.005) = 2.000~\mathrm{H}\]

   Check: \(\alpha = 1/(2\times10\times0.005) = 10\) and \(\omega_{0} = 1/\sqrt{(2)(0.005)} = 10\) \(\checkmark\)

5. P8.5 A 20 V step is applied to a series circuit with \(R = 6~\Omega\), \(L = 1~\mathrm{H}\), \(C = 40~\mathrm{mF}\), initially at rest. Find \(v(t)\) across the capacitor.

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   \[\alpha = 3, \quad \omega_{0} = 5 \;\Longrightarrow\; \text{underdamped}, \quad \omega_{d} = 4\]

   \[v(t) = 20 + e^{-3t}\left(A_{1}\cos 4t + A_{2}\sin 4t\right)\]

   With \(v(0)=0\): \(A_1 = -20\). With \(dv(0)/dt = i(0)/C = 0\): \(-3(-20)+4A_2 = 0\), so \(A_2 = -15\).

   \[v(t) = 20 - e^{-3t}\left(20\cos 4t + 15\sin 4t\right)~\mathrm{V}\]

   At \(t = 0.5~\mathrm{s}\): \(v = 18.81~\mathrm{V}\).

6. P8.6 For P8.5, find the damping ratio and the percentage overshoot.

   Show answer

   \[\zeta = \frac{3}{5} = 0.6000\]

   \[\%\,\text{OS} = 100\,e^{-\pi(0.6)/\sqrt{1-0.36}} = 100\,e^{-\pi(0.6)/0.8} = 100\,e^{-2.356} = 9.478\,\%\]

7. P8.7 A circuit has \(v(0^{-}) = 6~\mathrm{V}\) across a 0.2 F capacitor and \(i(0^{-}) = 3~\mathrm{A}\) through a 0.5 H inductor. At \(t=0^{+}\) the capacitor current is 3 A and the inductor voltage is \(-4~\mathrm{V}\). Find both initial derivatives.

   Show answer

   \[\frac{dv(0^{+})}{dt} = \frac{i_{C}(0^{+})}{C} = \frac{3}{0.2} = 15.00~\mathrm{V/s}\]

   \[\frac{di(0^{+})}{dt} = \frac{v_{L}(0^{+})}{L} = \frac{-4}{0.5} = -8.000~\mathrm{A/s}\]

8. P8.8 Construct the dual element values for a parallel RLC circuit with \(R = 4~\Omega\), \(L = 0.5~\mathrm{H}\), \(C = 0.25~\mathrm{F}\), and verify that \(\alpha\) and \(\omega_{0}\) match.

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   Original: \(\alpha = 1/(2\times4\times0.25) = 0.5\), \(\omega_{0} = 1/\sqrt{0.125} = 2.828\).
   Dual series circuit: \(R' = 1/4 = 0.25~\Omega\), \(L' = 0.25~\mathrm{H}\), \(C' = 0.5~\mathrm{F}\).

   \[\alpha' = \frac{0.25}{2(0.25)} = 0.5 \;\checkmark, \qquad \omega_{0}' = \frac{1}{\sqrt{(0.25)(0.5)}} = \frac{1}{\sqrt{0.125}} = 2.828 \;\checkmark\]

9. P8.9 An underdamped circuit has \(\alpha = 2\) and \(\omega_{d} = 6~\mathrm{rad/s}\). Find \(\omega_{0}\), \(\zeta\), and the interval between successive zero crossings.

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   \[\omega_{0} = \sqrt{\alpha^{2}+\omega_{d}^{2}} = \sqrt{4+36} = \sqrt{40} = 6.325~\mathrm{rad/s}\]

   \[\zeta = \frac{2}{6.325} = 0.3162, \qquad \Delta t = \frac{\pi}{\omega_{d}} = \frac{\pi}{6} = 0.5236~\mathrm{s}\]

10. P8.10 An ignition coil has \(L = 8~\mathrm{mH}\) carrying 4 A, with \(C = 0.5~\mu\mathrm{F}\) across the points. Find the stored energy and the peak voltage.

    Show answer

    \[w = \tfrac{1}{2}\left(8\times10^{-3}\right)(16) = 64.00~\mathrm{mJ}\]

    \[V_{\text{peak}} = I\sqrt{\frac{L}{C}} = 4\sqrt{\frac{8\times10^{-3}}{0.5\times10^{-6}}} = 4\sqrt{16\,000} = 4(126.5) = 505.9~\mathrm{V}\]