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A series circuit consists of a resistance of 6 Ω and an inductive reactance of 8 Ω. A potential difference of 141.4 V (r.m.s.) is applied to it. At a certain instant the applied voltage is + 100 V, and is increasing. Calculate at this current, (i) the current (ii) the voltage drop across the resistance and (iii) voltage drop across inductive reactance. |
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Answer» Z = R + jX = 6 + j8 = 10 ∠ 53.1° It shows that current lags behind the applied voltage by 53.1°. Let V be taken as the reference quantity. Then v = (141.4 x √2) × sinωt = 200 sin ω t; i = (Vm/Zsinωt) − 30° = 20sin(ωt − 53.1°). (i) When the voltage is + 100 V and increasing; 100 = 200 sin ωt; sin ωt = 0.5 ; ω t = 30° At this instant, the current is given by i = 20 sin (30° − 53.1°) = − 20 sin 23.1° = − 7.847A. (ii) drop across resistor = iR = − 7.847 × 6 = − 47 V. (iii) Let us first find the equation of the voltage drop VL across the inductive reactance. Maximum value of the voltage drop = ImXL = 20 × 8 = 160 V. It leads the current by 90°. Since current itself lags the applied voltage by 53.1°, the reactive voltage drop across the applied voltage by (90° − 53.1°) = 36.9°. Hence, the equation of this voltage drop at the instant when ωt = 30° is VL = 160 sin(30° + 36.9°) = 160sin66.9° = 147.2V. |
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