Question-151828

Question Number 151828 by DELETED last updated on 23/Aug/21

Answered by DELETED last updated on 23/Aug/21

R_(s1) =10Ω+10Ω=20Ω R_(s2) =10Ω+10Ω=20Ω (1/R_p )=(1/R_(s1) ) +(1/R_(s2) )=(1/(20))+(1/(20))=(2/(20)) R_p = ((20)/2)=10Ω R_(pengganti) =R_A +R_p +R_E (10+10+10)Ω=30Ω

$$\mathrm{R}_{\mathrm{s1}} =\mathrm{10}\Omega+\mathrm{10}\Omega=\mathrm{20}\Omega \\ $$$$\mathrm{R}_{\mathrm{s2}} =\mathrm{10}\Omega+\mathrm{10}\Omega=\mathrm{20}\Omega \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{s1}} }\:+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{s2}} }=\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{20}}=\frac{\mathrm{2}}{\mathrm{20}} \\ $$$$\mathrm{R}_{\mathrm{p}} =\:\frac{\mathrm{20}}{\mathrm{2}}=\mathrm{10}\Omega \\ $$$$\mathrm{R}_{\mathrm{pengganti}} =\mathrm{R}_{\mathrm{A}} +\mathrm{R}_{\mathrm{p}} +\mathrm{R}_{\mathrm{E}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{10}+\mathrm{10}+\mathrm{10}\right)\Omega=\mathrm{30}\Omega \\ $$$$ \\ $$

Answered by DELETED last updated on 23/Aug/21

•Sebelum R putus 58). R_s =R+R=2R (1/R_p )=(1/R)+(1/(2R))=(3/(2R)) →R_t =((2R)/3) I_t =(E/(2R/3))=((3E)/(2R)) Daya sebelum R putus: P=E.i_t =E.((3E)/(2R))=((3E^2 )/(2R)) •Setelah R putus i=(E/(2R)) →P=E.i=E.(E/(2R))=(E^2 /(2R)) P_1 :P_2 =((3E^2 )/(2R)) : (E^2 /(2R)) = 3:1

$$\bullet\mathrm{Sebelum}\:\mathrm{R}\:\mathrm{putus} \\ $$$$\left.\:\mathrm{58}\right).\:\mathrm{R}_{\mathrm{s}} =\mathrm{R}+\mathrm{R}=\mathrm{2R}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }=\frac{\mathrm{1}}{\mathrm{R}}+\frac{\mathrm{1}}{\mathrm{2R}}=\frac{\mathrm{3}}{\mathrm{2R}}\:\rightarrow\mathrm{R}_{\mathrm{t}} =\frac{\mathrm{2R}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\mathrm{I}_{\mathrm{t}} =\frac{\mathrm{E}}{\mathrm{2R}/\mathrm{3}}=\frac{\mathrm{3E}}{\mathrm{2R}} \\ $$$$\:\:\:\:\:\mathrm{Daya}\:\mathrm{sebelum}\:\mathrm{R}\:\mathrm{putus}: \\ $$$$\:\:\:\:\:\:\mathrm{P}=\mathrm{E}.\mathrm{i}_{\mathrm{t}} =\mathrm{E}.\frac{\mathrm{3E}}{\mathrm{2R}}=\frac{\mathrm{3E}^{\mathrm{2}} }{\mathrm{2R}} \\ $$$$\:\:\:\:\:\bullet\mathrm{Setelah}\:\mathrm{R}\:\mathrm{putus} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{i}=\frac{\mathrm{E}}{\mathrm{2R}}\:\rightarrow\mathrm{P}=\mathrm{E}.\mathrm{i}=\mathrm{E}.\frac{\mathrm{E}}{\mathrm{2R}}=\frac{\mathrm{E}^{\mathrm{2}} }{\mathrm{2R}} \\ $$$$\:\:\:\:\:\mathrm{P}_{\mathrm{1}} :\mathrm{P}_{\mathrm{2}} =\frac{\mathrm{3E}^{\mathrm{2}} }{\mathrm{2R}}\::\:\frac{\mathrm{E}^{\mathrm{2}} }{\mathrm{2R}}\:=\:\mathrm{3}:\mathrm{1} \\ $$

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