Your answer is partially correct. Try again. The resistance and the magnitude of

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Your answer is partially correct. Try again.

The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in which the current takes different paths through a piece of material. Each of the rectangular pieces is made from a material whose resistivity is ? = 1.50 x 10-2 ?*m, and the unit of length in the drawing is L0 = 7 cm. Each piece of material is connected to a 3.00-V battery. Find (a) the resistance and (b) the current in each case.

(a)

(b)

This is how I did it:

Ra = p L/A = (1.5 x 10-2)(4)(0.07m)/ 2(0.07)2  = 0.42857 Ohms

Rb = p L/A = (1.5 x 10-2)(0.07m)/ 8(0.07m)2 = 0.02678 Ohms

Rc = p L/A = (1.5 x 10-2)(2)(0.07m)/ 4(0.07m)2 = 0.10714 Ohms

Not sure where I went wrong, but please show all steps.

Your answer is partially correct. Try again.

Explanation / Answer

Ra= [(1.5e-2)(4)(.07)]/[(2)(.07^2)]= .428 ohm
Rb= [(1.5e-2)(.07)]/[(2)(4)(.07^2)]= .0267 ohm
Rc= [(1.5e-2)(2)(.07)]/[(4)(.07)^2]= .1071 ohm
Ia= 3/.0.428= 7.0093 A
Ib= 3/.0267 = 112.35 A
Ic= 3/.1071 = 28.011 A

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