Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 8550 by tawakalitu last updated on 15/Oct/16

Answered by ridwan balatif last updated on 16/Oct/16

Commented by tawakalitu last updated on 16/Oct/16

Wow. God bless you sir.

$$\mathrm{Wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by sandy_suhendra last updated on 16/Oct/16

Commented by sandy_suhendra last updated on 16/Oct/16

answer for pic A loop 1 ⇒ ΣE + ΣI.R=0 −E_1 +I_1 .R_1 +(I_1 −I_2 ).R_2 =0 −2+3I_1 +(I_1 −I_2 ).3 =0 3I_1 +3I_1 −3I_2 =2 6I_1 −3I_2 =2 .....(1) loop 2 ⇒ ΣE+ΣI.R=0 −E_2 +(I_1 −I_2 ).R_2 −I_2 .R_3 =0 −1+(I_1 −I_2 ).3−4I_2 =0 3I_1 −3I_2 −4I_2 =1 3I_1 −7I_2 =1 .....(2) (1)−2×(2) ⇒ 11I_2 =1 I_2 =(1/(11)) A subs to (1) we get ⇒ I_1 =(6/(11)) A ΔV at R_1 =I_1 .R_1 =(6/(11))×3=((18)/(11)) volt P at R_1 =ΔV.I_1 =((18)/(11))×(6/(11))=((108)/(121)) watt ΔV at R_2 =(I_1 −I_2 ).R_2 =((6/(11))−(1/(11)))×3=((15)/(11)) volt P at R_2 =ΔV.(I_1 −I_2 )=((15)/(11))×(5/(11))=((75)/(121)) watt ΔV at R_3 =I_2 .R_3 =(1/(11))×4=(4/(11)) volt P at R_3 =ΔV.I_2 =(4/(11))×(1/(11))=(4/(121)) watt

$$\mathrm{answer}\:\mathrm{for}\:\mathrm{pic}\:\mathrm{A} \\ $$$$\mathrm{loop}\:\mathrm{1}\:\Rightarrow\:\Sigma\mathrm{E}\:+\:\Sigma\mathrm{I}.\mathrm{R}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{E}_{\mathrm{1}} +\mathrm{I}_{\mathrm{1}} .\mathrm{R}_{\mathrm{1}} +\left(\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \right).\mathrm{R}_{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}+\mathrm{3I}_{\mathrm{1}} +\left(\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \right).\mathrm{3}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3I}_{\mathrm{1}} +\mathrm{3I}_{\mathrm{1}} −\mathrm{3I}_{\mathrm{2}} =\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6I}_{\mathrm{1}} −\mathrm{3I}_{\mathrm{2}} =\mathrm{2}\:.....\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{loop}\:\mathrm{2}\:\Rightarrow\:\Sigma\mathrm{E}+\Sigma\mathrm{I}.\mathrm{R}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{E}_{\mathrm{2}} +\left(\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \right).\mathrm{R}_{\mathrm{2}} −\mathrm{I}_{\mathrm{2}} .\mathrm{R}_{\mathrm{3}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}+\left(\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \right).\mathrm{3}−\mathrm{4I}_{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3I}_{\mathrm{1}} −\mathrm{3I}_{\mathrm{2}} −\mathrm{4I}_{\mathrm{2}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3I}_{\mathrm{1}} −\mathrm{7I}_{\mathrm{2}} =\mathrm{1}\:.....\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\mathrm{2}×\left(\mathrm{2}\right)\:\Rightarrow\:\mathrm{11I}_{\mathrm{2}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{11}}\:\mathrm{A}\:\:\mathrm{subs}\:\mathrm{to}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\Rightarrow\:\mathrm{I}_{\mathrm{1}} =\frac{\mathrm{6}}{\mathrm{11}}\:\mathrm{A} \\ $$$$\Delta\mathrm{V}\:\mathrm{at}\:\mathrm{R}_{\mathrm{1}} =\mathrm{I}_{\mathrm{1}} .\mathrm{R}_{\mathrm{1}} =\frac{\mathrm{6}}{\mathrm{11}}×\mathrm{3}=\frac{\mathrm{18}}{\mathrm{11}}\:\mathrm{volt} \\ $$$$\mathrm{P}\:\mathrm{at}\:\mathrm{R}_{\mathrm{1}} =\Delta\mathrm{V}.\mathrm{I}_{\mathrm{1}} =\frac{\mathrm{18}}{\mathrm{11}}×\frac{\mathrm{6}}{\mathrm{11}}=\frac{\mathrm{108}}{\mathrm{121}}\:\mathrm{watt} \\ $$$$\Delta\mathrm{V}\:\mathrm{at}\:\mathrm{R}_{\mathrm{2}} =\left(\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \right).\mathrm{R}_{\mathrm{2}} =\left(\frac{\mathrm{6}}{\mathrm{11}}−\frac{\mathrm{1}}{\mathrm{11}}\right)×\mathrm{3}=\frac{\mathrm{15}}{\mathrm{11}}\:\mathrm{volt} \\ $$$$\mathrm{P}\:\mathrm{at}\:\mathrm{R}_{\mathrm{2}} =\Delta\mathrm{V}.\left(\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \right)=\frac{\mathrm{15}}{\mathrm{11}}×\frac{\mathrm{5}}{\mathrm{11}}=\frac{\mathrm{75}}{\mathrm{121}}\:\mathrm{watt} \\ $$$$\Delta\mathrm{V}\:\mathrm{at}\:\mathrm{R}_{\mathrm{3}} =\mathrm{I}_{\mathrm{2}} .\mathrm{R}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{11}}×\mathrm{4}=\frac{\mathrm{4}}{\mathrm{11}}\:\mathrm{volt} \\ $$$$\mathrm{P}\:\mathrm{at}\:\mathrm{R}_{\mathrm{3}} =\Delta\mathrm{V}.\mathrm{I}_{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{11}}×\frac{\mathrm{1}}{\mathrm{11}}=\frac{\mathrm{4}}{\mathrm{121}}\:\mathrm{watt} \\ $$

Commented by tawakalitu last updated on 16/Oct/16

Wow. God bless you sir.

$$\mathrm{Wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com