Worked examples: stored energy, kickback spike, combining a network, and Ψ=LI per inductor

Four problems. The first two cement $v = L \, dI/dt$ and the energy formula. The third reduces a small series-parallel inductor network. The fourth works through a combination network with a given supply current, finding the flux linkage $\Psi = LI$ and stored energy on every individual inductor.

Example 1 — energy stored in an RF choke

A 100 µH RF choke carries a steady 3 A bias current. How much energy is stored in its magnetic field?

1. Apply W = ½ L I².
   $$W = \tfrac{1}{2} L I^2 = \tfrac{1}{2} \cdot 100 \times 10^{-6} \cdot 9 = 4.5 \times 10^{-4}\,\text{J}$$
2. Put it in perspective. 0.45 mJ is tiny — about the energy it takes to lift a paperclip one centimetre. But release it in microseconds (say, when a transistor switch opens) and the power is huge (450 mJ / µs = 450 kW instantaneous). The voltage required to dissipate that power across a stray resistance is the kickback spike, covered next.

Example 2 — a kickback spike

A 10 mH relay coil is carrying 100 mA. A MOSFET switch in series with the coil snaps open in 1 µs. What's the peak voltage induced across the coil?

1. Estimate dI/dt. The current falls from 100 mA to 0 in 1 µs, so
   $$\frac{dI}{dt} \approx \frac{0.1\,\text{A}}{1 \times 10^{-6}\,\text{s}} = 10^5\,\text{A/s}$$
2. Apply v = L · dI/dt.
   $$v = L \frac{dI}{dt} = 10 \times 10^{-3} \cdot 10^5 = 1000\,\text{V}$$
   A kilovolt across the coil — plenty to punch through the MOSFET's breakdown voltage and destroy it. This is why a flyback diode across every inductive load is standard practice; it gives the collapsing field somewhere benign to dump its energy.

Example 3 — combining a small inductor network

Three inductors: $L_1 = 10\,\text{mH}$, $L_2 = 22\,\text{mH}$, $L_3 = 100\,\mu\text{H}$. $L_1$ and $L_2$ are wired in parallel; that combination is in series with $L_3$. Find the equivalent inductance.

1. Collapse the parallel pair first. Inductors in parallel reciprocate:
   $$\frac{1}{L_{12}} = \frac{1}{10} + \frac{1}{22} \approx 0.1455\,\text{/mH}$$
   $$L_{12} \approx 6.87\,\text{mH}$$
   Parallel equivalent is smaller than the smaller element — same direction as resistors.
2. Add L₃ in series. $L_3 = 0.1\,\text{mH}$.
   $$L_{eq} = L_{12} + L_3 \approx 6.87 + 0.1 \approx 6.97\,\text{mH}$$
3. Sanity check. Series adds, so the total should be bigger than the parallel pair alone (6.87 mH). It is — by just the 0.1 mH contribution of the tiny series element. The small component barely moves the needle in series with a much bigger one. Opposite of the cap case (where the smallest dominates in series).

Example 4 — Ψ and W across a combination network

A 5 A current source drives the following network: $L_1 = 40\,\text{mH}$ in series with a parallel pair $L_2 = 60\,\text{mH}$ and $L_3 = 30\,\text{mH}$. Find the equivalent inductance, then the current, flux linkage, and stored energy for every inductor.

1. Reduce the parallel pair.
   $$\frac{1}{L_{23}} = \frac{1}{60} + \frac{1}{30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60}$$
   $$L_{23} = 20\,\text{mH}$$
2. Combine L₁ and L₂₃ in series.
   $$L_{eq} = L_1 + L_{23} = 40 + 20 = 60\,\text{mH}$$
3. Current through L₁ (series group — carries the full supply current):
   $$I_1 = I_s = 5\,\text{A}$$
   $$\Psi_1 = L_1 I_1 = 40 \times 10^{-3} \times 5 = 200\,\text{mWb}$$
   $$W_1 = \tfrac{1}{2} L_1 I_1^2 = \tfrac{1}{2} \times 40 \times 10^{-3} \times 25 = 500\,\text{mJ}$$
4. Current division in the parallel pair. Current divides inversely with L (current divider rule):
   $$I_2 = I_s \frac{L_{23}}{L_2} = 5 \times \frac{20}{60} = \frac{5}{3} \approx 1.67\,\text{A}$$
   $$I_3 = I_s \frac{L_{23}}{L_3} = 5 \times \frac{20}{30} = \frac{10}{3} \approx 3.33\,\text{A}$$
   Check: $I_2 + I_3 = \frac{5}{3} + \frac{10}{3} = 5\,\text{A}$ ✓
5. Flux linkage of L₂ and L₃. Parallel inductors share the same flux linkage — the dual of parallel capacitors sharing voltage:
   $$\Psi_2 = L_2 I_2 = 60 \times 10^{-3} \times \tfrac{5}{3} = 100\,\text{mWb}$$
   $$\Psi_3 = L_3 I_3 = 30 \times 10^{-3} \times \tfrac{10}{3} = 100\,\text{mWb}$$
   Both equal $L_{23} \times I_s = 20 \times 10^{-3} \times 5 = 100\,\text{mWb}$ ✓
6. Stored energy in L₂ and L₃.
   $$W_2 = \tfrac{1}{2} \times 60 \times 10^{-3} \times \left(\tfrac{5}{3}\right)^2 \approx 83.33\,\text{mJ}$$
   $$W_3 = \tfrac{1}{2} \times 30 \times 10^{-3} \times \left(\tfrac{10}{3}\right)^2 \approx 166.67\,\text{mJ}$$
   Total: $W = W_1 + W_2 + W_3 = 500 + 83.33 + 166.67 = 750\,\text{mJ}$
   Check: $\tfrac{1}{2} L_{eq} I_s^2 = \tfrac{1}{2} \times 60 \times 10^{-3} \times 25 = 750\,\text{mJ}$ ✓