Question 7 of 8: Mechanical power to drive a loop generating into a complex load
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because the set is a study resource rather than a sat examination. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, from which $\eta_0=\sqrt{\mu_0/\varepsilon_0}\approx 377\ \Omega$ and $c = 3\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, waveguides); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (statics, Faraday's law, plane waves); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short elements, image theory).
Question 7: Mechanical power to drive a loop generating into a complex load (20 marks)
Given. A single-phase elementary alternator: a multi-turn loop spun about a vertical axis in a steady horizontal field, feeding a fixed complex load. The loop's own inductance and resistance are to be neglected, so the loop is an ideal voltage source.
Given data
Quantity
Symbol
Value
Loop area
$A$
$400\ \text{cm}^{2}=0.0400\ \text{m}^{2}$
Number of turns
$N$
$45$
Rotational speed
—
$3600\ \text{rev/min}$
Flux density
$B$
$0.200\ \text{T}$ (horizontal, DC)
Load impedance
$Z_L$
$40+j40\ \Omega$ at 60 Hz
Loop self-inductance
—
neglected
Find. The time-averaged mechanical power that must be supplied to the shaft.
The rotating loop and its load. The rotation axis is vertical and perpendicular to $\mathbf{B}$, so the flux linkage varies sinusoidally at the mechanical frequency.
Approach. Convert the speed to an electrical frequency, differentiate the flux linkage to get the emf, drive the load with that source to find the rms current, and take the real power — which, for a lossless loop, is exactly the mean mechanical input.
Electrical frequency and angular velocity. One revolution of a two-pole loop produces one electrical cycle, so $$f=\frac{3600}{60}=60\ \text{Hz},\qquad \omega=2\pi f=377.0\ \text{rad/s},$$ which is exactly the frequency at which the load impedance is quoted — a useful internal consistency check on the data.
Flux linkage and induced emf. With $\theta=\omega t$ measured from the position of maximum linkage, $\Lambda(t)=NBA\cos\omega t$ with $NBA=(45)(0.200)(0.0400)=0.360\ \text{Wb}$. Faraday's law then gives $$e(t)=-\frac{d\Lambda}{dt}=NBA\,\omega\sin\omega t,\qquad E_{pk}=(0.360)(377.0)=135.7\ \text{V}.$$
Load current. Since the loop is treated as an ideal source, the load alone limits the current: $$|Z_L|=\sqrt{40^{2}+40^{2}}=56.57\ \Omega,\qquad I_{rms}=\frac{95.97}{56.57}=1.697\ \text{A},$$ at a phase angle of $45^{\circ}$ lagging, i.e. a power factor of 0.707.
Real power absorbed by the load. Only the resistive part converts: $$P=I_{rms}^{2}R=(1.697)^{2}(40)\;\Rightarrow\;\boxed{P=115.1\ \text{W}} .$$ The reactive power is $Q=I_{rms}^{2}X=115.1\ \text{var}$, equal to $P$ because $X=R$ here.
Close the energy balance to get the mechanical power. The loop has no resistance and its self-inductance is disregarded, so it stores and dissipates nothing on average and the shaft must supply exactly the electrical real power: $$\overline{P_{mech}}=P=115.1\ \text{W},\qquad \overline{\tau}=\frac{P}{\omega}=\frac{115.1}{377.0}=0.305\ \text{N}\cdot\text{m}.$$
It is worth being explicit about why the reactive power does not appear in the answer. The 40 $\Omega$ reactance exchanges energy with the source twice per cycle but returns all of it, so over a full cycle it draws no net mechanical work; it does, however, raise the current for a given real output and so increases the peak torque the shaft must withstand. The instantaneous mechanical power is not constant either: the product $e(t)i(t)$ pulsates at $120$ Hz about its 115.1 W mean, which is the classic single-phase torque ripple that polyphase machines exist to remove. Numerically averaging $e(t)i(t)$ over one cycle reproduces 115.1 W, confirming the phasor result.
