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Question-152723




Question Number 152723 by Tawa11 last updated on 31/Aug/21
Answered by Olaf_Thorendsen last updated on 31/Aug/21
V_(5Ω)  = 5Ω×I_(5Ω)  = 5Ω×2A = 10V  V_(10Ω)  = V_(5Ω)  = 10V  I_(10Ω)  = (V_(10Ω) /(10Ω)) = ((10V)/(10Ω)) = 1A  I_(20Ω)  = I_(5Ω) +I_(10Ω)  = 2A+1A = 3A  V_(20Ω)  = 20Ω×I_(20Ω)  = 20Ω×3A = 60V
$$\mathrm{V}_{\mathrm{5}\Omega} \:=\:\mathrm{5}\Omega×\mathrm{I}_{\mathrm{5}\Omega} \:=\:\mathrm{5}\Omega×\mathrm{2A}\:=\:\mathrm{10V} \\ $$$$\mathrm{V}_{\mathrm{10}\Omega} \:=\:\mathrm{V}_{\mathrm{5}\Omega} \:=\:\mathrm{10V} \\ $$$$\mathrm{I}_{\mathrm{10}\Omega} \:=\:\frac{\mathrm{V}_{\mathrm{10}\Omega} }{\mathrm{10}\Omega}\:=\:\frac{\mathrm{10V}}{\mathrm{10}\Omega}\:=\:\mathrm{1A} \\ $$$$\mathrm{I}_{\mathrm{20}\Omega} \:=\:\mathrm{I}_{\mathrm{5}\Omega} +\mathrm{I}_{\mathrm{10}\Omega} \:=\:\mathrm{2A}+\mathrm{1A}\:=\:\mathrm{3A} \\ $$$$\mathrm{V}_{\mathrm{20}\Omega} \:=\:\mathrm{20}\Omega×\mathrm{I}_{\mathrm{20}\Omega} \:=\:\mathrm{20}\Omega×\mathrm{3A}\:=\:\mathrm{60V} \\ $$
Commented by Tawa11 last updated on 31/Aug/21
Wow, thanks sir. I appreciate. God bless you.
$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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