Question

Consider the two cylindrical wires of lengths ly=4.1m and lz=2.6m are merged back to back as shown in the figure below to for

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Answer #1

Rio il 364X100 4.1 AL TX (1) ²x 10-6 R = 475.282 R2 = 2 li Az 261x10 & 2.6 ax(8)? xie R2 = 3.372 6 R - R. & R2 478.65u R Resi

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Answer #2

SOLUTION :


Resistance is directly proportional to length of wire and inversely proportional to cross sectional area of wire. The constant of proportionality (p) is termed as resistivity of wire with respect length and area.


So,


R1 

= p1 *  l1 / A1

= 364 * 10^(-6) * 4.1 / (pi * (1*10^(-3))^2)

= 475.046 Ω 


R2 

= p2 *  l2 / A2

= 261 * 10^(-6) * 2.6 / (pi * (8*10^(-3))^2)

= 3.375 Ω 


R1 an R2 are in series.


So,


Total resistance of the joint system, R

= R1 + R2

= 475.046 + 3.375

= 478.421

= 478.4  Ω (ANSWER)

answered by: Tulsiram Garg
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