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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A balanced delta-connected resistive load takes power P. If one of its three phase resistors burns out (open circuit), the power taken becomes:
सही उत्तर: क) 2P/3
The remaining two phases still have full line voltage across them, so each continues to take P/3: total 2P/3.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The total instantaneous power in a balanced three-phase circuit is:
सही उत्तर: ख) constant, equal to 3V_ph I_ph cos φ
The double-frequency pulsating terms of the three phases cancel, leaving constant power 3V_ph I_ph cos φ, which gives smoother motor torque.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A voltage v = 100 sin ωt + 50 sin 3ωt V drives a current i = 10 sin ωt + 5 sin(3ωt − 60°) A. The average power is:
सही उत्तर: क) 562.5 W
Only like harmonics give power: P = (100 × 10/2) cos 0° + (50 × 5/2) cos 60° = 500 + 62.5 = 562.5 W.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A current contains a fundamental of 10 A (RMS), a 3rd harmonic of 3 A and a 5th harmonic of 4 A. Its total harmonic distortion (THD) is:
सही उत्तर: ख) 50%
THD = √(3² + 4²)/10 = 5/10 = 50%.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Which of the following can be used to improve the power factor of an industrial plant, besides static capacitors?
सही उत्तर: क) an over-excited synchronous motor (synchronous condenser)
An over-excited synchronous machine draws leading current and supplies reactive power, like a capacitor bank.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A factory has a 10 kW resistive load and a 10 kW motor load at 0.6 power factor lagging. The overall power factor is about:
सही उत्तर: घ) 0.83 lagging
Total P = 20 kW; the motor's Q = 10 × 0.8/0.6 = 13.33 kVAR; pf = 20/√(20² + 13.33²) ≈ 0.83 lagging.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A series RLC circuit has half-power frequencies of 900 Hz and 1100 Hz. Its resonant frequency is about:
सही उत्तर: ख) 995 Hz
For a series RLC circuit f₀ = √(f₁f₂) = √(900 × 1100) ≈ 995 Hz (the arithmetic mean is only an approximation).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A series circuit has R = 10 Ω, L = 0.1 H and C = 100 μF on a 50 Hz supply. Its impedance is about:
सही उत्तर: ग) 10 Ω
X_L = 2π × 50 × 0.1 = 31.42 Ω and X_C = 1/(2π × 50 × 10⁻⁴) = 31.83 Ω nearly cancel, so Z = √(10² + 0.41²) ≈ 10 Ω (near resonance).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A symmetrical square wave of peak 10 V is measured with an average-responding meter calibrated to read RMS for sine waves. The meter reads about:
सही उत्तर: क) 11.1 V
The meter senses the average (10 V) and multiplies by the sine form factor 1.11, so it shows 11.1 V although the true RMS is 10 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The peak factor (crest factor) of a symmetrical triangular wave is:
सही उत्तर: ग) √3 ≈ 1.73
For a triangular wave V_rms = V_m/√3, so peak/RMS = √3.
💬 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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