# Voltage drop across an inductor [closed]

An inductor should oppose the current right, so shouldn't the voltage drop across the inductor be in the opposite direction as that shown in the figure?

## closed as off-topic by Bill N, freecharly, Jon Custer, ZeroTheHero, rob♦Apr 2 '18 at 23:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Bill N, freecharly, Jon Custer, ZeroTheHero, rob
If this question can be reworded to fit the rules in the help center, please edit the question.

An inductor should oppose the current right

No, that's not right. An (ideal) inductor 'opposes' a change in current but offers no opposition to a fixed current, i.e., the voltage $e$ across the inductor is zero for constant $I$.

so shouldn't the voltage drop across the inductor be in the opposite direction as that shown in the figure?

The $+$ and $-$ signs are there to indicate the reference polarity 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 $I$.

But the inductor opposes the change in current by inducing an emf opposite to the direction of flow of current, isn't it?

Why? If $I$ is increasing (decreasing), the voltage across the inductor $e$ is positive (negative) regardless of 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 positive for $I_0 < \frac{\mathcal{E}}{R}$ and strictly negative for $I_0 > \frac{\mathcal{E}}{R}$.

So, if the current is initially negative, the current will increase through 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 signs even though $I(t)$ does.

• But the inductor opposes the change in current by inducing an emf opposite to the direction of flow of current, isn't it? – Siva Raja Ganesh Mar 28 '18 at 13:16
• @SivaRajaGanesh, No. The induced EMF will not oppose the flow of current. It will oppose the change in current flow. Assuming that the arrow indicates the direction of "positive" current flow, then the voltage at the left end of the inductor will be positive with respect to the right end whenever the current is increasing, and the voltage at the left end will be negative with respect to the right end when the current is decreasing. – Solomon Slow Mar 28 '18 at 13:27