आधारभूत विद्युत् (Basic Electrical) – Basic Electrical Engineering MCQ
आधारभूत विद्युत् (Basic Electrical) – इलेक्ट्रिकल र टेलिकम
Basic Electrical Engineering · Technical – Electrical, Mechanical & Telecom · २,२९९ प्रश्न
२,२९९ प्रश्न उपलब्ध। मोड छान्नुहोस्:
आधारभूत विद्युत् (Basic Electrical): प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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The current in a conductor varies as i = (3 + 2t) A. The charge that flows through it between t = 0 and t = 4 s is:
सही उत्तर: क) 28 C
q = ∫₀⁴(3 + 2t)dt = 12 + 16 = 28 C.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A mobile charger delivers 2 A at 12 V. Its output power is:
सही उत्तर: घ) 24 W
P = VI = 12 × 2 = 24 W.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 230/12 V transformer has 460 primary turns. The number of secondary turns is:
सही उत्तर: ग) 24
N₂ = N₁ × V₂/V₁ = 460 × 12/230 = 24.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A current of 50 mA flows through a 220 ohm resistor. The voltage across it is:
सही उत्तर: ख) 11 V
V = IR = 0.05 × 220 = 11 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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An air-filled parallel-plate capacitor of 10 μF has half of its plate area (the full gap thickness) filled with a dielectric of relative permittivity 5. Its new capacitance is:
सही उत्तर: क) 30 μF
The two halves act in parallel: 5 μF × 1 + 5 μF × 5 = 30 μF.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A capacitor of 100 μF charges through a 10 kilo-ohm resistor from a DC supply. The time for its voltage to reach half the supply voltage is about:
सही उत्तर: ग) 0.69 s
v = V(1 − e^{−t/RC}) = V/2 gives t = RC ln 2 = 1 × 0.693 ≈ 0.69 s.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A DC shunt motor runs at 1000 rpm with a back emf of 200 V. If its back emf falls to 180 V with the flux unchanged, its speed becomes:
सही उत्तर: घ) 900 rpm
With constant flux N ∝ E_b, so N = 1000 × 180/200 = 900 rpm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The air-gap power of a three-phase induction motor is 10 kW and it runs with a slip of 4%. Its rotor copper loss is:
सही उत्तर: क) 400 W
Rotor copper loss = s × air-gap power = 0.04 × 10000 = 400 W.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A transformer has an iron loss of 400 W and a full-load copper loss of 900 W. Its maximum efficiency occurs at a load equal to:
सही उत्तर: ख) 2/3 of full load
Maximum efficiency when x²P_cu = P_i: x = √(400/900) = 2/3.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 20 kVA transformer has an iron loss of 200 W and a full-load copper loss of 400 W. Its efficiency at half load and unity power factor is about:
सही उत्तर: ख) 97.1%
Output = 10 kW; copper loss at half load = 400/4 = 100 W; η = 10000/(10000 + 200 + 100) ≈ 97.1%.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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