Transformer Magnetic Circuit and Maximum Power Transfer: 25-Nav-B6 Ocean Engineering and Offshore Structures, Undated paper, Q4

Question 4: Transformer Magnetic Circuit and Maximum Power Transfer (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The core carries both windings on one limb, and the flux returns through two parallel air paths whose lengths and cross-sections are tabulated. Because the iron is assumed to have infinite permeability, it contributes no reluctance and the entire magnetic circuit is these two paths.

Given data
Quantity	Symbol	Value
Gap length, centre limb	$d_1$ ($L_1$)	$3.77 \times 10^{-2}$ m
Gap length, outer limb	$d_2$ ($L_2$)	$7.54 \times 10^{-2}$ m
First path cross-section	$A_1$	0.02 m$^2$
Second path cross-section	$A_2$	0.02 m$^2$
Primary turns	$N_1$	200
Secondary turns	$N_2$	20
Operating frequency	$f$	1000 Hz
Primary dc test	—	10 mV gives 100 mA
Secondary dc test	—	0.1 mV gives 100 mA
Transducer (load) impedance	$Z_L$	0.078 $\Omega$

Find. Every element of the equivalent circuit referred to the primary — the two winding resistances and the magnetising reactance — and then the amplifier output impedance that maximises power delivered to the transducer.

Approach. A dc test excites no flux, so it isolates the winding resistance alone; the ac magnetising branch comes instead from the magnetic circuit, with the two air paths in parallel as seen by the wound limb. Referring the secondary quantities through $a^{2}$ then reduces the whole device to a simple series–shunt network, and the maximum-power-transfer theorem asks for the conjugate of the impedance seen looking in at the primary.

Extract the winding resistances from the dc tests. With direct current there is no rate of change of flux and hence no induced voltage, so the applied voltage divides across resistance only: $$R_1 = \frac{10\ \text{mV}}{100\ \text{mA}} = 0.100\ \Omega, \qquad R_2 = \frac{0.1\ \text{mV}}{100\ \text{mA}} = 0.001\ \Omega.$$
Form the turns ratio and refer the secondary resistance. $a = N_1/N_2 = 200/20 = 10$, and resistances refer across the ideal transformer as the square of the ratio: $$a^{2}R_2 = 10^{2} \times 0.001 = 0.100\ \Omega.$$ The two windings therefore contribute equally once referred — the design is deliberately balanced.
Compute the reluctance of each air path. With the iron of infinite permeability the only reluctance is that of the two paths, $\mathcal{R} = L/(\mu_0 A)$: $$\mathcal{R}_1 = \frac{3.77\times10^{-2}}{(4\pi\times10^{-7})(0.02)} = 1.50\times10^{6}\ \text{A}\,\text{Wb}^{-1},$$ and since $L_2$ is exactly twice $L_1$ over the same area, $\mathcal{R}_2 = 3.00\times10^{6}\ \text{A}\,\text{Wb}^{-1}$.
Combine the paths and find the magnetising inductance. The two paths offer alternative routes for the flux leaving the wound limb, so their reluctances combine as elements in parallel: $$\mathcal{R}_{eq} = \frac{\mathcal{R}_1\mathcal{R}_2}{\mathcal{R}_1+\mathcal{R}_2} = \frac{(1.5)(3.0)}{4.5}\times10^{6} = 1.00\times10^{6}\ \text{A}\,\text{Wb}^{-1}.$$ The magnetising inductance seen from the primary follows as $$L_m = \frac{N_1^{2}}{\mathcal{R}_{eq}} = \frac{200^{2}}{1.00\times10^{6}} = \boxed{40\ \text{mH}}$$
Convert to a reactance at the operating frequency. $$X_m = 2\pi f L_m = 2\pi(1000)(0.040) = 251.3\ \Omega,$$ so the magnetising branch is $j251.3\ \Omega$. This completes part [a]: the equivalent circuit is $R_1 = 0.1\ \Omega$ in series with the parallel combination of $jX_m = j251.3\ \Omega$ and the referred secondary branch $a^{2}R_2 = 0.1\ \Omega$ in series with the referred load.

Part [a] — equivalent circuit referred to the primary. Leakage reactances and core losses are neglected as the question directs, so only the magnetising branch shunts the series path.

Part [b] — matching the amplifier. With the transducer connected, the amplifier looks into the whole network:

Refer the load to the primary. $$a^{2}Z_L = 100 \times 0.078 = 7.80\ \Omega,$$ so the referred secondary branch beyond the magnetising node totals $$a^{2}R_2 + a^{2}Z_L = 0.1 + 7.8 = 7.90\ \Omega.$$
Combine with the magnetising branch, then add $R_1$. The magnetising reactance shunts the referred secondary branch, and the primary resistance is in series ahead of both: $$Z_{par} = \frac{jX_m\,(7.90)}{7.90 + jX_m} = \frac{j251.3 \times 7.90}{7.90 + j251.3} = 7.892 + j0.248\ \Omega,$$ $$Z_{in} = R_1 + Z_{par} = 0.1 + 7.892 + j0.248 = 7.99 + j0.248\ \Omega = 8.00\angle 1.78^\circ\ \Omega.$$ Because $X_m$ is some thirty times the series resistance, the magnetising branch is very nearly an open circuit and adds only a small inductive term.
Apply the maximum-power-transfer theorem. Maximum power is delivered when the source impedance is the complex conjugate of the load it drives, so the amplifier output impedance must be $$Z_{amp} = Z_{in}^{*} = \boxed{7.99 - j0.248\ \Omega \;\approx\; 8\ \Omega}$$ In practice the amplifier would simply be specified as an 8 $\Omega$ output — the small capacitive term needed to cancel the magnetising reactance is within the tolerance of any real output stage.

Final results — Question 4
Quantity	Result
Primary winding resistance $R_1$	0.100 $\Omega$
Secondary winding resistance $R_2$	0.001 $\Omega$
Turns ratio $a = N_1/N_2$	10
Referred secondary resistance $a^{2}R_2$	0.100 $\Omega$
Path reluctances $\mathcal{R}_1,\ \mathcal{R}_2$	$1.50\times10^{6}$ and $3.00\times10^{6}$ A/Wb
Equivalent reluctance (parallel)	$1.00\times10^{6}$ A/Wb
Magnetising inductance $L_m$	40 mH
Magnetising reactance $X_m$ at 1 kHz	251.3 $\Omega$
Referred load $a^{2}Z_L$	7.80 $\Omega$
Input impedance $Z_{in}$	$7.99 + j0.248\ \Omega$
Amplifier output impedance	$7.99 - j0.248\ \Omega \approx 8\ \Omega$

Reading the figure. $d_1$ is the air gap in the centre limb and $d_2$ the gap in the right-hand limb; both windings sit on the left limb, so the flux leaving that limb returns through the two gapped limbs in parallel. With $d_2 = 2d_1$ over equal areas the reluctances are exactly $1.5 imes10^{6}$ and $3.0 imes10^{6}$ A/Wb, and $L_m$ is exactly 40 mH.

Concept & Theory

Topic. Magnetic circuits, the origin of each element in the transformer equivalent circuit, and impedance matching through a turns ratio.

Key relations. Reluctance $\mathcal{R} = \ell/(\mu_0\mu_r A)$, combining in series and parallel like resistance; inductance $L = N^{2}/\mathcal{R}$; impedance referral $Z' = a^{2}Z$ with $a = N_1/N_2$; reactance $X_m = 2\pi f L_m$; and maximum power transfer at $Z_{source} = Z_{load}^{*}$.

Why this works. Each element of the equivalent circuit is measured by a test that isolates it. A dc test produces no changing flux, so no voltage is induced and the winding resistance stands alone — which is precisely why the question supplies dc data rather than ac. The magnetising branch, by contrast, is a property of the magnetic path, and assuming infinite permeability for the iron concentrates all of it in the air paths, so the reluctance calculation is exact rather than an estimate. Referring impedances through $a^{2}$ works because voltage scales as $a$ while current scales as $1/a$, so their ratio scales as $a^{2}$; this is the whole reason a transformer can match a fraction-of-an-ohm transducer to an amplifier expecting several ohms.

Common pitfall. Combining the two air paths in series rather than in parallel. They are alternative routes for the same flux leaving the wound limb, so the reluctances add reciprocally; treating them as series would give $4.5\times10^{6}$ A/Wb and an $L_m$ of 8.9 mH, wrong by a factor of 4.5. A second trap is referring the load by $a$ rather than $a^{2}$.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., Ch. 1 (magnetic circuits, reluctance) and Ch. 2 (transformer equivalent circuit, impedance transformation).

Question 4 of 8: Transformer Magnetic Circuit and Maximum Power Transfer

Question 4: Transformer Magnetic Circuit and Maximum Power Transfer (equal value)