Question about inductor and resistor in circuits

Question

When a pure DC source is connected with a pure inductor the change in current in the inductor is linear in the inductor with respect to time

But when a pure DC source is connected with a pure inductor and a resistor in series (RL Circuit) with it we get an exponential change in the current in that (RL) circuit.

My question is that why adding a resistance in series with inductor and DC source makes the current change exponential although we get a linear change in current when we connect DC source with a pure inductor.

What is in the resistor that makes the current change exponential when the resistance is added with a pure inductor in series with a dc source?

Answer 1

Well, for just an inductor connected to a DC voltage source:$$V = L\frac{dI}{dt},$$ so $$I=\frac{V}{L}t.$$ With a resistor, the voltage across the inductor is reduced by an amount proportional to the current: \begin{align} V_{DC}&=V_R+V_L,\\ V_R&=IR,\\ V_L&=L\frac{dI}{dt},\\ \frac{V_{DC}-IR}{L}&=\frac{dI}{dt} \end{align}

The resistor is represented by that extra $IR$ term in the final equation, which wasn't there before. It introduces feedback into the circuit: the current increase is due to the voltage across the inductor, but the resistor makes it so that that voltage (and thus the current increase) gets smaller as the current rises. The $I-t$ curve thus curves more and more strongly away from a line as time goes on, producing the exponential curve. $$I=\frac{V_{DC}}{R}\left(1-\exp\left(-\frac{R}{L}t\right)\right).$$ For a short enough time right when you turn on the circuit, the $I-t$ curve looks linear, as if the resistor weren't there (as the current through/voltage across the resistor hasn't developed much yet). After a long enough time after you turn on the circuit, the $I-t$ curve looks constant, as if the inductor weren't there (as the voltage across the inductor has dropped to basically zero).

Answer 2

It is useful to remember the analogies mechanics x electricity in cases like this, because they are more intuitive, and at the same time rigorous, because based on the same differential equations:

Inductor is mass, Voltage is force, current is velocity and resistance is friction.

Without friction the speed increases unbounded and linearly if the force is constant: $$F = ma = m\frac{dv}{dt} \implies v = (F/m)t$$

With friction, the linear behaviour is modified because the friction force tends to be proportional to velocity.

The result is an exponential growth until terminal velocity. Like the increase of current until that determined by Ohms Law.

$$F - kv = m\frac{dv}{dt}$$

$$m\frac{dv}{F - kv} = dt$$

$$v = \frac{F(1 - e^{-\frac{k}{m}t})}{k}$$