Circle the letter corresponding to the correct answer choices. If you are tore t

Question

Circle the letter corresponding to the correct answer choices. If you are tore that your answer it correct and it does not match any of those given, then write your answer on the blank space in f. For problems where you need to write in answers, correct units (SI system) must accompany the units Leaving off the units will cost you 1/3 rd the points for that answer, if the number is correct. An electrical current of 8.70 mA exists in a solid cylindrical wire whose diameter is 1.20 mm. Calculate the magnitude of the current density in the wire. Assume that electrons are the charge carriers and the conduction election density is 2.35 Times 10^29/m^3. (Note that this is the charge earner density and could also be stated as 2.35 Times 10^29 charge carriers/m^3 or electrons/m^3) Then, calculate the electron drift speed in the wire. A wire with a resistance of 0.80 Ohm is drawn out through a die so that its new length is 6 times its original length. Find the resistance of the longer w ire, assuming that the resistivity and density of the material are not changed during the drawing process. A potential difference of 18 V is maintained between the ends of a 2700 cm length of wire whose diameter is 0.66 mm. The conductivity of the wire is 5.80 Times 10^7 (ohm.m)^-1. Determine the rare at which energy in the wire is transformed from kinetic to thermal energy. A solar cell generates a potential difference of 5.62 V when a 1.20 k Ohm external resistance is connected across it, i.e., from one terminal of the cell to the other terminal of it, and a potential difference of 6.84 V w hen a 2 90 k Ohm external resistance is used. What is the internal resistance r, of the solar cell? (Give your answer in the form of "abc" Ohm.) What is the emf (V) of the solar cell? For the circuit diagram as Figure 4 of Assignment 7. - 15 kV, R = 4.5 k Ohm, and C = 7800.0 mu F. Determine the time constant for the circuit Determine the maximum charge that will appear on the capacitor. Assume that the capacitor is Initially uncharged, how long will it take for the capacitor to charge to half the maximum charge? A pent mass consisting of 34 charged particles travels at a velocity of magnitude 4.80 Times 10^7 m/s through a uniform: field of magnitude 4 8 T. Each charged particle has a charge of -12e and a mass of 3.60 Times 10^-29 kg. The angle between the

Explanation / Answer

1)

current density = I/A

A = pi*r^2

r = radius = 1.2/2 = 0.6 mm

density = 8.7*10^-3/(pi*(0.6*10^-3)^2) = 7.69*10^3 A/m^2

(2)

drift speed = I/(n*e*A)

v = (8.7*10^-3)/(2.35*10^29*1.6*10^-19*(pi*(0.6*10^-3)^2))

v = 2.04*10^-7 m/s

+++++++

2)

resistance = rho*L/A = rho*L^2/(A*L)

A*L = volume

R = rho*L^2/V

R2/R1 = (L2/L1)^2

R2/0.8 = (6)^2

R2 = 28.8 ohm <<<<<---answer

++++++

3)

resistance R = L/SA

S = conductivity

A = pi*r^2 = pi*(0.33*10^-3)^2

R = 27/( 5.8*10^7*pi*(0.33*10^-3)^2)

R = 1.36 ohms

power = V^2/R = 1.8^2/1.36

P = 2.38 W <<<--------answer

+++++++++++

(4)

i1 = E/(R1+r)

E - i1*r = V1

E - Er/(R1+r) = v1

E - (E*r)/(1200+r) = 5.62.........(1)

i2 = E/(R2+r)

E - i2*r = V2

E - E*r/(R2+r) = v2

E - (E*r)/(2900+r) = 6.84.......(2)

solving 1 & 2

r = 5.25*10^2 ohms   <<<------answer

E = 8.08 V <<<-----answer

time constant T = Rc = 4.5*10^3*7800*10^-6 = 3.51*10^1

Q = C*E = 7800*10^-6*15000 = 1.17*10^2 C

q =Q*(1-e^t/T)

Q/2= Q *(1-e^-(t/35.1))

1/2= (1-e^-(t/35.1))

t = 24.3 s <---------answer

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.