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phy 2 sheet 2

# phy 2 sheet 2 - In this metal the average random(thermal...

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Unformatted text preview: In this metal the average random (thermal) velocity of a free electron is 1.70×10 6 m s-1 . At room temperature these free electrons go, on average, a distance 20.00 nm before suffering a collision. What is the average time between two successive collisions?1.176×10-14 s d = distance v = velocity of a free electron convert distance to meters, then: ans = d/v UNITS = s Suppose we now apply a weak electric field of 5.0 V/m in our metal. In the time between collisions found above, what is the gain in velocity along the direction of the electric field? -0.0103441596177439 m/s Ef = Electric Field E = charge of an electron = 1.602E-19 t = time = answer from part 1 Me = Mass of an electron = 9.109E-31 ans = (Ef)*E*t/(Me) UNITS = m/s What is the magnitude of the current density J if the number of atoms per unit volume is 8.20×10 28 m-3 and we use the answer to part 2 as an estimate of the drift velocity?135885018.4025 A/m/m E = charge of an electron = 1.602E-19 DV = drift velocity = ans from part 2 #e = number of electrons (given) ans = (#e)*E*(DV) UNITS = A/m^2 *The question asks for a direction as well, so if its not positive, it's negative Finally, what is the resistivity ρ? The unit Ω can be typed as "ohm"3.680×10-8 ohm m Ef = Electric Field Cd = current density = answer from part 3 ans = (Ef)/(Cd) UNITS = (V*m/A) OR (ohms*m) Consider the circuit shown above. The battery in the circuit has a rated voltage of 6.00 V. When connected to a test resistance of R = 160.0 Ω, the measured voltage drop across R is 5.941 V.What is the internal resistance R i of the battery? 1.6000 ohm V = Voltage vd = voltage drop R = resistance given i = (Vd)/R ans = (V/i) - R UNITS = ohms Now suppose we want to replace our test resistance with a new one, such that the power dissipated in the resistance is maximized. What value of the resistance should we choose?1.6000 ohm ans = answer from part 1 UNITS = ohms For this optimal resistance, what is the power dissipated?5.6250 W i = current V = voltage R = resistance = answer from part 2 i = V/(2*R) ans = (i*V)/2 UNITS = W What is the resistance (in Ω) of a spool of copper wire (ρ = 1.69·10-8 Ω · m)? It has a length L = 108 m and a diameter of d = 2 mm. The formula for resistance is:R = ρL / A In the case of the circular wire:A = π r 2 A = π (d/2) 2 A = π/4 d 2 Substituting this into the resistance equation:R = ρL / (π/4 d 2 )= 4 ρL / (π d 2 )= 4 · 1.69·10-8 Ωm · 108 m / (π (2·10-3 m) 2 )= 0.580979204262655 Ω In the circuit below R 1 = 12, R 2 = 63, R 3 = 118 Ω, and V = 69 V. What is the magnitude of the voltage drop (V) across the R 2 resistor? i = current R = Resistance i = V/(R1 + R2 + R3) ans = i*(R2) UNITS = none In the circuit below R 1 = 10, R...
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