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Question Number 148915 by DELETED last updated on 01/Aug/21

Answered by DELETED last updated on 01/Aug/21

(1/(Rp)) = (1/8)+(1/8)=(2/8)=(1/4)  Rp=4 Ω    V_p =V_(rs1) =2 Vol  i_(Rs1) =(V_(Rs1) /R_(s1) ) = (2/8) =0.25 A  i_(Rs1) =i_(4Ω)  = 0.25 A  V_(4Ω) =i_(4Ω) ×R_(4Ω) =0.25×4=1 Vol//

$$\frac{\mathrm{1}}{\mathrm{Rp}}\:=\:\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{2}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{Rp}=\mathrm{4}\:\Omega \\ $$$$ \\ $$$$\mathrm{V}_{\mathrm{p}} =\mathrm{V}_{\mathrm{rs1}} =\mathrm{2}\:\mathrm{Vol} \\ $$$$\mathrm{i}_{\mathrm{Rs1}} =\frac{\mathrm{V}_{\mathrm{Rs1}} }{\mathrm{R}_{\mathrm{s1}} }\:=\:\frac{\mathrm{2}}{\mathrm{8}}\:=\mathrm{0}.\mathrm{25}\:\mathrm{A} \\ $$$$\mathrm{i}_{\mathrm{Rs1}} =\mathrm{i}_{\mathrm{4}\Omega} \:=\:\mathrm{0}.\mathrm{25}\:\mathrm{A} \\ $$$$\mathrm{V}_{\mathrm{4}\Omega} =\mathrm{i}_{\mathrm{4}\Omega} ×\mathrm{R}_{\mathrm{4}\Omega} =\mathrm{0}.\mathrm{25}×\mathrm{4}=\mathrm{1}\:\mathrm{Vol}// \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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