Conceptually understanding RL circuits

I'm struggling to conceptually understand the current-time profile of an RL circuit. Specifically, what causes the rate of change of current, $$\frac{\partial i}{\partial t}$$, to start off high when you first connect the battery and decrease with time?

I understand that the current through the inductor induces a magnetic field, $$\vec{B}$$, around the inductor which itself induces an emf, $$\epsilon_L$$, which acts to oppose the change in magnetic flux, $$\phi_B$$. This is a statement of Faraday's induction and Lenz's law.

To me, as soon as the battery is connected, the rate of change of current is at its maximum which means the induced emf across the inductor is at its maximum. is that right?

If it is, is it also correct to state that this reduces the rate of change in current and in so doing the induced emf decreases?

Would this not allow the current in the circuit to increase again?

• AH! I cracked it for myself. The rate of change of current is $\frac{\partial i}{\partial t} = \frac{\epsilon - iR}{L}$ and so at $i=0$ (also $t=0$) the rate of change of current is maximum at $\frac{\epsilon}{L}$ but as $i$ increases, this rate of change necessarily decreases leading to your standard logarithmic growth. It may seem trivial but this was genuinely what I was conceptually stuck on. – Jamie Smith Oct 11 '19 at 10:38

3 Answers

Assuming you are talking about a series RL circuit with an ideal inductor, it is correct that $$\frac{di}{dt}$$ is maximum and the voltage across the inductor is a maximum equal to the battery voltage when the battery is first connected to the circuit. The current is initially zero.

It is also correct that the rate of change in current then decreases. However, the amount of current is at the same time increasing until eventually it reaches a maximum of V/R where V is the battery voltage. Then the voltage across the inductor is zero and all the battery voltage is across the resistor.

Here are the relevant equations for the inductor current and inductor voltage as a function of time (assuming no initial current in the inductor):

$$i(t)=\frac{V}{R}(1-e^{-Rt/L})$$

$$v(t)=Ve^{-Rt/L}$$

Where $$V$$ is the battery voltage.

Hope this helps

When you say the $$di/dt$$ is initially "high" you are comparing it with the wrong thing, to understand what is going on.

If there was no inductor and just a resistor, the initial value of $$di/dt$$ would be (in theory) "infinite" as the current changed instantaneously from $$i = 0$$ to $$i = V/R$$.

The "back EMF" $$e_L$$ is initially equal and opposite to the applied voltage and reduces the rate of change of current to $$V/L$$. As the current increases, the voltage across the resistor increases and therefore the rate of change of current reduces even more, to $$(V-iR)/L$$. In the steady state condition, $$V = iR$$ and the rate of change of current is $$0$$.

In the circuit $$\mathcal E_{\rm battery}-\mathcal E_{\rm inductor} = V_{\rm resistor} \Rightarrow \mathcal E_{\rm battery} - L\dfrac{dI}{dt} =IR$$

On switch on $$I=0$$ as the current cannot changer instantaneously.
At this time $$\mathcal E_{\rm battery} = L\left [\dfrac{dI}{dt}\right]_{\rm maximum}$$ as $$\mathcal E_{\rm inductor}$$ cannot be larger than $$\mathcal E_{\rm battery}$$.

As time progresses and the current increases, hence $$IR$$ increases and so as $$\mathcal E_{\rm battery} -IR =\mathcal E_{\rm inductor}=L\dfrac{dI}{dt}$$ this means that $$\dfrac{dI}{dt}$$ decreases.

The rate of increase in current will decrease with time and the current will tend towards a steady value of $$\dfrac{\mathcal E_{\rm battery}}{R}$$.