Question

Part A How long must a 0.70-mm-diameter aluminum wire be to have a 0.46 A current when connected to the terminals of a 1.5 V flashlight battery? Express your answer to two significant figures and include the appropriate units. L = Value Units Submit Request Answer Part B What is the current if the wire is half this length? Express your answer to two significant figures and include the appropriate units. I= Value Units Submit Request Answer

Answer 1

Solution: PART A:- Given data:- Diameter of aluminum wire (D) = 0.70mm or o.foxiom 30 Radius of aluminum wire (R) - D o R = 0.70 x169m (1mm= 103m) 07 R = 0.35x10m Cwrent in wire (I) = 0.4GA voltage of battery (V) 71.5 V • specific Resistance of aluminum wire (8) = 2.65 x 108 rom (room standard data) from ohms Law VIR Here R = 4 PART:! V-I.PL Here A=KR? (cross-section area) I. SL 2 put all given data 0.46 x 2.65X168 XL TX(0.35 x103)2 L = -5XX(0-35 x103)2 0.46 * 2.65*168 L=47.36 m uplo two significant L= 47 m Answer If length of wire is half, them Resistance - R=st - RaL so 'R' also become half - due to proportional Relation and vIR I = x or Iata (for constant voltage) if 'Ř becomes half the current becomes double due to inverse relation Current (I') = 21 07 Il = 210.46A) 00 I'=032 A Answer So