Which of the following is the layer of a computer system between the hardware and the user program?

A 16 A, Type B Miniature Circuit Breaker (MCB) will disconnect a fault in 0.1s when the fault current is

Class B trip curve:
The MCB with class B trip characteristics trips instantaneously when the current flowing through it reaches between 3 to 5 times rated current. These MCBs are suitable for cable protection.
Hence, For 16 A current Type B MCB disconnect fault at 3*16 = 48 A to 5*16 = 80 A.

Insulation Power Factor testing on an oil-filled transformer is not helpful in determining ___

A form of AC testing that applies voltage and measures the leakage/loss current of electrical insulation.
It is a type of insulation testing used to evaluate the integrity of electrical insulation.

A thermistor is inherently suitable for temperature measurement involving

Thermistors are ideal when measuring a single point temperature that is within 50°C of ambient.
If the temperatures are excessively high or low, a thermistor will not work.
Most thermistors work best in the range between -55°C and +114°C.
Sensitivity of thermistor is high.
Span value of thermistor is low.

An alternator is supplying load of 300 kW at a power factor of 0.5 lagging. If the power factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading?

P = S*cosφ
S = P/cosφ
S = 300/0.5
S = 600 kVA
For same kVA rating and unity power factor,
P = 600*1
P = 600 kW
Pextra = 600 - 300
Pextra = 300 kW

Three equal star-connected inductors take 8 kW at 0.8 power factor when connected to a 460 V three - phase 3-wire supply. If one of the inductor in phase – Y is short circuited, then the line current in phase-Y is ___ Assume RYB sequence.

Here, Y phase is shorted,
Equation of 3-phase power = √3*VL*IL*cosφ
IL = P/(√3*VL*cosφ)
= 8000/(√3*400*0.8)
= 12.55 A
Phase impedance = Vp/Ip
= 460/(√3*12.55)
= 21.16 ohm
cosφ = 0.8, So φ = 37 degree.
Here Impedance of Y phase is shorted,
So, Neutral point and Y phase at same potential,
Three line voltages are given by,
Vry = 460∟0°
Vyb = 460∟120°
Vbr = 460∟-120°
Here N and Y phase at same potential,
Ir = (460∟-120°)/(21.16∟37°)
= -21.7∟83° A
Ib = (-460∟120°)/(21.16∟37°)
= 21.7∟-157° A
Ib = -(Ir + Iy)
= -21.7∟83° + 21.7∟-157°
= 37.6∟53° A