Worked examples: choke on the mains, and an 11 kV → 230 V transformer

Three problems. First: X_L for a real choke on 50 Hz mains, and how it changes at 20 kHz. Second: a magnetism sanity check — find B inside an iron-cored solenoid. Third: size a distribution transformer the way a protection engineer does, then verify with the playground.

Example 1 — 100 mH choke on 50 Hz mains vs 20 kHz SMPS

A 100 mH iron-cored choke sits in series with a 230 V RMS 50 Hz mains supply through a 10 Ω resistor. Later the same choke is re-used on the output of a switching power supply running at 20 kHz. Find X_L at each frequency and the current it limits.

1. Reactance at 50 Hz.
   $$X_L = 2\pi f L = 2\pi \cdot 50 \cdot 0.1 \approx 31.4\,\Omega$$
   Comparable to the 10 Ω resistor in series, so the choke significantly limits mains-frequency current.
2. Approximate RMS current. Ignore the resistor's phase contribution for a quick estimate (we'll do this properly once we reach impedance in Topic 8):$I \approx V / X_L = 230 / 31.4 \approx 7.3\,\text{A}$.
3. Reactance at 20 kHz (400× higher).
   $$X_L = 2\pi \cdot 20{,}000 \cdot 0.1 \approx 12.6\,\text{k}\Omega$$
   Four hundred-fold jump. A choke that barely troubled 50 Hz mains looks like a near-open at SMPS frequencies — exactly what you want in a power-supply filter.

Drag the frequency marker to 20 kHz and watch the line climb by 2.6 decades. Move the inductance slider down to 1 mH and the line slides down by two decades — all the qualitative rules about chokes fall out of that one sweep.

Example 2 — B inside an iron-cored solenoid

A solenoid has $N = 1000$ turns wound on an iron core $l = 0.2\,\text{m}$ long. Relative permeability $\mu_r = 2000$ (typical for silicon steel). The coil carries 0.5 A. Find the magnetic field intensity H, the flux density B, and the total flux Φ through a 1 cm² cross-section.

1. Field intensity H.
   $$H = \frac{N I}{l} = \frac{1000 \times 0.5}{0.2} = 2500\,\text{A·t/m}$$
2. Flux density B.
   $$B = \mu_0 \mu_r H = (4\pi \times 10^{-7})(2000)(2500) \approx 6.28\,\text{T}$$
   Hot — real iron saturates around 1.5 – 2 T, so a practical design would either drop the current or accept that the core is now deep in saturation and $\mu_r$ collapses. The formula still works; it just stops telling the truth once the iron gives up.
3. Total flux Φ through 1 cm².
   $$\Phi = B \cdot A = 6.28 \cdot (10^{-4}) \approx 628\,\mu\text{Wb}$$
   (Again, pre-saturation value — real core would give roughly 150 – 200 µWb after saturation takes hold.)

Example 3 — Distribution transformer: 11 kV → 230 V

A step-down distribution transformer takes an 11 kV primary down to 230 V for domestic use. If the primary has 2 000 turns, how many turns are on the secondary? Under a 10 kW load on the secondary, what currents flow on each side (treat as ideal)?

1. Turns ratio from voltage ratio.
   $$\frac{N_1}{N_2} = \frac{V_1}{V_2} = \frac{11{,}000}{230} \approx 47.8$$
   So $N_2 = 2000 / 47.8 \approx 42$ turns. Dozens, not thousands — the secondary is a short, fat winding of thick wire.
2. Secondary current under load.
   $$I_2 = P / V_2 = 10{,}000 / 230 \approx 43.5\,\text{A}$$
   That's why the low-voltage winding needs chunky copper.
3. Primary current (ideal).
   $$I_1 = P / V_1 = 10{,}000 / 11{,}000 \approx 0.91\,\text{A}$$
   Same power, 50× less current — the whole reason we transmit at high voltage.