Magnetic Circuit — Induced Voltages and Input Impedance: 25-Nav-B6 Ocean Engineering and Offshore Structures, May 2013, Q8

Question 8: Magnetic Circuit — Induced Voltages and Input Impedance (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.

Symbol	Quantity
$N_1$, $N_2$	primary and secondary turns
$\ell$	mean magnetic path length (m) — written $L$ on the paper
$A$	core cross-sectional area (m$^2$)
$\mu_R$	relative permeability of the core
$\mu_0$	$4\pi\times10^{-7}$ H m$^{-1}$ (front page)
$i_1(t)$	$I_p \sin \omega t$

The secondary is open-circuited into a voltmeter, so it carries no current; winding resistance, leakage flux and core losses are neglected, and the permeability is taken as constant (no saturation).

Find. [a] $v_1(t)$ and $v_2(t)$ in terms of $i_1(t)$; [b] the input impedance seen at the primary; [c] the waveforms of $v_{AB}$ and $v_{XY}$ relative to $i_1(t)$.

Figure 8 — the magnetic circuit. A single flux $\varphi(t)$, driven by the primary mmf, links both windings; the open-circuited secondary carries no current and contributes no mmf.

Approach. Compute the reluctance of the core from its geometry, get the flux from the primary mmf, then apply Faraday’s law to each winding; the ratio of primary voltage to primary current gives the impedance.

Reluctance of the magnetic circuit. For a uniform core of mean length $\ell$, area $A$ and permeability $\mu = \mu_0\mu_R$, $$\mathcal{R} = \frac{\ell}{\mu_0 \mu_R A}\ \ \text{A⋅turns/Wb}.$$ This is the magnetic analogue of $R = \rho\ell/A$ for a resistor.
Flux from the primary mmf. The secondary is open, so the primary supplies the entire mmf $\mathcal{F} = N_1 i_1(t)$ and $$\varphi(t) = \frac{\mathcal{F}}{\mathcal{R}} = \frac{N_1 i_1(t)}{\mathcal{R}} = \frac{\mu_0 \mu_R A N_1}{\ell}\,i_1(t).$$ Flux is therefore in phase with, and proportional to, the primary current.
[a] Primary voltage by Faraday’s law. With $N_1$ turns linking that flux, and with winding resistance neglected so the terminal voltage equals the induced emf, $$v_1(t) = N_1 \frac{\mathrm{d}\varphi}{\mathrm{d}t} = \frac{N_1^{2}}{\mathcal{R}}\,\frac{\mathrm{d}i_1}{\mathrm{d}t} = L_1 \frac{\mathrm{d}i_1}{\mathrm{d}t}, \qquad L_1 = \frac{N_1^{2}}{\mathcal{R}} = \frac{\mu_0\mu_R A N_1^{2}}{\ell}.$$ Substituting $i_1 = I_p \sin\omega t$, $$\boxed{\;v_1(t) = \frac{\mu_0\mu_R A N_1^{2}}{\ell}\,\omega I_p \cos\omega t = \omega L_1 I_p \cos\omega t\;}$$
Secondary voltage. The same flux links $N_2$ turns, so $$v_2(t) = N_2 \frac{\mathrm{d}\varphi}{\mathrm{d}t} = \frac{N_1 N_2}{\mathcal{R}}\,\frac{\mathrm{d}i_1}{\mathrm{d}t} = M\,\frac{\mathrm{d}i_1}{\mathrm{d}t}, \qquad M = \frac{N_1 N_2}{\mathcal{R}},$$ that is $$\boxed{\;v_2(t) = \frac{\mu_0\mu_R A N_1 N_2}{\ell}\,\omega I_p \cos\omega t = \frac{N_2}{N_1}\,v_1(t)\;}$$ The ideal-transformer turns ratio drops straight out, as it must when leakage is neglected.
[b] Input impedance. In phasor form $i_1 \to I_p$ and $v_1 \to j\omega L_1 I_p$, so $$\boxed{\;Z_{\text{in}} = \frac{V_1}{I_1} = j\omega L_1 = j\,\frac{\mu_0\mu_R A N_1^{2}\,\omega}{\ell}\;}$$ with magnitude $\omega L_1$ and phase exactly $+90^\circ$. With losses neglected and the secondary open, the circuit presents a pure magnetising inductance — it consumes no real power, only reactive.
[c] Waveform relationships. The current is a sine and both voltages are cosines of the same frequency, so each voltage leads the current by $90^\circ$ (a quarter cycle): the voltages peak as the current passes through zero, and are zero as the current peaks. Their peak magnitudes are $$\hat V_{AB} = \omega L_1 I_p = \frac{\mu_0\mu_R A N_1^{2}\omega I_p}{\ell}, \qquad \hat V_{XY} = \frac{N_2}{N_1}\,\hat V_{AB},$$ and the two voltages are in phase with each other (for the winding sense shown, with $A$ and $X$ as the corresponding dotted terminals).

Part [c] — $i_1(t)$ sinusoidal, with $v_{AB}$ and $v_{XY}$ both cosinusoidal and therefore leading the current by 90°; the secondary amplitude is scaled by $N_2/N_1$.

Check: numerical scale check The question is symbolic. Substituting a representative core — $N_1 = 500$, $N_2 = 250$, $\ell = 0.60$ m, $A = 2.0\times10^{-3}$ m$^2$, $\mu_R = 2500$, $I_p = 1.5$ A at 60 Hz — gives $\mathcal{R} = 9.55\times10^{4}$ A⋅t/Wb, $L_1 = 2.62$ H, $M = 1.31$ H, and hence $\hat V_{AB} = 1481$ V peak with $\hat V_{XY} = 740$ V peak. These are plausible magnitudes for such a core and confirm the algebra dimensionally.

Quantity	Result
Core reluctance	$\mathcal{R} = \ell/(\mu_0\mu_R A)$
Core flux	$\varphi(t) = N_1 i_1(t)/\mathcal{R}$
Self-inductance	$L_1 = \mu_0\mu_R A N_1^{2}/\ell$
Mutual inductance	$M = \mu_0\mu_R A N_1 N_2/\ell$
[a] Primary voltage	$v_1 = L_1\,\mathrm{d}i_1/\mathrm{d}t = \omega L_1 I_p\cos\omega t$
[a] Secondary voltage	$v_2 = M\,\mathrm{d}i_1/\mathrm{d}t = (N_2/N_1)v_1$
[b] Input impedance	$Z_{\text{in}} = j\omega L_1$ (purely inductive, $\angle +90^\circ$)
[c] Phase relation	both voltages lead $i_1$ by 90°; amplitude ratio $N_2/N_1$

Topic. Magnetic circuits, Faraday’s law, self and mutual inductance, and the open-circuit transformer.

Key relations. $\mathcal{R} = \ell/(\mu_0\mu_R A)$; $\varphi = Ni/\mathcal{R}$; $v = N\,\mathrm{d}\varphi/\mathrm{d}t$; $L = N^{2}/\mathcal{R}$; $M = N_1N_2/\mathcal{R}$; $v_2/v_1 = N_2/N_1$; $Z_{\text{in}} = j\omega L_1$.

Why this works. The magnetic-circuit analogy lets a distributed field problem be solved like a simple dc circuit: mmf $Ni$ plays the role of emf, flux $\varphi$ the role of current, and reluctance $\ell/\mu A$ the role of resistance. With the secondary open there is only one mmf source and one flux path, so a single division gives the flux, and Faraday’s law converts it to voltage at each winding by simply counting turns — which is precisely why the voltage ratio is the turns ratio and nothing else. The derivative in Faraday’s law is what puts the voltages a quarter cycle ahead of the current, and it is also the reason the input impedance is purely imaginary: with no resistance and no secondary current there is no mechanism to absorb real power, only to store and return magnetic energy each half cycle. Note that the high $\mu_R$ of the iron is what makes $\mathcal{R}$ small and $L_1$ large; introducing even a small air gap would raise the reluctance dramatically and cut the inductance.

Common pitfall. Writing $v_2/v_1 = N_1/N_2$ — the voltage ratio follows the turns ratio directly, while it is the current ratio that inverts. Equally common is losing the $N_1^{2}$ in $L_1$: one factor of $N_1$ comes from producing the flux and the second from linking it. Finally, sketching the voltages as lagging the current (as they would in a resistive-inductive load) rather than leading it by exactly 90° misses the point of the loss-free assumption.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., §1.3–1.5 (magnetic circuits, Faraday’s law) and §2.1–2.4 (the ideal transformer and the magnetising branch).

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

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