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
-
Three 100 W lamps are used for 5 hours a day for 30 days. At Rs. 12 per unit, the energy cost is:
सही उत्तर: क) Rs. 540
Energy = 0.3 kW × 5 h × 30 = 45 kWh; cost = 45 × 12 = Rs. 540.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A star network has R_A = 2 ohm, R_B = 3 ohm and R_C = 6 ohm. In the equivalent delta, the resistance between terminals A and B is:
सही उत्तर: ख) 6 ohm
R_AB = R_A + R_B + R_A R_B/R_C = 2 + 3 + 6/6 = 6 ohm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
Sources of 12 V (series 2 ohm) and 6 V (series 3 ohm) both feed one node, which is connected to ground through 6 ohm. By Millman's theorem, the node voltage is:
सही उत्तर: ग) 8 V
V = (12/2 + 6/3)/(1/2 + 1/3 + 1/6) = 8/1 = 8 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A 20 V source with a 5 ohm series resistance has a 20 ohm resistor across its output terminals. The Norton equivalent of this network is:
सही उत्तर: ग) 4 A in parallel with 4 ohm
Short-circuit current = 20/5 = 4 A; with the source shorted, R_N = 5 ohm parallel 20 ohm = 4 ohm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A source of emf 24 V and internal resistance 6 ohm feeds a load of 12 ohm. The power delivered to the load is about:
सही उत्तर: घ) 21.3 W
I = 24/(6 + 12) = 4/3 A, so P = (4/3)² × 12 ≈ 21.3 W (slightly below the 24 W obtained with a matched 6 ohm load).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A battery of emf 6 V and internal resistance 0.5 ohm delivers 2 A. The power wasted inside the battery is:
सही उत्तर: क) 2 W
Internal loss = I²r = 4 × 0.5 = 2 W.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
An 18 V supply feeds a 6 ohm resistor in series with a parallel pair of 12 ohm and 4 ohm. The current through the 12 ohm resistor is:
सही उत्तर: ख) 0.5 A
Total R = 6 + 3 = 9 ohm, so I = 2 A; the 12 ohm branch takes 2 × 4/16 = 0.5 A.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A balanced Wheatstone bridge has P = 2 ohm, Q = 4 ohm, R = 3 ohm and S = 6 ohm, with a 5 ohm galvanometer between the bridge midpoints. The resistance seen by the supply is:
सही उत्तर: ख) 10/3 ohm
Being balanced, the galvanometer carries no current and can be removed: (2 + 3) in parallel with (4 + 6) = 5 × 10/15 = 10/3 ohm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A total current of 12 A divides among three parallel resistors of 2 ohm, 3 ohm and 6 ohm. The current in the 2 ohm resistor is:
सही उत्तर: ग) 6 A
The parallel combination is 1 ohm, so the voltage is 12 V and the 2 ohm branch carries 12/2 = 6 A.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A 24 V supply is connected across a 3 kilo-ohm and a 9 kilo-ohm resistor in series. The voltage across the 9 kilo-ohm resistor is:
सही उत्तर: घ) 18 V
By the voltage divider rule, V = 24 × 9/(3 + 9) = 18 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।
नमस्ते! पाठ्यक्रम, सामान्य ज्ञान, भाषा परीक्षा वा कुनै प्रश्नको उत्तर बुझ्न सोध्नुहोस्।
AI ले गल्ती गर्न सक्छ। महत्वपूर्ण कुरा आधिकारिक स्रोतमा जाँच्नुहोस्।