Wires A and B , having equal lengths of 43.1 m and equal diameters of 2.84 mm, a

Question

Wires A and B, having equal lengths of 43.1 m and equal diameters of 2.84 mm, are connected in series. A potential difference of 62.3 V is applied between the ends of the composite wire. The resistances are RA = 0.115 and RB = 0.659 . For wire A, what are (a) magnitude J of the current density and (b) potential difference V? (c) Of what type material is wire A made (see the table below)? For wire B, what are (d) J and (e) V? (f) Of what type material is B made?

Resistivities of Some Materials at Room Temperature (20°C) Resistivity, p Temperature Coefficient Material (2 m) of Resistivity, a (KT) Typical Metals 1.62 x 10 Silver 4.1 x 10 1.69 x 10 Copper 4.3 x 10 2.35 x 10-8 4.0 x 10 Gold 2.75 x 10 4.4 x 10 Aluminum 0.002 x 10 4.82 x 10 Manganina 5.25 x 10-8 Tungsten 4.5 x 10 9.68 x 10-8 Iron 6.5 x 10 3.9 x 10 10.6 x 10 Platinum Typical Semiconductors 70 X 10-3 2.5 x 103 Silicon, pure 8.7 x 10 Silicon, n-type 2.8 x 10 Silicon, p-typ Typical Insulators 1010 10 14 Glass 16 Fused quartz 10 "An alloy specifically designed to have a small value of a. bPure silicon doped with phosphorus impurities to a charge carrier density of 1023 m 3. 'Pure silicon doped with aluminum impurities toa charge carrier density of 102 m

Explanation / Answer

here,

the current through the resistors , I = V /( RA + RB)

I = 62.3 /( 0.115 + 0.659) A

I = 80.5 A

for wire A

a)

the current density , J = I/(pi * r^2)

J = 80.5 /( pi * (1.42 * 10^-3)^2)

J = 1.27 * 10^7 A/m^2

b)

the potential difference across A , Va = I * Ra

Va = 80.5 * 0.115 = 9.26 V

c)

let the resistivity be p

R = p * l/area

0.115 = p * 43.1 /( pi * (1.42 * 10^-3)^2)

p = 1.69 * 10^-8 ohm.m

so the material is copper

d)

the current density , J = I/(pi * r^2)

J = 80.5 /( pi * (1.42 * 10^-3)^2)

J = 1.27 * 10^7 A/m^2

e)

the potential difference across B , Vb = I * Rb

Vb = 80.5 * 0.659 = 53.05 V

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