Exercises — Understand inductance and the henry

The henry is defined by "1 volt from a current changing at 1 amp per second." So $1\text{H}=\frac{1\text{V}}{1\text{A/s}}=1\frac{\text{V}\cdot \text{s}}{\text{A}}.$ Answer: (b).

• (a) is upside-down.
• (c) $\text{C/V}$ is the farad (a capacitor unit — the dual, see Capacitance and the Farad).
• (d) has no per-second, so it can't be an inductance unit.

The one linking voltage to rate of change of current: $V=L\frac{dI}{dt}.$ The symbol $\frac{dI}{dt}$ just means "how many amps the current climbs each second." It is not the current $I$ itself — that distinction is the whole point of an inductor.

Constant current means the current is not changing, so its rate of change is zero: $\frac{dI}{dt}=0\text{A/s}\Rightarrow V=L\cdot 0=0\text{V}.$ An ideal inductor in steady DC is just a plain piece of wire. Answer: 0 V.

"Steadily" means the slope is constant, so $\frac{dI}{dt}=\frac{\Delta I}{\Delta t}=\frac{12-0}{3}=4\text{A/s}.$ Then $V=L\frac{dI}{dt}=0.4\times 4=1.6\text{V}.$

First unit hygiene: $250\text{mH}=0.250\text{H}$. $E=\frac{1}{2}L{I}^2=\frac{1}{2}(0.250)(6^2)=\frac{1}{2}(0.250)(36)=4.5\text{J}.$

Rearrange $E=\frac{1}{2}L{I}^2$ for $I$: $I=\sqrt{\frac{2E}{L}}=\sqrt{\frac{2(0.90)}{0.020}}=\sqrt{90}\approx 9.49\text{A}.$ We take the positive root because current magnitude is what's asked (a negative root would just mean current in the other direction storing the same energy — energy depends on $I^2$).

$L={\mu }_0\frac{N^{2}A}{\ell }=4\pi \times 10^{-7}\cdot \frac{400^2\cdot 2\times 10^{-4}}{0.10}.$ Step by step: $400^2=160000$; times $2\times 10^{-4}$ gives $32$; divide by $0.10$ gives $320$. $L=4\pi \times 10^{-7}\times 320=(1.2566\times 10^{-6})(320)\approx 4.02\times 10^{-4}\text{H}\approx 0.40\text{mH}.$

Read the figure below to see why this answer is so sensitive to $N$: the curve is not a straight line but a parabola. The black dot marks this very problem ($N=400$), and the second black dot at $N=200$ sits at only one quarter of the height — that is the $L\propto N^2$ law made visible. Trace up from any $N$ on the horizontal axis to read its inductance off the red curve; notice how quickly the curve steepens as you move right, which is why adding a few turns to an already-large coil buys a lot of inductance.

The voltage tracks the slope of the current, not its value.

• Rise: slope $=\frac{8-0}{4}=+2\text{A/s}\Rightarrow V=2\times 2=+4\text{V}$ (constant, positive).
• Fall: slope $=\frac{0-8}{2}=-4\text{A/s}\Rightarrow V=2\times (-4)=-8\text{V}$ (constant, negative).

So a triangle current makes a square-wave voltage: flat at $+4$ V while climbing, flat at $-8$ V while falling. The fall is steeper (half the time, same height), so its voltage magnitude is double. In the figure, the black line is the current and the red line is the voltage: notice the red line is perfectly flat wherever the black line is a straight ramp, and it jumps at each corner of the triangle — those corners are where the slope changes abruptly.

The induced voltage always opposes the change in current (Lenz's Law).

• While current rises, the coil fights the increase — it pushes back like a headwind, giving one polarity.
• While current falls, the coil fights the decrease — now it tries to keep the current going, so it pushes in the opposite direction, flipping the polarity. Mathematically the sign of $V=L\frac{dI}{dt}$ simply follows the sign of the slope $\frac{dI}{dt}$: positive slope → one sign, negative slope → the other. Zero slope → zero volts.

From $L=\mu \frac{N^{2}A}{\ell }$, inductance scales as $N^2$. Doubling $N$ gives $\frac{L_B}{L_A}=\frac{(2N{)}^2}{N^2}=4.$ Four times larger. Two effects stack: twice the turns make twice the field and give twice as many loops for that field to link — $2\times 2=4$.

(a) Use the definition $L=\frac{N\Phi }{I}$: $L=\frac{150\times 4\times 10^{-4}}{2}=\frac{0.06}{2}=0.03\text{H}=30\text{mH}.$ (b) Switching off means current goes from 2 A to 0 in $5\text{ms}=0.005$ s: $\left|\frac{dI}{dt}\right|=\frac{2}{0.005}=400\text{A/s},$ $|V|=L\left|\frac{dI}{dt}\right|=0.03\times 400=12\text{V}.$ This is why switching inductors off fast produces big voltage spikes — the fast $\frac{dI}{dt}$ amplifies.

(a) $E=\frac{1}{2}L{I}^2=\frac{1}{2}(0.5)(10^2)=\frac{1}{2}(0.5)(100)=25\text{J}.$ (b) $\frac{dI}{dt}=\frac{10}{4}=2.5\text{A/s}$, so $V=0.5\times 2.5=1.25\text{V}.$ (c) During a linear rise the average current is $\frac{0+10}{2}=5$ A. Energy delivered: $E=V\cdot I_{\text{avg}}\cdot t=1.25\times 5\times 4=25\text{J}.✓$ The two methods agree — energy is genuinely stored in the field, exactly $\frac{1}{2}L{I}^2$.

Same current, so the same $\frac{dI}{dt}=6$ A/s flows through both. Voltages add: $V_1=0.3\times 6=1.8\text{V},V_2=0.7\times 6=4.2\text{V},$ $V_{\text{total}}=1.8+4.2=6.0\text{V}.$ (b) Since $V_{\text{total}}=L_{\text{eff}}\frac{dI}{dt}$: $L_{\text{eff}}=\frac{6.0}{6}=1.0\text{H}=L_1+L_2.$ Series inductances add — just like series resistors (and unlike series capacitors, the dual).

Start from $L={\mu }_0\frac{N^{2}A}{\ell }$ and solve for $N$: $N=\sqrt{\frac{L\ell }{{\mu }_{0}A}}.$ Plug in ($L=1.0\times 10^{-3}$ H): $N=\sqrt{\frac{(1.0\times 10^{-3})(0.08)}{(4\pi \times 10^{-7})(5\times 10^{-4})}}=\sqrt{\frac{8.0\times 10^{-5}}{6.283\times 10^{-10}}}=\sqrt{1.273\times 10^5}\approx \mathrm{356.8.}$ You cannot wind a fraction of a turn, and you must not undershoot the spec, so round up: $N=357$ turns. Verify with $N=357$: $L=(4\pi \times 10^{-7})\frac{357^2\cdot 5\times 10^{-4}}{0.08}=(1.2566\times 10^{-6})\frac{127449\cdot 5\times 10^{-4}}{0.08}.$ $127449\times 5\times 10^{-4}=63.72$; divide by $0.08\Rightarrow 796.6$; times $1.2566\times 10^{-6}$: $L\approx 1.001\times 10^{-3}\text{H}=1.001\text{mH}\ge 1.0\text{mH}.✓$

(a) First $\mu =800\times 4\pi \times 10^{-7}=800\times 1.2566\times 10^{-6}=1.0053\times 10^{-3}\text{H/m}$ (keep the $\pi$ — this is the step people botch). $L=\mu \frac{N^{2}A}{\ell }=(1.0053\times 10^{-3})\frac{500^2\cdot 3\times 10^{-4}}{0.12}.$ $500^2=250000$; times $3\times 10^{-4}=75$; divide by $0.12=625$; times $\mu$: $L=1.0053\times 10^{-3}\times 625\approx 0.6283\text{H}\approx 0.628\text{H}.$ (b) $E=\frac{1}{2}L{I}^2=\frac{1}{2}(0.6283)(2^2)=\frac{1}{2}(0.6283)(4)=1.257\text{J}.$ (c) $\left|\frac{dI}{dt}\right|=\frac{2}{0.0005}=4000\text{A/s}$, so $|V|=L\left|\frac{dI}{dt}\right|=0.6283\times 4000\approx 2513\text{V}.$ A 2 A coil on a battery of a few volts can slam ~2500 V across the opening switch — this is the arc/spark that destroys contacts, and why real circuits add a "flyback" diode. The iron core ($\mu =800{\mu }_0$) is what made $L$ (and the spike) so large.

Constant current ($\frac{dI}{dt}=0$):

• Voltage: $V=L\cdot 0=0$ V — the inductor is invisible, just a wire.
• Energy: $E=\frac{1}{2}L{I}^2>0$ — energy is still parked in the magnetic field; it isn't being added to or removed, just held.

Zero current ($I=0$):

• Energy: $E=\frac{1}{2}L(0{)}^2=0$ J — nothing stored, the field has collapsed.
• Voltage: depends on how you reached zero. If you glide to zero smoothly ($\frac{dI}{dt}\to 0$ at the end) the voltage fades to 0. If you chop the current off instantly, $\frac{dI}{dt}$ is huge and momentarily so is $V$ — the L5.2 spike. So "current = 0" alone does not fix the voltage; the slope at that instant does.

These two degenerate cases show the core truth of the whole topic: energy answers to $I$, voltage answers to $\frac{dI}{dt}$ — never confuse the height with the slope.

An inductor's voltage answers to ? and its energy answers to ? . Voltage answers to the slope $\frac{dI}{dt}$; energy answers to the value $I$ (as $\frac{1}{2}L{I}^2$).