Question

The terminals of a 0.70 V watch battery are connected by a 60.0-m-long gold wire with a diameter of 0.200 mm. Part A You may want to review (Pages 755 - 759) For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Soil resistivity. What is the current in the wire? Express your answer using three significant figures. 80 AD OD? I= MA

Answer 1

V=0.70 V

Length of the wire(L)=60.0 m

Diameter of the wire(d)=0.200 mm

d = 0.200 x 10-3 m

Cross sectional area(A) = ar2 = 7(d/2)2 =

Resistivity of the gold (p) = 2.44 x 20-8 ohm – meter

Resistance of the gold wire :-

R=PL/A = 4pLad4

From ohm's Law:-

V=IR

I=V/R

V T=40L/πd2 πd2V 4DL

7 X (0.200 x 10-3)2 x 0.70 4 x (2.44 x 10-8) X 60.0

I = 0.0150 Amp

I = 15.0 m A Ans.