EE MCQ

MGVCL Exam Paper (30-07-2021 Shift 1)
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In a 3-phase, 4-wire, 400/230 V system, a lamp (L1) of 100 W is connected to one phase and neutral and a lamp (L2) of 150 W is connected to the second pahse and neutral. If the neutral wire is disconnected accidently, what will be the voltage across a 100 W lamp (L1)?

When neutral disconnected Lamp are in series connection,

So, for series connection current is same for both Lamp,

Resistance of Lamp (L1) = (230)²/P

= (230)²/100

= 529 ohm

Resistance of lamp (L2) = (230)²/150

= 352.67 ohm

So, current through the lamp = E/(R1 + R2)

= 400/(529 + 352.67)

= 0.454 A

Voltage across lamp (L1) = I*R1

= 0.454*230

= 240 V

Related Question

MGVCL Exam Paper (30-07-2021 Shift 1)
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The necessary equation to be solved during the load flow analysis using Fast-Decoupled methid is given below:

where B' and B'' are formed using the imaginary part of the bus admittance matrix Ybus, n is the total number of buses in the system, m is the number of voltage regulated buses. The size of matrix B' is

∆P/|Vi|= -B' ∆δ

∆Q/|Vi| = -B'' ∆|Vi|

where B' and B'' are formed using the imaginary part of the bus admittance matrix Ybus, n is the total number of buses in the system, m is the number of voltage regulated buses. The size of matrix B' is

∆P/|Vi|= -B' ∆δ

∆Q/|Vi| = -B'' ∆|Vi|

In FDLF method,

B' corresponds to susceptance of unknown of PQ and PV bus

Order of B' matrix is = (n -1)*(n - 1)

Where n is total no. of buses.

Order of B'' = (n - m - 1)*(n - m - 1)

MGVCL Exam Paper (30-07-2021 Shift 1)
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A THHN conductor will have 75°C termination on one end and a 60/75°C termination on the other, which ampacity column will be used?

Conductor with 75 °C temperature rating must have ampacity from the 60°C column or the 75°C column.

It is permissible to use the 75°C rating if the installed conductor is rated at least 75 °C.

Because all of the connection points in this example have at least a 75°C rating, the conductor’s ampacity can be based on the 75°C column