નીચેના રુઢિપ્રયોગોને અર્થ સાથે સાંકળી જોડી બનાવો.
1. કાન કરડવા - 1. બાતમી મળી જવી.
2. કાન ફૂંકવા - 2. વાત પર ધ્યાન ન આપવું.
૩. કાને વાત પહોચવી - ૩. કાન

In the following question, one part of the sentence may have an error. Find out which part of the sentence has an error and select the option corresponding to it. If the sentence contains no error, Select "No error'' option. (Avoid punctuation errors)

(A) Bubin together with / (B) her parents are coming / (C) to attend the party. / (D) NO ERROR

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

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

Determine maximum permissible load which a 15 MVA, ONAN cooled transformer to IS : 2026 can take for 6 hr, if the initial load on it was 12 MVA. The weighted ambient temperature is 20˚C. Assume the permissible load kVA as a fraction of rated kVA is 1.25.

Maximum permissible load = Rated MVA rating*fraction of rated kVA
Maximum permissible load = 15*1.25
Maximum permissible load = 18.75 MVA

જ્વાળા: ધુમાડો:: બરફ : ___