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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Resistors of 2 ohm, 3 ohm and 6 ohm are connected in parallel. The equivalent resistance is:
सही उत्तर: क) 1 ohm
1/R = 1/2 + 1/3 + 1/6 = 1, so R = 1 ohm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 20 V source with a 10 Ω series resistor feeds a 10 Ω shunt resistor; a further 5 Ω series resistor leads from the top of the shunt to the output terminals. The Thevenin equivalent at the output terminals is:
सही उत्तर: क) 10 V in series with 10 Ω
Open-circuit: no current in the 5 Ω, so V_Th = 20 × 10/(10 + 10) = 10 V. With the source shorted, R_Th = 5 + (10 ∥ 10) = 5 + 5 = 10 Ω.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Which first three colour bands identify a 3.3 kΩ resistor in the four-band code?
सही उत्तर: ग) orange, orange, red
Orange = 3, orange = 3 and red multiplier = ×100: 33 × 100 = 3300 Ω = 3.3 kΩ.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 12 V, 24 W resistive lamp is to be run from a 230 V, 50 Hz supply using a series capacitor (to avoid resistive loss). The capacitance needed is about:
सही उत्तर: ख) 28 μF
Lamp: I = 2 A, R = 6 Ω. Required |Z| = 230/2 = 115 Ω, so X_C = √(115² − 6²) ≈ 114.8 Ω and C = 1/(2π × 50 × 114.8) ≈ 27.7 μF.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 0.1 H inductor carries 2 A. To what voltage must a 50 μF capacitor be charged to store the same energy?
सही उत्तर: घ) about 89 V
Inductor energy = ½LI² = ½ × 0.1 × 4 = 0.2 J; ½CV² = 0.2 J gives V² = 0.4/(50 × 10⁻⁶) = 8000, so V ≈ 89.4 V.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 200 V DC supply feeds a 50 Ω load through a two-wire cable whose total (go-and-return) resistance is 2 Ω. The power lost in the cable is about:
सही उत्तर: घ) 29.6 W
I = 200/(50 + 2) ≈ 3.85 A; cable loss = I²R = 3.85² × 2 ≈ 29.6 W.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The electric field strength E at a point is related to the electric potential V by:
सही उत्तर: ग) E = −dV/dx (the negative of the potential gradient)
The field points in the direction of decreasing potential, and its magnitude is the potential gradient, so E = −dV/dx (unit V/m).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The difference between an overload and a short circuit is that:
सही उत्तर: क) an overload is excess current in a healthy circuit, whereas a short circuit is a very low-impedance fault between conductors causing a very large current
Overload currents are moderate and sustained (cleared by the thermal trip); short-circuit currents are many times rated current and must be cleared instantly.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a working series circuit of a battery, a closed (ideal) switch and a lamp, the voltage measured across the closed switch is:
सही उत्तर: ख) zero
An ideal closed switch has zero resistance, so V = IR = 0 across it; the whole supply voltage appears across the lamp.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A capacitor is marked '4n7'. Its capacitance is:
सही उत्तर: ख) 4.7 nF
The letter n stands for nanofarad and marks the decimal point: 4n7 = 4.7 nF (= 4700 pF).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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