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When a dc voltage source is connected with an ideal inductor we get a linear change/increase in current in inductor with respect to time.

But when we add resistance in series with inductor and apply dc voltage we get exponential increase in current in the circuit and so the change in current is not linear now.

What is the reason behind that?

I wonder how a resister can modify the rate of change of current in an inductor or circuit? Resistor just decreases the current in the circuit, what it has to do with the shape of the current or current change?

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  • \$\begingroup\$ Ideal Inductor shorts DC voltage source there is no linear increase in current \$\endgroup\$ – sarthak Apr 20 at 10:38
  • \$\begingroup\$ @sarthak not true at all. \$\endgroup\$ – Andy aka Apr 20 at 11:15
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    \$\begingroup\$ @sarthak I'm pretty sure that \$v = L\frac{di}{dt}\$ still applies in this case. \$\endgroup\$ – Elliot Alderson Apr 20 at 11:44
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But when we add resistance in series with inductor and apply dc voltage we get exponential increase in current

It's not an exponential rise but a reverse exponential rise: -

enter image description here

Picture from here.

As current rises, the volt drop across the resistor increases and this reduces the effective terminal voltage appearing across the inductor hence, the rate of change of current in the inductor (as defined by \$\dfrac{di}{dt} = \dfrac{V}{L}\$) has to fall.

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  • \$\begingroup\$ Actually i was looking for why its exactly exponential but not of any other shape. I guess to know that one has to dig deep into electromagnetics. \$\endgroup\$ – Alex Apr 20 at 14:57
  • \$\begingroup\$ Do you know how to develop the 1st order differential equation for this type of circuit? Do you know how to solve 1st order differential equations. You don't need to delve deeply into EM. \$\endgroup\$ – Andy aka Apr 20 at 14:59
  • \$\begingroup\$ Yes i have solved such circuits using differential equations. What i am trying to do is to solve the circuits philosophically. Lol \$\endgroup\$ – Alex Apr 20 at 16:15
  • \$\begingroup\$ So what i got is that when switch is closed we have equal and opposite voltage across the inductor. To maintain that voltage across the inductor current has to change at some constant rate. But as soon as the current changes with that rate. There will a voltage drop across the resistor and voltage across the inductor goes down. As the voltage across the inductor goes down then to maintain that lower voltage across inductor current has to change with the new lower rate and so on. Am i right @Andy aka? \$\endgroup\$ – Alex Apr 20 at 16:24
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    \$\begingroup\$ @Alex that sounds about right in a philosophical way. \$\endgroup\$ – Andy aka Apr 20 at 17:45

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