In a simple AC circuit with a source and an inductor, we have $$\Delta V + \Delta V_{L} = 0 $$ due to KVL where $\Delta V $ is the voltage source and $\Delta V _{L}$ is the voltage across the inductor.

If we assume that the AC source gives us the voltage of say $\Delta V = V_{\text{Max}} \sin(\omega t)$,

we have

$$ V_{\text{Max}} \sin(\omega t) + \Delta V_{L} = 0 $$


$$ \Delta V_{L} = - V_{\text{Max}} \sin(\omega t) = V_{\text{Max}}\sin(\omega t - \pi) $$ Now to calculate the current,

$$\Delta V_{L} = -L\frac{dI_L}{dt} $$

$$ V_{\text{Max}} \sin(\omega t) = L\frac{dI_L}{dt} $$

$$ \frac{dI_L}{dt} = \frac{V_{\text{Max}}}{L}\sin(\omega t) $$

$$I_{L} = -\frac{V_{\text{Max}}}{\omega L}\cos(\omega t) = I_{\text{Max}}\sin(\omega t - \frac{\pi}{2})$$

where $I_{\text{Max}} \equiv\frac{V_{\text{Max}}}{\omega L}$.

But in this case, the current leads the voltage as shown below where current is the green line and the voltage is the red line. Not too sure where I made a mistake. enter image description here

  • $\begingroup$ If the voltage source and inductance are connected at two points A and N, and the positive direction is Source - A - inductance - E - Source, then your graph correctly shows that when the current increases the voltage (in the forward direction SAIE) is negative. But often the voltage across a load is regarded as being in the opposite direction (especially in elementary DC stuff). If the voltage at E is regarded as 0V then the voltage at A will be the reverse of your graph. Don't know if this is the case, but maybe. $\endgroup$
    – Peter
    Feb 5, 2022 at 5:26

1 Answer 1


You are mixing two different things.

If you do KVL, the FIRST thing you need to do is to assign name AND a direction to every voltage and current in the circuit. Let's do this here.

enter image description here

The choice of direction is somewhat arbitrary. You can count $V_L$ positive from A to B or positive from B to A. The only difference is the numbers will come out with a different sign, i.e. $V_{AB} = - V_{BA}$.

This being said, a common convention is to choose directions for sources so that voltage is counted against the current and for passives to have voltage and current going in the same direction. This way all powers come out to be positive and you can use "standard" impedance equations. If you choose opposite directions for passive you need to flip the sign of the impedance equations, i.e. Ohm's law becomes $V = - R \cdot I$

So in this case we would get

$$V_1 = V_L = V_{AB}$$

At the inductor we have also

$$V_L = L_1 \frac{\partial I_1}{\partial t} \rightarrow I_1 = \frac{1}{L_1}\int V_L dt$$

So if we have $V_1(t) = V_{0}sin(\omega t)$, we get

$$ V_L = V_{0}sin(\omega t) \;\;\;\; I_1(t) = \frac{V_0}{L_1} \sin(\omega t - \pi/2)$$

So the inductor current will always lead the inductor voltage as long as you choose the direction of current and voltage to be the same.


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