प्राविधिक – इलेक्ट्रिकल, मेकानिकल र टेलिकम
प्राविधिक – इलेक्ट्रिकल, मेकानिकल र टेलिकम
Technical – Electrical, Mechanical & Telecom · ४,५०० प्रश्न
४,५०० प्रश्न उपलब्ध। मोड छान्नुहोस्:
इलेक्ट्रिकल र टेलिकम: 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 50 Hz induction motor runs with a slip of 4%. The frequency of the rotor currents is:
सही उत्तर: ग) 2 Hz
Rotor frequency f_r = s f = 0.04 × 50 = 2 Hz.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 4-pole wave-wound DC generator has 400 armature conductors and a flux of 25 mWb per pole, and runs at 1200 rpm. Its generated emf is:
सही उत्तर: घ) 400 V
E = PΦZN/(60A) with A = 2 for wave winding: 4 × 0.025 × 400 × 1200/120 = 400 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The secondary voltage of a transformer is 240 V at no load and 230 V at full load. Its voltage regulation (referred to full-load voltage) is about:
सही उत्तर: क) 4.35%
Regulation = (V_NL − V_FL)/V_FL = 10/230 ≈ 4.35% (it would be 4.17% if referred to the no-load voltage).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A transformer has an iron loss of 1 kW and a full-load copper loss of 4 kW. Maximum efficiency occurs at what fraction of full load?
सही उत्तर: ख) one half
Maximum efficiency occurs when copper loss = iron loss: x²(4) = 1, so x = 0.5.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 100 kVA transformer has an iron loss of 1 kW and a full-load copper loss of 2 kW. Its efficiency at half load and 0.8 power factor is about:
सही उत्तर: ख) 96.4%
Output = 0.5 × 100 × 0.8 = 40 kW; losses = 1 + (0.5)² × 2 = 1.5 kW; η = 40/41.5 ≈ 96.4%.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 2200 V, 50 Hz transformer core has a net cross-section of 0.02 m² and a maximum flux density of 1.2 T. The number of primary turns required is about:
सही उत्तर: क) 413
N = V/(4.44 f B_m A) = 2200/(4.44 × 50 × 1.2 × 0.02) ≈ 413.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Three identical resistors are star-connected (no neutral wire) to a balanced three-phase supply. If one resistor goes open-circuit, the total power becomes:
सही उत्तर: घ) one half of the original
The remaining two resistors are in series across a line voltage: P = V_L²/(2R), while originally P = V_L²/R; so the power halves.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A three-phase, 11 kV line carries 52.5 A. The apparent power transmitted is about:
सही उत्तर: ग) 1000 kVA
S = √3 V_L I_L = 1.732 × 11 kV × 52.5 A ≈ 1000 kVA.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a balanced star-connected system with positive (RYB) phase sequence, the line voltage V_RY:
सही उत्तर: ग) leads the phase voltage V_R by 30°
V_RY = V_R − V_Y = √3 V_ph∠30° relative to V_R, so it is √3 times larger and leads V_R by 30°.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A balanced three-phase load of 30 kW at 0.866 lagging power factor is measured by the two-wattmeter method. The two readings are:
सही उत्तर: क) 20 kW and 10 kW
With φ = 30°, W₁ ∝ cos 0° and W₂ ∝ cos 60°, so W₁ = 2W₂; with W₁ + W₂ = 30 kW, the readings are 20 kW and 10 kW.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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