Question

A number 16 copper wire has a diameter of 1.291 mm. Calculate the resistance of a 43.0 m long piece of such wire. (Use 1.72x 10-8 sm for the resistivity of copper.) 5.650x10-1 Ohm You are correct. Previous Tries For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor wiring, the limit is 6.0 A for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current? 2.034x101 w You are correct. Previous Tries What would be the voltage between the ends of the wire in the above problem? Submit Answer Tries 0/20 What is the current density in the wire when it is carrying the maximum allowable current? (Current density is the current in the wire divided by the cross sectional area of the wire.) Submit Answer Tries 0/20 What is the drift velocity of the electrons when the wire is carrying the maximum allowable current? (The density of electrons in copper is 8.47x1028 m-3.) Submit Answer Tries 0/20

Answer 1

The Resistance of conductor is given by R=1h A f-Resistivity of material to em length of material L A - Area of material = 782 m2 r-Radius Given: Length of the wire Ls 43 m Resistivity of the Wire P= 1.72 x10 nm Diameber'the wire d=27=1.2914m -8 = Radis of the wire ra 0.646mm -3 80.646x10 m cross sectional Area of wire A=ar2 A = 78²= 3.14x (0.816 x 10334 A = 3.14x (0.646x103) = 3.14% 0.417 X10 x 10-6 1-31 x 150 m2 A = CSScanned with CamScanner

Resistance of the wire R=PL -8 = 172410143 1:31 X10-6 R = 56.458 X10 0²2 OR R 5.646aio's o'n OR R = 0.5651 2 The current flowing through wire 1 GA AS we Inow, the polential can be defined as vo W q => W & qv; w-work done q- charge - 31+ W t q. V. Z W-power-P. P= IV t a current - from ohn's Law Y = IR Power P = I²R CS Scanned with CamScanner

Power dissipation P = I²R =66)*0.565) 36x 0.565 Po 20.325 n. *3) using Relation (Previous) Power = current & Holtage P = IV 20.325 = 6 x v. ㅋ Ve 3.387 volts Voltage between ends of the wire V: 3.387 rolls As defination *4 Current density Jas A from the data Ia 6A A = 1.31 x 100 m2 J- 6 1.31X106 J = 4.5810 A/m2 CS Scanned with CamScanner

chared particles <3 Drift velocity :- The Velocity acquired by uncler external Electric field. 11 het 00 000 is Let us consider l' small length in a wire a of Conductor n- eleckon's density Darren * The total charge present in length Q- he.(Al) e-charge on elecron A Area of the wire. Elechons are moving with ild dritt redacity this length & in time t distance Vd - l time t. Current flowing the wire ne al Elxd) and covers lva: vd = 1 t But I s I= neavd CSscanned with CamSçanner

Velocity [vd I he A. Im3 8.47*10*1.6 x '910 X10 Sony drift Given data: Current I = 6A, vd nea 28 6 8147* 1-6X1-3) x103 17.153 Total number of elecrons per unit volume = 8:47X10 ez16x1619c A = 1.31x106m 6 x 1031 X10 nd 0.338x103 m m/sec COM vd = 338x 100 m/sec CScanned with CamScanner