0
\$\begingroup\$

If this voltage is applied to an ideal inductor. How will be the current waveform look like?

Kindly any one show me the graph for inductor current and if possible explain it. I shall be very grateful! (Voltage is in blue)

enter image description here

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Sounds like homework. What have you tried so far? Do you know the formula for voltage across an inductor? \$\endgroup\$
    – winny
    Commented Apr 25, 2020 at 9:21
  • 2
    \$\begingroup\$ If you want a quantitive answer, then do the math, it's a very simple relation, dI/dt=V/L. If you want a qualitative answer, then it's in my answer to your previous very similar question \$\endgroup\$
    – Neil_UK
    Commented Apr 25, 2020 at 9:24
  • \$\begingroup\$ i just want the graph of current. @Neil_UK . \$\endgroup\$
    – Alex
    Commented Apr 25, 2020 at 9:30
  • 2
    \$\begingroup\$ I’m voting to close this question because it looks like homework and there is no attempt to solve it. \$\endgroup\$ Commented Apr 25, 2020 at 9:36

1 Answer 1

1
\$\begingroup\$

Kindly any one show me the graph for inductor current and if possible explain it. I shall be very grateful!

It's quite easy to do if you use a simulator (like Micro-cap): -

enter image description here

Voltage in blue and current through a 2 henry inductor in red. Timebase in seconds.

The current-voltage relationship for an inductor is: -

$$V = L\dfrac{di}{dt}$$

The above can be solved for current by integrating both sides.

$$i = \dfrac{1}{L}\int{v\: dt}$$

And, for a linearly rising voltage (\$v = V_{PK}\cdot t\$),

$$i = \dfrac{V_{PK}}{L}\int{t\: dt} = \dfrac{V_{PK}}{2L}t^2$$

Circuit: -

enter image description here

Slope of the current graph

enter image description here

This is just to show that visualizing the slope helps to understand the formula. I've got a green circle on the current graph and I've visually calculated the slope to be 5 amps per second. This corresponds with an applied voltage at that moment (blue circle) of 10 volts and, given that the inductance is 2 henry, the slope mathematically agrees because: -

$$\dfrac{V}{L} = \dfrac{di}{dt} = 5\text{ amps per second}$$

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Commented Jun 10, 2020 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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