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Question Number 152748 by Tawa11 last updated on 01/Sep/21

Answered by Olaf_Thorendsen last updated on 01/Sep/21

R_T  : total resistance  E = R_T I = (I/((1/R)+(1/X))) = ((3A)/((1/(5Ω))+(1/X))) = ((6A)/((1/(2Ω))+(1/X)))  3((1/2)+(1/X)) = 6((1/5)+(1/X))  1,5+(3/X) = 1,2+(6/X)  (3/X) = 1,5−1,2 = 0,3  X = 10Ω (Answer C)

$$\mathrm{R}_{\mathrm{T}} \::\:\mathrm{total}\:\mathrm{resistance} \\ $$$$\mathrm{E}\:=\:\mathrm{R}_{\mathrm{T}} \mathrm{I}\:=\:\frac{\mathrm{I}}{\frac{\mathrm{1}}{\mathrm{R}}+\frac{\mathrm{1}}{\mathrm{X}}}\:=\:\frac{\mathrm{3A}}{\frac{\mathrm{1}}{\mathrm{5}\Omega}+\frac{\mathrm{1}}{\mathrm{X}}}\:=\:\frac{\mathrm{6A}}{\frac{\mathrm{1}}{\mathrm{2}\Omega}+\frac{\mathrm{1}}{\mathrm{X}}} \\ $$$$\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{X}}\right)\:=\:\mathrm{6}\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{X}}\right) \\ $$$$\mathrm{1},\mathrm{5}+\frac{\mathrm{3}}{\mathrm{X}}\:=\:\mathrm{1},\mathrm{2}+\frac{\mathrm{6}}{\mathrm{X}} \\ $$$$\frac{\mathrm{3}}{\mathrm{X}}\:=\:\mathrm{1},\mathrm{5}−\mathrm{1},\mathrm{2}\:=\:\mathrm{0},\mathrm{3} \\ $$$$\mathrm{X}\:=\:\mathrm{10}\Omega\:\left(\mathrm{Answer}\:\mathrm{C}\right) \\ $$

Commented by Tawa11 last updated on 01/Sep/21

Thanks for your time sir. God bless you. I appreciate.

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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