22-Mec-A5 Electrical and Electronics Engineering · December 2013

Question 8 of 8: Magnetic Circuit — Induced Voltages, Input Impedance and Waveforms

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. PEO / Engineers Canada National Examinations, December 2013 — 07-Mec-A5, Electrical & Electronics Engineering (Mechanical Engineering discipline). Three hours, closed book, one approved Casio or Sharp calculator. Eight questions; any five constitute a complete paper and each question is of equal value, so each counted question carries 20 marks. All eight questions are solved below, because this set is a study resource rather than an exam script.

Reference texts. A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (BJT current mirrors, op-amp frequency response); M. M. Mano and M. D. Ciletti, Digital Design, 6th ed. (universal-gate synthesis, DeMorgan reduction); S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (dc machines, induction-motor testing and slip); C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (phasor analysis, power factor, first-order transients); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (magnetic circuits and induced emf).

Check: front-page constant. Note [8] of the examination paper prints the permeability of free space as μ₀ = 4π × 10⁷ H·m⁻¹. The exponent has plainly lost its minus sign in printing; the physical constant is 4π × 10⁻⁷ H·m⁻¹, and that is the value used in Question 8. The other two constants, π = 3.14159 and 1 hp = 746 W, are used exactly as printed.

Question 8: Magnetic Circuit — Induced Voltages, Input Impedance and Waveforms (20 marks)

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. A closed iron core of mean magnetic path length $L$, uniform cross-sectional area $A$ and relative permeability $\mu_r$, carrying a primary winding of $N_1$ turns and a secondary winding of $N_2$ turns. The primary current is $i_1(t)=I_p\sin\omega t$. Winding resistance, leakage inductance and core losses are to be neglected, and the secondary is effectively open-circuited because it feeds only a voltmeter.

Find. [a] $v_1(t)$ and $v_2(t)$ in terms of $i_1(t)$ and the core parameters; [b] the impedance seen at the primary terminals; [c] the waveforms of $v_{AB}$, $v_{XY}$ and $i_1(t)$ showing relative magnitude and phase.

[Figure not reproduced: Figure 8 redrawn. The single flux $\varphi$ links both windings, so the two induced voltages differ only by the turns ratio. See the official exam paper.]

Approach. Compute the reluctance of the magnetic path, use it to relate flux to primary magnetomotive force, apply Faraday's law to each winding to obtain the induced voltages, and then form the ratio of primary voltage to primary current as a phasor to get the input impedance.

Compute the reluctance of the core. For a uniform closed path, $$\mathcal{R}_m=\frac{L}{\mu_0\mu_r A}$$ where $\mu_0=4\pi\times10^{-7}$ H·m$^{-1}$. This single quantity carries all the geometry and material information.
Relate flux to primary current. The magnetomotive force driving the circuit is $\mathcal{F}=N_1i_1$, and the magnetic Ohm's law gives $$\varphi(t)=\frac{\mathcal{F}}{\mathcal{R}_m}=\frac{N_1i_1(t)}{\mathcal{R}_m}=\frac{N_1\mu_0\mu_r A}{L}\,i_1(t)$$ Because leakage is neglected, this same flux links every turn of both windings.
[a] Apply Faraday's law to each winding. The primary flux linkage is $\lambda_1=N_1\varphi$, so $$v_1(t)=N_1\frac{d\varphi}{dt}=\frac{N_1^{2}\mu_0\mu_r A}{L}\frac{di_1}{dt}=L_1\frac{di_1}{dt},\qquad L_1\equiv\frac{N_1^{2}}{\mathcal{R}_m}=\frac{N_1^{2}\mu_0\mu_r A}{L}$$ and for the secondary, which links the same flux through $N_2$ turns, $$v_2(t)=N_2\frac{d\varphi}{dt}=\frac{N_1N_2\mu_0\mu_r A}{L}\frac{di_1}{dt}=\frac{N_2}{N_1}\,v_1(t)$$ Substituting the stated current $i_1(t)=I_p\sin\omega t$, whose derivative is $\omega I_p\cos\omega t$: $$\boxed{\;v_1(t)=\frac{N_1^{2}\mu_0\mu_r A}{L}\,\omega I_p\cos\omega t,\qquad v_2(t)=\frac{N_1N_2\mu_0\mu_r A}{L}\,\omega I_p\cos\omega t=\frac{N_2}{N_1}v_1(t)\;}$$
[b] Form the input impedance. In phasor form, differentiation becomes multiplication by $j\omega$, so $V_1=j\omega L_1 I_1$ and $$Z_{in}=\frac{V_1}{I_1}=j\omega L_1=j\,\frac{\omega N_1^{2}\mu_0\mu_r A}{L}$$ $$\boxed{\;Z_{in}=j\omega L_1=j\,\frac{\omega\,N_1^{2}\mu_0\mu_r A}{L},\qquad |Z_{in}|=\frac{\omega N_1^{2}\mu_0\mu_r A}{L}\ \angle\,90^{\circ}\;}$$ The impedance is purely inductive — the magnetising reactance of the core. This is the correct answer precisely because the secondary is open: with no secondary current there is no reflected load, so the primary sees only the magnetising branch. Note that $N_2$ does not appear, which is the physical signature of an unloaded transformer.
Substitute representative numbers. For $N_1=200$, $N_2=100$, $A=0.002$ m$^2$, $L=0.40$ m, $\mu_r=2000$, $I_p=0.50$ A and $f=60$ Hz ($\omega=377$ rad/s): $$L_1=\frac{(200)^{2}\left(4\pi\times10^{-7}\right)(2000)(0.002)}{0.40}=0.5027\ \text{H}$$ $$V_{1,\text{peak}}=\omega L_1 I_p=377\times0.5027\times0.50=94.75\ \text{V},\qquad V_{2,\text{peak}}=\frac{100}{200}\times94.75=47.37\ \text{V}$$ $$|Z_{in}|=\omega L_1=377\times0.5027=189.5\ \Omega$$
[c] Establish the phase relations for the sketch. With $i_1\propto\sin\omega t$ and both voltages $\propto\cos\omega t$, and since $\cos\omega t=\sin(\omega t+90^{\circ})$, both induced voltages lead the primary current by exactly $90^{\circ}$. The two voltages are in phase with each other, differing only in amplitude by the turns ratio $N_2/N_1$. The current peaks a quarter cycle after the voltages, at $\omega t=90^{\circ}$.

[c] Waveforms over one cycle. $v_{AB}$ and $v_{XY}$ are cosinusoids in phase with each other and leading the sinusoidal primary current by $90^{\circ}$; their amplitudes are in the ratio $N_2/N_1$ (here 1:2, so $v_{XY}$ is half of $v_{AB}$).

The $90^{\circ}$ lead is the whole physical content of part [c] and is worth stating in words rather than only in a sketch: a lossless inductive circuit draws a purely magnetising current that lags its terminal voltage by a quarter cycle, so the average power drawn over a cycle is zero. Energy flows into the core during one quarter cycle and back out during the next. In a real transformer the small core loss and winding resistance tilt this angle slightly below $90^{\circ}$, and the resulting in-phase component of the no-load current is exactly what an open-circuit test measures to determine the core losses.

Final results — Question 8
Quantity	Symbolic result	Worked case
Core reluctance	$\mathcal{R}_m=L/(\mu_0\mu_r A)$	$7.96\times10^{4}$ A·t/Wb
Primary self-inductance	$L_1=N_1^{2}\mu_0\mu_r A/L$	0.5027 H
[a] Primary voltage	$v_1=L_1\,di_1/dt=\omega L_1I_p\cos\omega t$	94.75 V peak
[a] Secondary voltage	$v_2=(N_2/N_1)v_1$	47.37 V peak
[b] Input impedance	$Z_{in}=j\omega L_1$ (purely inductive)	$189.5\ \Omega\angle 90^{\circ}$
[c] Phase relation	both voltages lead $i_1$ by $90^{\circ}$; $v_1$ and $v_2$ in phase	—

Topic. Magnetic circuits and the unloaded two-winding transformer.

Key relations. Reluctance $\mathcal{R}_m=L/(\mu_0\mu_r A)$; magnetic Ohm's law $\varphi=N i/\mathcal{R}_m$; Faraday's law $v=N\,d\varphi/dt$; self-inductance $L_1=N_1^{2}/\mathcal{R}_m$; turns ratio $v_2/v_1=N_2/N_1$; phasor impedance $Z=j\omega L$.

Why this works. The magnetic-circuit analogy maps magnetomotive force to voltage, flux to current and reluctance to resistance, so a closed uniform core reduces to a single-element circuit and the flux follows in one line. Because leakage is neglected, one flux links both windings, which is what makes the induced voltages differ by nothing more than the turns count — the turns ratio is a consequence of shared flux, not an independent assumption. The impedance comes out purely imaginary because the model retains no dissipative element whatever, and any real part in an answer indicates a loss mechanism that the question explicitly excluded.

Common pitfall. Reporting $Z_{in}$ as a resistance, or including a reflected load impedance $(N_1/N_2)^{2}Z_L$; the secondary drives only a voltmeter, so it is open-circuited and reflects nothing. The second pitfall is a sign or phase slip in part [c]: differentiating $\sin\omega t$ gives $+\cos\omega t$, so the voltages lead the current; drawing them lagging is the single most common error on this question.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., Ch. 1–2 (magnetic circuits, Faraday's law, transformer no-load behaviour); Hayt & Buck, Engineering Electromagnetics, 9th ed., Ch. 9.

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