The sign in the case of an inductor is indeed easy to be uncertain about. I would say this is a good illustration of a more general difficulty with signs in physics. The way to get signs right is sometimes not to worry over one equation or another equation of opposite sign, but rather to be clear in your mind about what happens in a simple example case.

I find it very useful to consider the simple circuit with just a resistor and an inductor in it. The voltage around the circuit is zero, so we get the equation

$IR + L dI/dt = 0$

It is easy to see that the sign is correct in this equation, because then we get

$ dI/dt = - (R/L) I$

for which the solution is exponential decay. If we had the opposite sign we would get exponential growth of the current, which is clearly wrong. But you are free to consider the first equation either in the form I wrote it, or in the form

$IR = - L dI/dt$

First of all, don't mix up voltage with current. In your examples 1 and 2 it is certainly true that the voltages across the resistor and inductor are the same w.r.t. the source voltage. This is just Kirchhoff's voltage law. However, this still results in a current lag in the inductor compared to the resistor. Say the source voltage is

$\Delta V_S = V_0 \sin(\omega t)$

So the resistor voltage is

$\Delta V_R = - V_0 \sin(\omega t)$

so that $\sum \Delta V = 0$ as Kirchhoff's law requires. The same goes for the voltage across the inductor, $\Delta V_L$.

But for the resistor we have Ohm's Law

$\Delta V_R = IR$ so the current through the resistor is just

$I_R = -\frac{V_0}{R} \sin(\omega t)$

But for the inductor we have

$\Delta V_L = - L \frac{dI}{dt}$.

So to get the current, $I_L$, you need to **integrate** $\Delta V_L$ w.r.t. time so in this example you will get a cosine instead of a sine. Thus, we see a phase shift in the current (but not the voltage). It is worth going through the integral yourself, but it is also in most elementary circuits textbooks.

## Best Answer

No, that's not right. An (ideal) inductor 'opposes' a

changein current but offers no opposition to a fixed current, i.e., the voltage $e$ across the inductor is zero for constant $I$.The $+$ and $-$ signs are there to indicate the

referencepolarity of the voltage across the inductor. The voltage $e$ can be positive or negative and, according to the drawing, if $e$ is positive, the left-most inductor terminal is positive with respect to the right-most terminal. But, if $e$ is negative, the left-most terminal is negative with respect to the right-most terminal.As the circuit is drawn, if $I$ is

increasing, the inductor voltage $e$ should be positive to oppose the increase of $I$.From the comments:

Why? If $I$ is increasing (decreasing), the voltage across the inductor $e$ is positive (negative)

regardlessof whether the instantaneous value of $I$ is negative or positive.If you solve the differential equation for the current, you'll find that

$$I(t) = \frac{\mathcal{E}}{R} + \left(I_0 - \frac{\mathcal{E}}{R}\right)e^{-tR/L}$$

where $I_0 = I(0)$ is the instantaneous value of the current when $t = 0$.

The voltage across the inductor is then

$$e = L\frac{dI}{dt} = \left(\mathcal{E} - I_0R\right)e^{-tR/L}$$

If you stare at that a bit, you'll see that voltage across the inductor is strictly

positivefor $I_0 < \frac{\mathcal{E}}{R}$ and strictlynegativefor $I_0 > \frac{\mathcal{E}}{R}$.So, if the current is initially negative, the current will

increasethrough zero to become positive (asymptotically approaching the steady state value $I_\infty = \frac{\mathcal{E}}{R}$) but the voltage across the inductor will always be positive.In other words, $e$

will not change signseven though $I(t)$ does.