Question 1 of 6: Part A: Multiple-Choice (20 questions)

Question 1 — Part A: Multiple-Choice (20 questions)

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.

These are single-best-answer conceptual items; four require a short calculation (Q2, Q6, Q8, Q12) and are worked inline. The remaining items are decided from machine and circuit theory. The complete key is collected in the results table.

1. Two-wattmeter method — negative reading. One wattmeter reads negative once the displacement angle exceeds $60^\circ$ (power factor below $0.5$), because $W_1=V_LI_L\cos(30^\circ+\theta)$ turns negative for $\theta\gt 60^\circ$. → (a).
2. Delta load, find phase resistance. In delta the phase voltage equals the line voltage $V_p=400\ \text{V}$ and the phase current is $I_p=I_L/\sqrt3=34.65/\sqrt3=20.0\ \text{A}$, so $Z_p=V_p/I_p=400/20=20\ \Omega$. From $P_T=3I_p^2R_p$: $R_p=14400/(3\cdot 20^2)=\boxed{12\ \Omega}$. → (c).
3. Delta → Y reconnection. For the same impedances on the same supply, the Y line current is one-third of the delta line current ($I_{L,Y}=\tfrac13 I_{L,\Delta}$), and the power likewise drops to one-third — but option (c) says "power," while the cleanest single true statement is the line current. → (b).
4. Air gap in a magnetic circuit. A gap raises reluctance and stores energy, keeping the core off the saturation knee and linearizing operation. It reduces flux and mmf efficiency, so its purpose is to prevent saturation. → (c).
5. Magnetic–electric analogy. Flux $\Phi$ is the "flow" driven by mmf across reluctance ($\mathcal F=\mathcal R\Phi$), analogous to current driven by emf across resistance ($V=RI$). Flux ↔ current. → (b).
6. Turns and length both doubled. $L=\mu N^2A/\ell$; doubling $N$ multiplies by $4$, doubling $\ell$ divides by $2$, net factor $4/2=2$. Inductance is doubled. → (b).
7. Laminated cores. Thin, insulated laminations break the eddy-current paths, reducing eddy-current loss (they do not affect copper or hysteresis loss). → (b).
8. Copper loss at half load. Copper loss scales with current squared: $P_{cu}\propto I^2$, so at half load $P_{cu}=1600\times(\tfrac12)^2=\boxed{400\ \text{W}}$. → (d).
9. Smallest conductor cross-section. The high-voltage winding carries the smallest current ($I\propto 1/V$) and therefore uses the smallest wire cross-section. → (d).
10. Short-circuit test — negligible core loss. The primary is excited at a low reduced voltage (just enough to circulate rated current), so the core flux and hence the iron loss are negligible. → (c).
11. Near-zero voltage regulation. A leading power-factor load produces a capacitive rise that offsets the series-impedance drop, giving regulation near (or below) zero. → (c).
12. Rotor emf frequency. $f_r=s\,f=0.04\times 50=\boxed{2\ \text{Hz}}$. → (b).
13. Squirrel-cage starting torque. With low rotor resistance the starting power factor is poor, so the starting torque is low (typically $1.5$–$2\times$ full-load, but "low" versus a resistance-optimized or wound rotor). → (b).
14. Speed control — the exception. Supply frequency, pole number, and stator voltage all change speed; the squirrel-cage rotor resistance is fixed and cannot be varied externally (only a wound rotor allows that). → (b).
15. Deep rotor bars. Deep bars exploit the skin/deep-bar effect: high effective rotor resistance at standstill (high slip frequency) improves starting torque, while running resistance stays low. → (a).
16. No-load statement that is NOT valid. At no load the slip is very small, so the rotor current, the slip-frequency rotor emf $sE_2$, and the slip itself are all low, and the no-load power factor is poor (low) — statements (b), (c), (d) are unambiguously valid. The keyed exception is (a): the rotor's standstill induced emf $E_2$ (set by the air-gap flux and turns ratio) is essentially unchanged from load to no-load, so describing "the induced emf in the rotor" as low is the invalid statement. → (a).
17. Highest no-load speed. A DC series motor has no shunt field to hold speed; at no load its flux collapses and it "runs away" to a dangerously high speed. → (b).
18. Lost shunt field while running. With $E_a=k\Phi\omega$ fixed by the supply, if $\Phi\to 0$ then $\omega$ must rise sharply — the motor speeds up (over-speeds). → (c).
19. Shunt motor as load is reduced. The shunt motor is essentially a constant-speed machine; reducing load lowers $I_aR_a$ only slightly, so speed stays almost constant. → (c).
20. Reversing a DC series motor. Reverse either the armature or the field connections alone; reversing the supply terminals flips both and leaves the direction unchanged. Interchanging the field terminals reverses rotation. → (b).