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

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.

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 $i_1(t)$.

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.

1. 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.
2. 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.
3. [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 $i_1 = I_p \sin\omega t$, $$\boxed{\;v_1(t) = \frac{\mu_0\mu_R A N_1^{2}}{\ell}\,\omega I_p \cos\omega t = \omega L_1 I_p \cos\omega t\;}$$
4. 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.
5. [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.
6. [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).