0
\$\begingroup\$

Q. Lower the self inductance of a coil _.

Options :

  1. more will be the weber-turns.
  2. more will be the emf induced.
  3. lesser the flux produced by it.
  4. smaller the delay in establishing steady current through it.

Ans - option 4

But why is it so? why not the other options? For the first option, I can say that if we increase the weber-turns, then it will increase the inductance which is in contradiction to the question given.

\$\endgroup\$
2
\$\begingroup\$

Let's review the formulas that apply:

Weber-Turns (flux linkage): $$ \lambda=N\Phi $$

Flux: $$ \Phi=\frac {Ni} {\Re}$$

where \$\Re\$ is the reluctance, that depends on the coil's core material and geometry.

Voltage (emf)

$$ V=\frac {d\lambda}{dt}=\frac {N^2} {\Re} \frac {di}{dt}=L\frac {di}{dt}$$

The problem here is that it is not specified how the coil is excited.

Coil excited by a sinewave current

If it is excited by a sinewave current, then it is rather clear that lower inductance, implies lower flux linkage, lower emf. About the flux it is given by \$\Phi=\frac {Ni} {\Re}=\frac L N i\$. So reducing L alone doesn't give an indication of what \$\Phi\$ will do. For example, N can be reduced in more proportion than L if one choose a different coil material for example. Therefore \$\Phi\$ can be bigger or lesser than before.

Coil excited by a voltage

In contrast, if the coil excited by a voltage, then the flux linkage is given by the voltage alone (because \$V=\frac {d\lambda}{dt}\$, linkage is the voltage integration), and so doesn't depends on the inductance. Also the flux doesn't depends really on the inductance in this case, but just on the number of turns. (And because if wished you can have a big inductance with few turns by selecting a different core, then you might have a big or smaller flux by reducing L). Finally, again if excited by a voltage, the emf will be equal to it, so doesn't depends on the inductance.

Actually whatever excitation you consider, the first 3 options are wrong.

The last option asks about a steady current, so one can assume it is excited by a voltage in series with a resistor.

\$\endgroup\$
  • \$\begingroup\$ After referring certain books, I have come to the following conclusion : self-inductance is the property of the coil due to which it opposes any increase or decrease of current ( which produces flux) through it. If self-inductance is reduced, this implies that now the opposition to the current is less i.e. in less time the current can become steady through the coil. \$\endgroup\$ – Vibhu Mar 15 '15 at 15:58
  • 1
    \$\begingroup\$ Ok! I thought you were having problems to understand the 3 first options. The 4th option is rather straightforward. From V=Ldi/dt, you can see that i=1/L·integrate V. Therefore 1/L is the slope at which i increases assuming constant voltage. The bigger the L, the slower the rate (i.e bigger "delay"). \$\endgroup\$ – Roger C. Mar 15 '15 at 16:39

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.