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
-
A two-wire DC distributor has a resistance of 0.1 ohm in each conductor and carries 20 A. The voltage drop along the line is:
सही उत्तर: क) 4 V
Current flows through both conductors: drop = 20 × (0.1 + 0.1) = 4 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
An energy meter has a constant of 1500 revolutions per kWh. When a 2 kW load is supplied for 30 minutes, the number of revolutions of its disc should be:
सही उत्तर: क) 1500
Energy = 2 × 0.5 = 1 kWh, so revolutions = 1500 × 1 = 1500.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A voltmeter is connected through an 11000/110 V potential transformer and reads 105 V. The line voltage being measured is:
सही उत्तर: ख) 10.5 kV
The PT ratio is 100, so the line voltage = 105 × 100 = 10,500 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A voltmeter reads 98 V when the true voltage is 100 V. The percentage error of the reading is:
सही उत्तर: ग) −2%
Error = (measured − true)/true × 100 = (98 − 100)/100 × 100 = −2%.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
An ammeter of resistance 0.1 ohm reads up to 5 A. The shunt needed to extend its range to 50 A is about:
सही उत्तर: घ) 0.0111 ohm
The shunt must carry 45 A at the meter voltage 0.5 V: S = 0.5/45 ≈ 0.0111 ohm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
Three identical coils each of resistance 6 ohm and reactance 8 ohm are connected in star to a 400 V, three-phase supply. The line current is about:
सही उत्तर: घ) 23.1 A
Z = 10 ohm per phase; phase voltage = 231 V; line current = phase current = 231/10 ≈ 23.1 A.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A balanced star-connected load has 10 ohm resistance per phase and draws 23.1 A per line. The line voltage of the supply is about:
सही उत्तर: क) 400 V
Phase voltage = 23.1 × 10 = 231 V; line voltage = √3 × 231 ≈ 400 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A balanced three-phase load draws 50 A per line from a 400 V supply. Its apparent power is about:
सही उत्तर: ख) 34.6 kVA
S = √3 V_L I_L = 1.732 × 400 × 50 ≈ 34,640 VA.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
In the two-wattmeter method, the wattmeters read 4 kW and 2 kW for a balanced load. The power factor of the load is about:
सही उत्तर: ग) 0.866
tan φ = √3(W1 − W2)/(W1 + W2) = √3 × 2/6 = 0.577, so φ = 30° and pf = 0.866.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A balanced delta-connected load with an impedance of 20 ohm per phase at 0.8 pf lagging is fed from a 400 V three-phase supply. The total active power is:
सही उत्तर: ग) 19.2 kW
Phase current = 400/20 = 20 A; P = 3 × 400 × 20 × 0.8 = 19,200 W.
💬 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।
नमस्ते! पाठ्यक्रम, सामान्य ज्ञान, भाषा परीक्षा वा कुनै प्रश्नको उत्तर बुझ्न सोध्नुहोस्।
AI ले गल्ती गर्न सक्छ। महत्वपूर्ण कुरा आधिकारिक स्रोतमा जाँच्नुहोस्।