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Question Number 15881 by tawa tawa last updated on 15/Jun/17

Commented by sandy_suhendra last updated on 16/Jun/17

Answered by sandy_suhendra last updated on 16/Jun/17

V=240 v  f = 50 Hz  i = 10 A  V_R =140 v  V_L =50 v  1)V^2 =V_R ^(   2)  + (V_L −V_C )^2       240^2 =140^2 +(V_L −V_C )^2       (V_L −V_C )^2  = 38,000       50−V_C  = −195       V_C  = 245 v  2) V_C  = i.X_C        245 = 10 ((1/(2πfC)))       24.5 = (1/(2×3.14×50×C))        C = 0.00013 F = 130 μF  3) V_L  = i.X_L         50 = 10 (2πfL)           5 = 2×3.14×50×L           L = 0.016 H = 16 mH

$$\mathrm{V}=\mathrm{240}\:\mathrm{v} \\ $$$$\mathrm{f}\:=\:\mathrm{50}\:\mathrm{Hz} \\ $$$$\mathrm{i}\:=\:\mathrm{10}\:\mathrm{A} \\ $$$$\mathrm{V}_{\mathrm{R}} =\mathrm{140}\:\mathrm{v} \\ $$$$\mathrm{V}_{\mathrm{L}} =\mathrm{50}\:\mathrm{v} \\ $$$$\left.\mathrm{1}\right)\mathrm{V}^{\mathrm{2}} =\mathrm{V}_{\mathrm{R}} ^{\:\:\:\mathrm{2}} \:+\:\left(\mathrm{V}_{\mathrm{L}} −\mathrm{V}_{\mathrm{C}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{240}^{\mathrm{2}} =\mathrm{140}^{\mathrm{2}} +\left(\mathrm{V}_{\mathrm{L}} −\mathrm{V}_{\mathrm{C}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\left(\mathrm{V}_{\mathrm{L}} −\mathrm{V}_{\mathrm{C}} \right)^{\mathrm{2}} \:=\:\mathrm{38},\mathrm{000} \\ $$$$\:\:\:\:\:\mathrm{50}−\mathrm{V}_{\mathrm{C}} \:=\:−\mathrm{195} \\ $$$$\:\:\:\:\:\mathrm{V}_{\mathrm{C}} \:=\:\mathrm{245}\:\mathrm{v} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{V}_{\mathrm{C}} \:=\:\mathrm{i}.\mathrm{X}_{\mathrm{C}} \\ $$$$\:\:\:\:\:\mathrm{245}\:=\:\mathrm{10}\:\left(\frac{\mathrm{1}}{\mathrm{2}\pi\mathrm{fC}}\right) \\ $$$$\:\:\:\:\:\mathrm{24}.\mathrm{5}\:=\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}.\mathrm{14}×\mathrm{50}×\mathrm{C}} \\ $$$$\:\:\:\:\:\:\mathrm{C}\:=\:\mathrm{0}.\mathrm{00013}\:\mathrm{F}\:=\:\mathrm{130}\:\mu\mathrm{F} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{V}_{\mathrm{L}} \:=\:\mathrm{i}.\mathrm{X}_{\mathrm{L}} \\ $$$$\:\:\:\:\:\:\mathrm{50}\:=\:\mathrm{10}\:\left(\mathrm{2}\pi\mathrm{fL}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{5}\:=\:\mathrm{2}×\mathrm{3}.\mathrm{14}×\mathrm{50}×\mathrm{L} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{L}\:=\:\mathrm{0}.\mathrm{016}\:\mathrm{H}\:=\:\mathrm{16}\:\mathrm{mH} \\ $$

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