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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Three 10 ohm heaters are star-connected across a 400 V (line), three-phase supply. Their total power consumption is:
सही उत्तर: घ) 16 kW
Phase voltage = 400 / sqrt 3 = 231 V; power per phase = 231^2 / 10 = 5.33 kW; total = 3 x 5.33 = 16 kW.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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For a star (Y) connection, how does the current in each supply line compare with the current in the phase winding it feeds?
सही उत्तर: घ) They are equal
In a star connection each line is in series with one phase, so line current = phase current.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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If the capacitance of a series LC circuit is made four times larger, its resonant frequency:
सही उत्तर: ख) Becomes half
f0 = 1 / (2 pi sqrt(LC)), so multiplying C by 4 divides f0 by sqrt 4 = 2.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A factory draws 400 kW at 0.6 lagging power factor. What capacitor rating is needed to raise the power factor to 0.8 lagging?
सही उत्तर: ग) About 233 kVAR
Q1 = 400 x tan(acos 0.6) = 400 x 4/3 = 533 kVAR; Q2 = 400 x 0.75 = 300 kVAR; capacitor = 533 - 300 = 233 kVAR.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A capacitor of reactance 23 ohm is connected to a 230 V AC supply. The reactive power it supplies is:
सही उत्तर: क) 2300 VAR
Q = V^2 / X_C = 230^2 / 23 = 52900 / 23 = 2300 VAR.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Using the voltage phasor V and current phasor I of a load, the complex power S = P + jQ is calculated as:
सही उत्तर: ख) V multiplied by the complex conjugate of I
S = V I* gives P + jQ with Q positive for an inductive (lagging) load.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The admittance of an impedance Z = 3 + j4 ohm is:
सही उत्तर: घ) 0.12 - j0.16 S
Y = 1/Z = (3 - j4) / (3^2 + 4^2) = (3 - j4)/25 = 0.12 - j0.16 S.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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From the same AC supply, a resistor branch draws 3 A and a pure inductor branch connected in parallel draws 4 A. The total supply current is:
सही उत्तर: क) 5 A
The branch currents are 90 degrees apart, so I = sqrt(3^2 + 4^2) = 5 A.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Two 50 Hz sinusoidal voltages differ in phase by 90 degrees. The time between their positive peaks is:
सही उत्तर: ग) 5 ms
One cycle lasts 20 ms (360 degrees), so 90 degrees corresponds to 20 / 4 = 5 ms.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A coil of 100 turns and area 0.01 m^2 rotates at 100 rad/s in a uniform magnetic field of 0.1 T. The peak emf induced is:
सही उत्तर: ग) 10 V
E_m = N B A omega = 100 x 0.1 x 0.01 x 100 = 10 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
यी अभ्यास प्रश्न हुन्, आधिकारिक प्रश्न होइनन्। पाठ्यक्रम, विज्ञापन र नतिजाका लागि सम्बन्धित निकायको आधिकारिक वेबसाइट हेर्नुहोस्। उत्तरमा गल्ती भेटे प्रश्नमुनिको "टिप्पणी / गल्ती रिपोर्ट" मा लेख्नुहोस्।
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