2. What is the resistance of 100 meters of silver wire which is 2.25 mm in diame

Question

2. What is the resistance of 100 meters of silver wire which is 2.25 mm in diameter?

3. A length of gold wire is connected to a precision 5-Volt power supply and a current of 0.2604 A is measured at 20°C. The wire is then immersed in a new environment with a different temperature where the measured current is 0.2417 A is measured. What is the temperature of the new environment?

5. Calculate the peak current in a 3.4 resistor connected to a 120-V rms ac source.

6. A wire 4.2 m in length and 0.5 mm in diameter carries 50 mA of current when connected to a 1.2 V battery. If the drift velocity is 1.2 x 10-5 m/s, calculate

a) the resistance of the wire R,

b) the resistivity,

c) the current density j,

d) the electric field inside the wire, and,

e) the number n of free electrons per unit volume.

Explanation / Answer

2. use from the definition of resistance R = rho L/A

where rho is ressitivity of silver = 1.6 e -8

L is length = 100 m

Area = pi r^2 = 3.14 * (2.25 e -3/2)^2 = 4.90625e-6 m ^2

so

Resistance R = 1.6 e-8 * 100/(4.90625e-6)

R = 0.326 ohms

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3. use Rt = Ro *(1+ ADT)

where A is temp coefficent of resistance = of gold = 0.0034 perdeg C

as V = iR

1/i = (1+ADT)/i1

0.2604 = 0.2417 * (1+ 0.0034 * T2-T1)

0.2604 *(1+ 0.0034 (T2-T1) = 0.0187

1+ 0.0034 DT = 0.0718

dT = (1-0.0718 )/(0.0034)

T2-T1 = 273

T2 = 273+ 20 = 293 degC

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5. applyh Vrms = irms *R

irms = 120/3.4

irms = 35.29

ipeak = 1.414 * 35.29

ipeak = 50 Amps

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6. use the formula for drifft velcoity Vd = i/neA

resistance R = V/i = 1.2/0.050 = 24 Ohms

Resistivity rho = RA/L = (24 * 3.14* 0.5 e-3* 0.5 e-3/(4 *4.2)

rho = 1.12 e -6 ohm-m

Current density J = current/area = i/A

J = 0.05*4 /(3.14* 0.5 e-3* 0.5 e-3) =2 .54 e 5 A/m^2

use J = E/rho

so E = j * rho = 2.54 e 5 * 1.13 e-6

E = 0.287 V/m

n = i/(Vd * eA)

n = (0.05)* 4 /(1.2 e-5 *1.6 e-19*3.14* 0.5 e-3* 0.5 e-3)

n = 1.32 e 29 electrons

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