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AL phy (urgent!!)
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1. A 1000 uF capacitor is joined in series with a 2.5V, 0.30A lamp and 50 Hz supply. Calculate (a) the p.d. (r.m.s.) of the supply to light the lamp to its normal brightness, and (b) the p.d. across the capacitor and the resistor respectively. 2. A 2.0H inductor of resistance 80Ω is connected in series with a 120Ω... 顯示更多 1. A 1000 uF capacitor is joined in series with a 2.5V, 0.30A lamp and 50 Hz supply. Calculate (a) the p.d. (r.m.s.) of the supply to light the lamp to its normal brightness, and (b) the p.d. across the capacitor and the resistor respectively. 2. A 2.0H inductor of resistance 80Ω is connected in series with a 120Ω resistor and a 240V, 50Hz supply. Find (a) the current in the circuit, and (b) the phase angle between the applied p.d. and the current. 3. A circuit consists of an inductor of 200uH and resistance 10Ω in series with a variable capacitor and a 0.10V, 1.0MHz supply. Calculate (a) the capacitor to give resonance (b) the p.d.s across the inductor and the capacitor at resonance, and (c) the Q-factor of the circuit at resonance (Q-factor=VL/V)
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1a) Resistance of the lamp = 2.5/0.3 = 8.33 Ω Reactance of the capacitor = 1/(2π x 50 x 1000 x 10-6) = 3.18 Ω Hence impedance = √(8.332 + 3.182) = 8.92 Ω So the required supply r.m.s. p.d. = 8.92 x 0.3 = 2.67 V b) Since the lamp is under its normal brightnes, p.d. across it = 2.5 V P.d. across the capacitor = 3.18 x 0.3 = 0.95 V 2a) Total resistance = 200 Ω Reactance = 2π x 50 x 2 = 628.3 Ω Hence impedance = √(2002 + 628.32) = 659.4 Ω Current = 240/659.4 = 0.303 A b) Since the reactive component of the impedance is inductive, the current should suffer a phase lag to the applied p.d. with the angle: tan-1 (628.3/200) = 72.3 deg. 3a) For resonance: 1/(2πfC) = 2πfL C = 1/[(2πf)2L] = 1.267 x 10-10 F b) At resonance, current = 0.1/10 = 0.01 A P.d. across L = 0.01 x 2π x 106 x 200 x 10-6 = 12.57 V Since resonance occurs, p.d. across C is also 12.57 V c) Q-factor = 12.57/0.1 = 125.7
其他解答:E2A5F59BAA12C031
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