Explain too so that I can correct my mistakes! Thanks 1. Building an Inductor. a

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Explain too so that I can correct my mistakes! Thanks

1. Building an Inductor. a. Find a possible set of values for the number of turns, length, and cross-sectional area of a solenoid with a self-inductance of 1.00 H. b. What is the back emf produced by such an inductor when the current through it is increasing at the rate of 100 A/s? 2. Finding R and L At t = 0 an inductor is connected to a 9.0 V battery. At t = 2.0 s, the current through the inductor is 0.37 A, and after about 10 s, the current reaches a stable value of 1.0 A. Find the resistance and the inductance of the inductor. 3. Transformer. A transformer has 500 primary turns and 10 secondary turns. a. If Vp is 120 V, what is Vs with an open circuit? b. If the secondary now has a resistive load of 15 ohm, what is the current in the (i) primary and (ii) secondary? 4. Hair Dryer in Europe. You plan to take your hair dryer to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The dryer puts out 1700 W at 120 V. a What could you do to operate your blower via the 240 V line in Europe? i. Use a step-up transformer with N2/N1 = 2. ii. Use a step-down transformer with N2/N1 = 1/2. b. What current will be drawn from the 240 V line when the dryer is connected to the 120 V side of the transformer? c. What resistance connected directly to the 240 V line would draw the same power as the dryer in part B? 5. Saving on Your Electric Bill. a. Fluorescent bulbs deliver the same amount of light using much less power. If one kW-hr^-1 costs 10c, estimate the amount of money you would save each month by replacing all the 75 W incandescent bulbs in your house by 15 W fluorescent ones. b. A typical large power plant might generate 1000 Megawatts (1 GW) of electrical power. If all the households in the US were using incandescent bulbs and switched to fluorescent, estimate the number of power plants that would no longer be needed.

Explanation / Answer

1)

L = N*A*uo*N/l

given L = 1 H

N^2A/l = L/uo

N^2A/l = 1/(4*3.14*10^7) = 796178

(b)

back emf = L*dI/dt

emf = 1*100 = 100 v

++++++

2)

in RL circuit

i = I*( 1-e^-tR/L )

I = V/R

given V = 9v

at t = 2s current i = 0.37

0.37 = I*(1-e^-2R/L) ....(1)

at t= 10s .... i = 1A is steady

t is the time constant = L/R = 10 .........(2)

from 1

0.37 = 9/R*(1-e^-2/10)

R = (9/0.37)*(1-e^-(1/5))

R = 4.41 ohms   <<<--------answer

L = 10*R = 44.1 H <<<--------answer

++++++++++++++

3)

transformer ratio = Vp/Vs = Np/Ns

Vs = (Ns*Vp)/Np

Vs = (10*120)/500

Vs = 2.4 volts

b)

current Is = Vs/Rs = 10/15 = 0.16

power Pp = Ps

Vp*Ip = Vs*Is

Ip = (Vs*Is)/Vp

Ip = (10*0.16)/500

Ip = 0.032 A <<-----answer

+++++++++++++

4)

Vin = 240

Vout = 120

N1/N2 = 120/240 = 1/2

here the voltage has to be reduced

so use a step down transformer of ratio 1/2

+++++

b)

Power = V*I

i = P/V = 1700/240 = 7.08 A

c)

cuurent req = i = 1700/120 = 14.16 A

resistane required R = V/i = 240/14.16 = 16.9 ohms

5)

energy consumed by incandescent bulb = E1 = P1*t = (75*30*24)/1000 = 54 k w hr

energy consumed by flourescent bulb = E2 = P2*t = (15*30*24)/1000 = 10.8 k w hr

amoun saved = (E1 - E2)10 = $432 <<<--------answer

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