Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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 ?

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

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

?????

$$????? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com