EE MCQ

MGVCL Exam Paper (30-07-2021 Shift 1)
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An RC series circuit with R = 150 Ω and C = 2 μF is excited by a DC source of 15 V by closing the switch at t = 0 with an initially uncharged capacitor. The initial and final voltages across the capacitor respectively are

For t = 0 in RC circuit,

After t = 0+ condition, capacitor is behave as short circuit.

Voltage across the it 0.

At final time,

Capacitor is behave as open circuit, voltage across the capacitor at final time is equal to 15 V.

Capacitor opposes the change of voltage.

Related Question

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

MGVCL Exam Paper (30-07-2021 Shift 1)
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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.

MGVCL Exam Paper (30-07-2021 Shift 1)
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A given solar cell has the following specifications:

Open-circuit voltage, Voc: 0.7 V

Short-circuit current, Isc: 3.5 A

Short-circuit current, Isc: 3.5 A

Short-circuit current, Isc: 3.5 A

If the solar cell operates at maximum power point, calculate the efficiency.

Open-circuit voltage, Voc: 0.7 V

Short-circuit current, Isc: 3.5 A

Short-circuit current, Isc: 3.5 A

Short-circuit current, Isc: 3.5 A

If the solar cell operates at maximum power point, calculate the efficiency.

Equation is given by:

Efficiency = (F.F*Voc*Isc)/Pin

= (0.6*3.5*0.7)/10

= 0.14702

% efficiency = 14.7 %

MGVCL Exam Paper (30-07-2021 Shift 1)
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Calculate the capacitance per meter of a 50 Ω load cable that has an inductance of 50 nH/m.

Z = √(L/C)

C = L/Z²

C = 50*10⁻⁹/(50*50)

C = 0.02 nF

C = 20 pF