Your Answer : (Not Answered)

Correct Answer : D

The emf reports the difference between the junctions, so if the reference is warmer than the compensation assumes, the computed hot junction temperature comes out low by exactly that amount. Lead resistance hardly matters, since the measuring circuit draws almost no current.

Your Answer : (Not Answered)

Correct Answer : B

It is a transformer and needs an alternating excitation, typically a few kilohertz, so that the coupling to the secondaries can induce anything at all. A phase-sensitive detector then converts the output to a signed direct voltage.

Your Answer : (Not Answered)

Correct Answer : A

A shaft encoder and a turbine flow meter with a pulse pickup are examples: the output is a count or a code rather than a magnitude. No analogue to digital conversion is needed afterwards, which is a considerable advantage in a digital system.

Your Answer : (Not Answered)

Correct Answer : D

A longer integration averages the interference over more of its cycles and so suppresses it further, at the cost of a slower reading rate. Twisted or screened leads with the guard properly connected would help further still.

Your Answer : (Not Answered)

Correct Answer : B

Mains hum superimposed on the input averages exactly to nothing over a whole number of its own cycles, so the integrator sees only the wanted signal. This is the entire reason for the twenty millisecond period and the reason given is the correct one.

Your Answer : (Not Answered)

Correct Answer : A

A moving pointer conveys direction and rate of change immediately, whereas a flickering set of digits conveys neither. This is why many digital instruments now include a bar graph alongside the numerals, to recover what the pointer gave.

Your Answer : (Not Answered)

Correct Answer : C

A faster electron spends less time between the plates and is deflected less, so sensitivity falls as the accelerating voltage rises. Brightness does improve, which is the conflict that post-deflection acceleration was devised to resolve.

Your Answer : (Not Answered)

Correct Answer : A

Metering wants early saturation to protect the instruments and high accuracy at normal current; protection wants no saturation until far above rated current and accepts a looser ratio. No single core can do both, so two cores share one primary conductor.

Your Answer : (Not Answered)

Correct Answer : B

There is nothing to zero: the register counts from wherever it stands and only the difference between two readings is used. An indicating instrument needs a zero adjustment because its pointer must rest at the mark when nothing is applied.

Your Answer : (Not Answered)

Correct Answer : C

The energy is \( 20/1200 = 0.01667 \) kWh, and dividing by the period expressed in hours, \( 30/3600 = 0.00833 \) h, gives 2 kW. The general form is \( P = 3600n/(Kt) \) with n revolutions in t seconds and K in rev/kWh.

Your Answer : (Not Answered)

Correct Answer : C

For three wires Blondel's minimum is two, so the third instrument adds nothing to the total. What the two-wattmeter method cannot give is the power of an individual phase, since neither reading is the power of any one phase by itself.

Your Answer : (Not Answered)

Correct Answer : C

The scale is uniform and the instantaneous multiplication makes the reading true power whatever the waveform. Damping, however, is by air friction, because a damping magnet would interfere with the weak working field of the coils.

Your Answer : (Not Answered)

Correct Answer : A

Magnitude balance requires the products of opposite pairs to be equal, so \( 200 \times Z = 400 \times 100 \) and \( Z = 200\ \Omega \). The phase condition is a separate equation and must be satisfied at the same time before the detector goes silent.

Your Answer : (Not Answered)

Correct Answer : B

The product \( RC \) is \( 10^4 \times 10^{-8} = 10^{-4} \) second, and dividing that into \( 1/2\pi \) gives about 1592 hertz. The same network, placed round an amplifier, fixes the frequency of the Wien bridge oscillator.

Your Answer : (Not Answered)

Correct Answer : C

Ohm's law gives \( R = 2 \times 10^{-3} / 10 = 2 \times 10^{-4} \) ohm, that is 0.2 milliohm or 200 microhms. Such a value could not be measured at all without a large current and separate potential leads, which is exactly what the four-terminal construction provides.

Your Answer : (Not Answered)

Correct Answer : C

Leakage creeping over the surface of the insulation would otherwise be added to the wanted volume current. A conducting band round the specimen, connected to the guard, intercepts that surface current and returns it to the source without letting it pass through the measuring element.

Your Answer : (Not Answered)

Correct Answer : A

The scale reads from the zero mark, but the electrical arm includes the thick copper end strips and the soldered joints beyond it. These add a small equivalent length at each end, found by a measurement with two known resistances and thereafter allowed for.

Your Answer : (Not Answered)

Correct Answer : C

Traceability is what connects a reading to the definition of the unit, and without it no amount of resolution or repetition establishes correctness. Repetition reduces the random component only, and extra digits may show nothing but noise.

Your Answer : (Not Answered)

Correct Answer : A

Correct answer: False, since resolution and accuracy are different properties

Extra digits let the instrument discriminate finer changes, which is resolution. Whether those digits are correct is accuracy, and that depends on calibration. A five-digit meter that is 1 per cent out displays its error to five figures - the additional digits make the wrong answer more precise, not more right.

Your Answer : (Not Answered)

Correct Answer : D

Correct answer: 0.16

\[ \sigma_{\bar{x}} = \frac{0.5}{\sqrt{10}} = \frac{0.5}{3.162} = 0.158. \] The mean is therefore far better determined than any single reading - but only against random error. If the instrument reads systematically high, all ten readings are high and this figure says nothing about it.