Inductors

The current follows from the integral form. Since \(v = 0\) for \(t < 0\), the inductor starts with \(i(0) = 0\):

Route 1 — integrate the power. The instantaneous power absorbed is

Route 2 — use the state directly. The stored energy depends only on the present current:

Agreement is guaranteed here because \(i(0) = 0\). In general the two differ by the initial stored energy — Problem 5.

Why the second route is better. Integrating the power requires knowing \(v\) and \(i\) over the whole interval; the energy formula needs only the current at the one instant asked about. The inductor's state is its current, and everything about its stored energy follows from that alone.

Note the growth rates. A voltage rising as \(t^2\) integrates to a current rising as \(t^3\), and the energy — being quadratic in current — rises as \(t^6\). Integration always smooths: the current is a degree gentler than the voltage that produced it.

Faraday's law states that a changing magnetic flux linkage \(\lambda\) induces a voltage:

For an \(N\)-turn coil, \(\lambda = N\Phi\) — the flux \(\Phi\) counted once for every turn it links.

Inductance is defined as the constant of proportionality between flux linkage and the current producing it:

So \(L\) measures how much flux linkage a coil produces per ampere — a purely geometric and magnetic property, fixed by turns, area, length and core material.

Combining the two, and assuming \(L\) is constant:

The constancy of \(L\) is exactly the linearity assumption of Sets 9 to 13. A saturable iron core has \(L\) falling with current, and every theorem in this book fails for it.

The energy. Integrate the power absorbed from a state of zero current:

The change of variable from \(t\) to \(i\) is the key step, and it is why the result depends only on the final current and not on how the current got there.

The unit. From \(L = \lambda/i\), one henry is one weber-turn per ampere; equivalently, from \(v = L\,di/dt\), one volt-second per ampere. A 1 H inductor develops 1 V when its current changes at 1 A/s.

The voltage requires the product rule:

The stored energy:

When does absorption stop? The inductor absorbs while its stored energy is rising, so set \(dw/dt = 0\) — or equivalently note that the voltage changes sign when

which is also where the current peaks, since \(v = L\,di/dt = 0\) there.

The three quantities are locked together:

The peak energy is \(5(0.2)^2e^{-2} = 27.07\ \text{mJ}\).

Note the sign of the power. Negative power does not mean anything is broken — it means the inductor is delivering energy back to the circuit, drawing on what it stored earlier. Problem 15 shows that over the whole waveform it returns every joule.

Integrate, keeping the initial condition explicit:

Apply the condition at \(t_0 = -\pi/2\), where \(\sin(-5\pi/2) = -1\):

Check by substituting \(t = -\pi/2\): \(0.6(-1) + 1.6 = 1\ \text{A}\;\checkmark\), and by differentiating: \(2 \times 0.6 \times 5\cos 5t = 6\cos 5t\;\checkmark\)

Reading the result. The current has two parts, and they are physically different:

The sinusoidal part is determined entirely by \(v\) and \(L\); the constant is determined entirely by history. A purely sinusoidal voltage does not guarantee a purely sinusoidal current — any DC offset present at the start persists for ever, since an ideal inductor has no resistance to decay it.

Amplitude relation. The current amplitude is \(6/(2 \times 5) = 0.6\ \text{A}\), that is \(V_m/\omega L\). The quantity \(\omega L = 10\ \Omega\) is the inductive reactance of Problem 17 — the inductor opposes high-frequency current more strongly.

The current:

The current has reversed. The voltage changes sign at \(t = 1\ \text{s}\), after which it drives the current down through zero and negative. The current peaks where \(v = 0\):

The energy stored at \(t = 4\):

Positive, despite the negative current — energy depends on \(i^2\) and cannot be negative. The direction of the current is irrelevant to how much magnetic energy is stored.

The energy absorbed over the interval is a different quantity:

Why they differ — by exactly 4 J. The inductor already held \(\tfrac12(2)(2)^2 = 4\ \text{J}\) at \(t = 0\). The circuit supplied 320 J during the interval; the total present at the end is 324 J:

The two questions to keep apart are therefore:

First interval, \(0 < t < 1\), with \(1/L = 100\):

Second interval, \(1 < t < 2\). The new starting value is the previous interval's final value — not zero:

Verify continuity and the end value:

The current returns exactly to zero, because the voltage waveform has zero net area — the positive area from 0 to 1 s exactly cancels the negative area from 1 to 2 s.

Collecting the result:

Two checks worth making on any piecewise integration: that the pieces agree at every boundary (current is continuous — Problem 14), and that a zero-net-area voltage returns the current to its starting value. Both hold here.

Compute the slopes in A/s, not A/ms — this is where the arithmetic usually goes wrong:

Apply the element law with \(L = 5 \times 10^{-3}\ \text{H}\):

The voltage is a rectangular waveform even though the current is trapezoidal — differentiation of straight-line segments gives constants, and the corners of the current become steps in the voltage:

Note the flat portion. With 10 A flowing steadily the inductor holds \(\tfrac12(0.005)(100) = 0.25\ \text{J}\), yet develops no voltage whatever. A constant current, however large, produces nothing — this is the DC steady state of Problem 13, arrived at from the waveform side.

And note the corners. The voltage jumps instantaneously at \(t = 2\) and \(4\ \text{ms}\) because the current's slope changes discontinuously. Inductor voltage may jump freely; only its current may not.

Take each interval in turn, computing \(di/dt\) and multiplying by \(L = 3\ \text{H}\).

a\(t < -1\ \text{s}\): the current is zero and constant, so \(v = 0\).

b\(-1 < t < 0\): the current rises 1 A in 1 s, a slope of \(+1\ \text{A/s}\):

c\(0 < t < 2\): constant current, so \(v = 0\) — even though 1 A is flowing and 1.5 J is stored.

d\(2 < t < 3\): the current falls 1 A in 1 s, a slope of \(-1\ \text{A/s}\):

e\(t > 3\): zero and constant again, so \(v = 0\).

Check the areas. The voltage waveform has area \(3 \times 1 = 3\ \text{V·s}\) positive and \(3 \times 1\) negative, summing to zero — consistent with the current starting and ending at zero, since \(\Delta i = \frac1L\int v\,dt\).

Integrate from the instant where the current is known:

The \(0.01\) is \((-0.1)^2\) — the lower limit, which is easy to lose.

aAt \(t = 0\):

bAt \(t = 1.5\ \text{ms} = 0.0015\ \text{s}\):

Barely different from \(i(0)\), because \(t^2\) is still negligible against 0.01 at these times.

cAt \(t = 45\ \text{ms} = 0.045\ \text{s}\):

Reading the signs. All three currents are negative, and the reason is the interval before \(t = 0\): there \(v = 4.3t\) is negative, so it drove the current downwards from its initial 100 µA to \(-10.65\ \text{mA}\). After \(t = 0\) the voltage is positive and the current climbs back, reaching \(-8.47\ \text{mA}\) at 45 ms — still negative, because it has not yet recovered.

The passive sign convention means \(i\) is taken entering the terminal at which \(v\) is positive. With that choice \(p = vi\) is the power absorbed, and the element law is \(v = +L\,di/dt\). Reversing either reference reverses the sign of the whole answer.

aConstant 5 V:

A ramp — and one that grows without limit. Holding a constant voltage across an ideal inductor drives its current up for ever, at 833 A per second here.

This is not a paradox but a warning. An ideal inductor has no resistance, so nothing opposes the steady growth. In any real circuit the winding resistance, the source's internal resistance, or magnetic saturation intervenes long before the current becomes large — which is the subject of Set 18.

bSinusoidal drive:

Read the two parts of this result. The current is a cosine of amplitude 44.21 A plus a DC offset of the same size:

The current never goes negative — it oscillates between 0 and 88.4 A. The offset appears because the voltage was switched on at a zero crossing of the sine, so the first half-cycle contributes a full positive area with nothing before it to cancel.

The amplitude is \(V_m/\omega L = 100/(120\pi \times 0.006) = 44.21\ \text{A}\), confirming the reactance relation of Problem 17.

The current is the running integral of the voltage. Over each half-period the voltage is constant, so the integral is a straight line:

Over the first half-period, \(V = +1\ \text{V}\) for \(\Delta t = 0.5\ \text{ms}\):

A constant voltage gives a constant slope, so the current ramps linearly. Over the second half-period the voltage reverses and the current ramps back down by the same amount.

So the current is a triangular wave of 0.5 A peak, rising and falling in step with the sign of the voltage:

Why the current is periodic here. The square wave has zero average over a full period, so the net area is zero and the current returns exactly to where it began. Any DC component in the applied voltage would instead make the current drift upward cycle after cycle, as in Problem 10(a).

An engineering reading. Peak current is \(V\Delta t/L\) — proportional to the applied volt-seconds and inversely proportional to inductance. Doubling the switching frequency halves \(\Delta t\) and so halves the current ripple, which is precisely why switching converters run at high frequency and why their inductors can then be small.

Series rule. Inductors in series carry the same current, and KVL adds their voltages:

Parallel rule. Inductors in parallel share the same voltage, and KCL adds their currents:

Identical in form to resistors — series add, parallel reciprocals add. Capacitors, as Set 17 will show, do the opposite.

Apply them. The parallel pair first:

Then the three series elements:

A caution the rules conceal. Both derivations assumed each inductor's flux links only itself. If two coils share magnetic flux they are mutually coupled, and the series rule becomes \(L_1 + L_2 \pm 2M\) — Set 26's subject. Physical separation or orthogonal orientation is what makes the simple rules valid.

And a caution about initial conditions. These rules describe the terminal behaviour of the combination. Replacing two parallel inductors by one \(L_{eq}\) discards any information about how the initial current divided between them — which is the substance of Problem 16.

What "a long time" means. All transients have died away, so every current is constant and

Zero volts across the inductor for any current — which is exactly the definition of a short circuit. In DC steady state, replace every inductor by a wire.

Redraw with the short in place. The 6 Ω resistor is now connected between node \(A\) and the reference — but so is the short, so node \(A\) sits at 0 V:

The 6 Ω carries nothing — it has been short-circuited by the inductor, exactly as in Set 7, Problem 6.

So the whole source current flows through the inductor, limited only by the 3 Ω:

The stored energy:

Two things worth noticing. First, the inductance never entered the current calculation — only the energy. In steady state the value of \(L\) is irrelevant to every voltage and current in the circuit. Second, 16 J is now stored, and it has to go somewhere if the source is disconnected. That is the subject of Problem 19, and of Set 18.

The proof. Consider the current change across an interval of length \(\varepsilon\) straddling any instant \(t_0\):

If the voltage is bounded — say \(|v| \le V_{\max}\) — then the integral is bounded too:

The change vanishes as the interval shrinks, so \(i(t_0^+) = i(t_0^-)\). The current is continuous.

The exact condition for a jump. The bound fails only if \(v\) is unbounded — specifically, if it contains an impulse:

An impulsive voltage of strength \(\Lambda\) volt-seconds produces an instantaneous current step of \(\Lambda/L\). Nothing less will do it.

Why the energy argument gives the same answer. Since \(w = \tfrac12Li^2\), a jump in current means a jump in stored energy — which requires infinite power for an instant. Finite sources cannot supply that.

Why voltage is different. The element law is \(v = L\,di/dt\), and nothing constrains \(di/dt\) to be continuous. Problem 7's trapezoidal current has corners, so its voltage steps instantaneously between \(+25\), \(0\) and \(-25\ \text{V}\):

The practical statement used throughout Sets 18 and 19:

The current an instant after a switch operates equals the current an instant before, however violently the rest of the circuit changes.

The instantaneous power from Problem 3:

Positive for \(t < 0.2\ \text{s}\), negative thereafter — for example \(p(0.1) = +0.184\ \text{W}\) and \(p(0.3) = -0.075\ \text{W}\).

Integrating over all time:

Exactly zero. Everything absorbed during the first 0.2 s — a peak store of 27.07 mJ — is returned to the circuit afterwards.

The general proof needs no integration at all. Since the energy is a function of the state,

and here \(i(0) = 0\) while \(i(\infty) = 0\), so the difference vanishes. The waveform in between is irrelevant.

The contrast with a resistor:

What this means for AC circuits. An ideal inductor consumes no average power however hard it is driven. It exchanges energy with the source twice per cycle, and that exchange is what Set 27 will call reactive power — real current, real voltage, no net energy transfer.

The contradiction. Series elements must carry the same current, but Problem 14 says neither inductor's current can change instantaneously:

All three cannot hold. Something in the idealisation must give.

What gives is the boundedness of the voltage. Problem 14 showed a jump requires an impulse, so the connection generates an impulsive voltage across each inductor. The conserved quantity is flux linkage, \(\lambda = Li\), because the two impulses are equal and opposite around the loop:

The common current afterwards:

Between the two original values, weighted by inductance — the larger inductor dominates because it holds more flux per ampere.

The energy accounting. Before:

Where did 2.67 J go? The ideal model cannot say — it has nowhere to put it. In a real circuit the answer depends on what was neglected:

All four are excluded by the ideal model, which is precisely why it predicts an impossible instantaneous transition and loses energy without accounting for it.

Conservation of flux linkage is the right principle here, not conservation of energy. Kirchhoff's voltage law around the loop survives the impulse and forces \(\lambda\) to be conserved; energy is not conserved because the model has an unmodelled dissipation path.

The voltage:

The phase. The voltage leads the current by exactly a quarter cycle. Physically: the voltage is largest where the current is changing fastest, which is at the current's zero crossings, and zero where the current is stationary, at its peaks — the observation already made in Problem 3.

The amplitude ratio defines the inductive reactance:

An inductor passes DC freely and blocks high frequencies — the basis of every choke and filter.

The average power. The instantaneous power is

A sinusoid at twice the supply frequency, symmetric about zero. The inductor absorbs energy for a quarter cycle and returns it the next — Problem 15's losslessness in periodic form.

Note the doubled frequency. Power in a reactive element pulses at \(2\omega\), which is why single-phase supplies produce 100 Hz hum from a 50 Hz mains, and why three-phase systems — whose total power is constant — are preferred for machines.

A real inductor is three elements, not one: the inductance itself, the resistance of the wire in series with it, and the capacitance between adjacent turns in parallel with the whole.

Quality factor compares the reactance to the loss:

Equivalently, \(Q = 2\pi \times\) (energy stored)/(energy lost per cycle). A \(Q\) of 126 means the coil behaves as an inductor to about 1% accuracy at this frequency.

Note that \(Q\) depends on frequency. At 50 Hz the same coil has \(X_L = 31.4\ \Omega\) and \(Q = 6.3\) — no longer a good inductor at all, since a fifth of its impedance is resistive.

Self-resonance. The stray capacitance resonates with the inductance at

Above \(f_0\) the capacitance dominates and the component behaves as a capacitor. A coil used above its self-resonant frequency is not an inductor in any useful sense.

The usable band is therefore bounded at both ends:

Two further effects not captured even by this three-element model: the skin effect raises \(R\) with frequency, so \(Q\) rises more slowly than \(X_L\) suggests; and a ferromagnetic core adds hysteresis and eddy-current losses, and saturates at high current, making \(L\) itself current-dependent — which destroys linearity and with it every theorem of Sets 9 to 13.

The estimate. If the current really did fall from 1 A to zero in 1 µs:

A hundred kilovolts from a coil that was running on a few volts.

Why the ideal model says infinity. An ideal switch opens in zero time, so \(di/dt \to \infty\). The model is being asked to force a discontinuity in inductor current, which Problem 14 showed requires an impulsive voltage — and it obliges.

What actually happens. The stored energy

must go somewhere in a few microseconds. The voltage rises until it finds a path — typically by breaking down the air gap between the opening contacts, producing an arc. The arc is the circuit's solution: it maintains a conducting path so the current can decay at a rate the coil permits, rather than being forced to zero.

The consequences are entirely practical: contacts erode and eventually weld; nearby semiconductors see voltages far beyond their ratings; and the fast edge radiates broadband interference.

The remedies all work by providing a deliberate path for the current:

A flyback diode reduces 100 kV to under a volt, at the cost of a slower release of the relay — the 50 mJ is then dissipated slowly in the coil's own resistance.

The effect is also useful. A boost converter, an ignition coil and a camera flash all work by interrupting an inductor's current deliberately, using exactly this mechanism to produce a voltage far higher than the supply.

What survives unchanged. Everything derived from Kirchhoff's laws and linearity — which is nearly everything:

What is genuinely new. Three things: the element law involves a derivative; the circuit has a memory, held in the inductor current; and energy can be stored and recovered rather than only dissipated.

The duality with the capacitor, which Set 17 will make explicit. Every statement about an inductor becomes a true statement about a capacitor under the substitutions of Set 8:

Problem 16's flux-linkage argument and Set 17's charge-conservation argument are the same proof read through this dictionary.

Why both elements matter together. An inductor resists changes in current and a capacitor resists changes in voltage. Put both in one circuit and each fights the other, producing the oscillation that Set 19 studies — energy sloshing between magnetic and electric storage at a rate set by \(1/\sqrt{LC}\).