1.

Find the resistance if all of these parts are connected in :(a) Parallel(b) Series

Answer»

SOLUTION :`because R = (rho L)/(A)`. When the conductor is cut into 4 equal parts, length of each PART `=(L)/(4)` and so resistance of each part `R_("part")=(rho(L//4))/(A)=(R )/(4)`.
(a) When these 4 parts are connected in parallel then net resistance `R_(p)` is GIVEN as :
`(1)/(R_(p))=(1)/((R_(1))_("part"))+(1)/((R_(2))_("part"))+(1)/((R_(3))_("part"))+(1)/((R_(4))_("part"))`
`=(4)/(R )+(4)/(R )+(4)/(R )+(4)/(R )=(16)/(R )`
`rArr "" R_(p)=(R )/(16)`
(b)If the 4 parts are connected in SERIES then net resistance `R_(s)` is given as :
`R_(s)=(R_(1))_("part")+(R_(2))_("part")+(R_(3))_("part")+(R_(4))_("part")`
`=(R )/(4)+(R )/(4)+(R )/(4)+(R )/(4)=R`


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