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 load of 30 Ω per phase is to be replaced by an equivalent balanced star load. The star impedance per phase is:
सही उत्तर: ख) 10 Ω
For a balanced network Z_Y = Z_Δ/3 = 30/3 = 10 Ω.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Three 20 Ω resistors are star-connected to a 400 V (line) three-phase supply. The total power consumed is:
सही उत्तर: क) 8 kW
Phase voltage = 400/√3 ≈ 231 V; P = 3 × 231²/20 = 400²/20 = 8 kW (in delta it would be 24 kW).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A load current of 20 A has a power factor of 0.6 lagging. Its active (in-phase) and reactive (quadrature) components are:
सही उत्तर: ख) 12 A and 16 A
Active component = I cos φ = 20 × 0.6 = 12 A; reactive component = I sin φ = 20 × 0.8 = 16 A.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A series RL circuit has R = X_L at 50 Hz. If the frequency is raised to 100 Hz, the power factor becomes about:
सही उत्तर: ग) 0.447
At 100 Hz X_L = 2R, so pf = R/√(R² + 4R²) = 1/√5 ≈ 0.447.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A coil of 0.1 H and 10 Ω resistance is in parallel with a 10 μF capacitor. The dynamic impedance at resonance is:
सही उत्तर: घ) 1000 Ω
Z_D = L/(CR) = 0.1/(10 × 10⁻⁶ × 10) = 1000 Ω.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The capacitance needed to make a 0.1 H inductor resonate at 1 kHz is about:
सही उत्तर: ख) 0.253 μF
C = 1/(ω²L) = 1/((2π × 1000)² × 0.1) ≈ 2.53 × 10⁻⁷ F.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A series RLC circuit at 100 V resonance has R = 5 Ω and X_L = X_C = 250 Ω. The voltage across the capacitor at resonance is:
सही उत्तर: घ) 5000 V
I = 100/5 = 20 A; V_C = I X_C = 20 × 250 = 5000 V (Q = 50 times the supply voltage).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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If the power factor of a load is improved from 0.6 to 1.0 at the same active power and voltage, the line current is reduced by:
सही उत्तर: क) 40%
I ∝ 1/pf at constant P and V: the new current is 0.6 of the old, a reduction of 40%.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 230 V, 50 Hz single-phase load takes 2 kW at 0.7 power factor lagging. The shunt capacitance needed to raise the power factor to unity is about:
सही उत्तर: ग) 123 μF
Q = P tan φ = 2000 × 1.0202 ≈ 2040 VAR; C = Q/(V²ω) = 2040/(230² × 314.16) ≈ 123 μF.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A load draws a current of 5∠−36.87° A from a 100∠0° V supply. Its active and reactive powers are:
सही उत्तर: घ) 400 W and 300 VAR (lagging)
S = VI* = 100 × 5∠36.87° = 400 + j300 VA: P = 400 W and Q = 300 VAR absorbed (lagging, inductive).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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