Question-146100

Question Number 146100 by tabata last updated on 10/Jul/21

Commented by tabata last updated on 10/Jul/21

$${in}\:{the}\:{following}\:{figure}\:,{calculate}\:{the}\:{equivalent}\:{resistance}\:{of}\:{the} \\ $$$${external}\:{circuit}\:{and}\:{then}\:{calculate}\:{the}\:{current}\:{through}\:{it}\:? \\ $$

Answered by Olaf_Thorendsen last updated on 10/Jul/21

R_(ab) = 2Ω ⇒ (1/R_(ab) ) = (1/2) Ω^(−1) (1/R_(cd) ) = (1/5)+(1/8)+(1/(12)) = ((49)/(120)) Ω^(−1) R_(ef) = (1/((1/7)+(1/6)))+3 = ((81)/(13)) Ω ⇒ (1/R_(ef) ) = ((13)/(81)) Ω^(−1) (1/R_(eq) ) = (1/R_(ab) )+(1/R_(cd) )+(1/R_(ef) ) = (1/2)+((49)/(120))+((13)/(81)) (1/R_(eq) ) = ((3463)/(3240)) ⇒ R_(eq) = ((3240)/(3463)) ≈ 0,94Ω U_(eq) = U_(ab) = U_(cd) = U_(ef) = 20V I_(eq) = (U_(eq) /R_(eq) ) = 20×((3463)/(3240)) = ((3463)/(162)) ≈ 21,38A

$$\mathrm{R}_{\mathrm{ab}} \:=\:\mathrm{2}\Omega\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{ab}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Omega^{−\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{cd}} }\:=\:\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{12}}\:=\:\frac{\mathrm{49}}{\mathrm{120}}\:\Omega^{−\mathrm{1}} \\ $$$$\mathrm{R}_{\mathrm{ef}} \:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{6}}}+\mathrm{3}\:=\:\frac{\mathrm{81}}{\mathrm{13}}\:\Omega\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{ef}} }\:=\:\frac{\mathrm{13}}{\mathrm{81}}\:\Omega^{−\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{eq}} }\:=\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{ab}} }+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{cd}} }+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{ef}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{49}}{\mathrm{120}}+\frac{\mathrm{13}}{\mathrm{81}} \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{eq}} }\:=\:\frac{\mathrm{3463}}{\mathrm{3240}}\:\Rightarrow\:\mathrm{R}_{\mathrm{eq}} \:=\:\frac{\mathrm{3240}}{\mathrm{3463}}\:\approx\:\mathrm{0},\mathrm{94}\Omega \\ $$$$\mathrm{U}_{\mathrm{eq}} \:=\:\mathrm{U}_{\mathrm{ab}} \:=\:\mathrm{U}_{\mathrm{cd}} \:=\:\mathrm{U}_{\mathrm{ef}} \:=\:\mathrm{20V} \\ $$$$\mathrm{I}_{\mathrm{eq}} \:=\:\frac{\mathrm{U}_{\mathrm{eq}} }{\mathrm{R}_{\mathrm{eq}} }\:=\:\mathrm{20}×\frac{\mathrm{3463}}{\mathrm{3240}}\:=\:\frac{\mathrm{3463}}{\mathrm{162}}\:\approx\:\mathrm{21},\mathrm{38A} \\ $$

Commented by tabata last updated on 10/Jul/21

sir R_(cd ) =(1/5)+(1/8)+(1/(1.2)) not (1/5)+(1/8)+(1/(12))

$${sir}\:{R}_{{cd}\:} =\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}\:{not}\:\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Commented by tabata last updated on 11/Jul/21

$$????? \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply