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Question Number 34954 by ajfour last updated on 13/May/18

Commented by ajfour last updated on 13/May/18

Find resistance of a cylindrical  conductor of radius b and thickness  t with a concentric hole of radius  a, across its inner and outer  surfaces.(conductivity  given σ ) .

$${Find}\:{resistance}\:{of}\:{a}\:{cylindrical} \\ $$$${conductor}\:{of}\:{radius}\:\boldsymbol{{b}}\:{and}\:{thickness} \\ $$$$\boldsymbol{{t}}\:{with}\:{a}\:{concentric}\:{hole}\:{of}\:{radius} \\ $$$$\boldsymbol{{a}},\:{across}\:{its}\:{inner}\:{and}\:{outer} \\ $$$${surfaces}.\left({conductivity}\:\:{given}\:\sigma\:\right)\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/May/18

dR=(1/σ)×(dl/dA)  let consider elementary shell  thickness dx,cross area 2Πx×t  so dl=dx  dA=2Πxt  =(1/σ)×(dx/(2Πxt))  R=(1/(2Πσt))∫_a ^b (dx/x)  =(1/(2Πσt))×ln((b/a))

$${dR}=\frac{\mathrm{1}}{\sigma}×\frac{{dl}}{{dA}}\:\:{let}\:{consider}\:{elementary}\:{shell} \\ $$$${thickness}\:{dx},{cross}\:{area}\:\mathrm{2}\Pi{x}×{t}\:\:{so}\:{dl}={dx} \\ $$$${dA}=\mathrm{2}\Pi{xt} \\ $$$$=\frac{\mathrm{1}}{\sigma}×\frac{{dx}}{\mathrm{2}\Pi{xt}} \\ $$$${R}=\frac{\mathrm{1}}{\mathrm{2}\Pi\sigma{t}}\int_{{a}} ^{{b}} \frac{{dx}}{{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\Pi\sigma{t}}×{ln}\left(\frac{{b}}{{a}}\right) \\ $$

Commented by ajfour last updated on 13/May/18

correct answer.  but  dA should be A and not dA.

$${correct}\:{answer}. \\ $$$${but}\:\:{dA}\:{should}\:{be}\:{A}\:{and}\:{not}\:{dA}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/May/18

true...thanks..

$${true}...{thanks}.. \\ $$

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