1
\$\begingroup\$

"A student tried to make an AC generator consists of a coil of one turn, and cross sectional area of (7.54 × 10^-3 m²), and rotates in magnetic flux density (2.5 × 10^-2 T), and when it was taken out of the magnetic flux in 1.256 × 10^-3 sec an induced emf of 1.5V has been generated." That was the description of a generator, then it asked whether that generator is practical or not? The answer was "No" without an explanation, so could you please explain it for me?

\$\endgroup\$
  • \$\begingroup\$ "1.5V has been generated." for how long? plus 1.5V AC is not a standard for any device. was he too cagey to not tell? \$\endgroup\$ – atmnt Apr 16 '17 at 15:06
  • \$\begingroup\$ @user134429, I don't know actually for how long, may be it is for (1.256 × 10^-3 sec) as supposed to be the time of the change in flux. And it didn't give any other information more than those given in the question above \$\endgroup\$ – Asmaa Apr 16 '17 at 15:14
  • \$\begingroup\$ do you really believe it can be practical when the duration is 1.256 × 10^-3 sec? \$\endgroup\$ – atmnt Apr 16 '17 at 15:25
  • \$\begingroup\$ Is the duration the only issue here? \$\endgroup\$ – Asmaa Apr 16 '17 at 15:37
  • 1
    \$\begingroup\$ How many practical generators use a coil of one turn? \$\endgroup\$ – JIm Dearden Apr 16 '17 at 15:41
0
\$\begingroup\$

It is not a description of a practical generator:

  1. The magnetic field is weaker than the field of a typical refrigerator magnet.
  2. There is no indication of a mechanism to sustain the magnetic field or maintain its direction.
  3. The coil of wire is initially described as rotating, but then described as having been taken out of the field. A generator requires continuous rotation or some other continuous motion.
  4. There is no indication of a method of supporting, rotating or otherwise moving the coil of wire with respect to the magnetic field.
  5. There is no indication of any means to connect any kind of electrical load to the coil. A practical generator continuously supplies electrical power through electrical conductors to a load that can utilize the power.
  6. There are practical generators that are only intended to detect how fast something is moving. There may be some minimum voltage and current necessary for that function, but the minimum could be extremely low. It is not necessary to prove any minimum to say that a generator is practical.
\$\endgroup\$
1
\$\begingroup\$

In order for it to be practical the frequency of current must yield an inductive impedance about 10x to 100x the DCR of the wire. Otherwise it will not work. even at 50MHz it won't work unless highly tuned with heavy copper thin hollow core conduit. Then skin effects and RPM , inertial mass, imbalance etc make it impossible.

Thus the core in a generator must have a magnetization flux to couple the magnetic field thru the commutation DCR to generate any appreciable V*I= power.

otherwise the inductive impedance will just shunt or short out the magnetic field and generate very little voltage or current. This can be done with permanent magnets, laminated iron core with an excitation field current to increase voltage etc.

Do an impedance ratio calculation with a 0.1uH single turn and see what Z(f) you get at 60 Hz with conductor and current needed to get 1V then conduction losses and magnet strength in Tesla to get that current. You would need a cryogenic cooler and a superconductor.

\$\endgroup\$
1
\$\begingroup\$

I think you (and unfortunately, us) are taking this question far to serious. This is a question of scale.

Spin a coil in a magnetic field or spin a magnet in a coil and all you get is 1.5V.

Although you can get into the theory, it is NOT practical because the generator does not have the capability of an AAA 1.5V battery. A lot of trouble for a weak flashlight, which cannot move.

So no practical application, but it does demonstrate the principle.

$$V_{Ind} = N B l v$$

Significantly, increase all of those and now you have a practical generator.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.