a wire of length 10 metre is cut into five equal parts what will be resistance of one part of that wire?
plss explain it properly
Answers
Answer:
Since the length x must be shaped into 4 equal sides, each side will have length x/4. The area of the square is s^2 = (x/4)^2 = (x^2)/16.
The remaining piece, with length 10-x must be shaped into three sides, each having a length of (10-x)/3. The area of an equilateral triangle is (s^2 sqrt3)/4 = ((10-x)/3)^2 sqrt 3)/4 =
(sqrt 3/36)(x^2 - 20x + 100)
The total area is the sum:
(x^2)/16 + (sqrt3/36)(x^2 - 20x + 100)
The minimum occurs either at an end point (x=0 or x = 10), or where the derivative = 0. Inspection of the function shows the minimum is not at either of the end points, so take the derivative and set is equal to zero.
2x/16 + (sqrt 3/36)(2x - 20) = 0
(1/8 + sqrt 3/18)x = (5 sqrt 3)/9
[(9 + 4 sqrt 3)/72]x = (5 sqrt 3/9)
x = (40 sqrt 3)/(9 + 4 sqrt 3)
Answer:
Resistance of a piece of wire is proportional to its length. A piece of wire has a resistance R. The wire is cut into five equal parts. Therefore, resistance of each part =R/5
Resistance of a piece of wire is proportional to its length. A piece of wire has a resistance R. The wire is cut into five equal parts. Therefore, resistance of each part =R/5All the five parts are connected in parallel.
Resistance of a piece of wire is proportional to its length. A piece of wire has a resistance R. The wire is cut into five equal parts. Therefore, resistance of each part =R/5All the five parts are connected in parallel.∴equivalent resistance (R') is given as。^‿^｡。^‿^｡
