Your laptop has wireless communications connectivity, and you might even have a

Question

Your laptop has wireless communications connectivity, and you might even have a wireless keyboard or mouse. But there's one wire you haven't been able to get rid of yet - the power cord.
Researchers are working on ways to circumvent the need for a direct electrical connection for power, and they are experiencing some success. Recently, investigators were able to use current flowing through a primary coil to power a 40 W lightbulb connected to a secondary coil 2.0 m away, with approximately 15%efficiency. The coils were large and the efficiency low, but it's a start.
The wireless power transfer system is outlined in(Figure 1) . An AC supply generates a current through the primary coil, creating a varying magnetic field. This field induces a current in the secondary coil, which is connected to a resistance (the lightbulb) and a capacitor that sets the resonance frequency of the secondary circuit to match the frequency of the primary circuit.

Part A

At a particular moment, the current in the primary coil is counterclockwise, as viewed from the secondary coil. At the center of the secondary coil, the field from the primary coil is

A- to the right

B- to the left

C- Zero

Part B

At a particular moment, the magnetic field from the primary coil points to the left and is increasing in strength. The field due to the induced current in the secondary coil is

A- to the right

B- to the left

C- Zero

Part C

The power supply drives the primary coil at 9.9 MHz. If this frequency is doubled, how must the capacitor in the secondary circuit be changed?

A- Increase by a factor of 2.

B- Increase by a factor of 2?.

C- Decrease by a factor of 2.

D- Decrease by a factor of 4.

Part D

What are the rms and peak currents for a 40 W bulb? (The rms voltage is the usual 120 V).

A- 0.33 A , 0.33 A .

B- 0.47 A , 0.47 A .

C- 0.47 A , 0.33 A .

D- 0.33 A , 0.47 A .

Explanation / Answer

part A )

as the current is counter clockwise ,

the magnetic field is to the right

A) to the right

part B)

according to Lenz's law , the induced current in secondary will oppose the change in primary coil

Hence, as the field is increasing inprimary ,

field due to secondary is A) to the right

part C )

as the frequency must be the same ,

f = 1/2pi*sqrt(L*C)

as fnew/f = sqrt(C/Cnew)

2^2 = C/Cnew

Cnew = C/4

hence , the capacitor is D- Decrease by a factor of 4

part D)

as Power = Vrms * Irms

40 = 120 * Irms

Irms = 0.33 A ,

Now, Ipeak = sqrt(2) * Irms

Ipeak = 0.47 A

Hence , the correct choice is D) 0.33 A , 0.47 A

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