# Is the voltage drop across an inductor in a purely inductive circuit equal to $X_L\times I$ or back EMF?

In a purely inductive circuit, due to Kirchoff's voltage law we say that voltage drop across the inductor is equal to (negative of) applied AC voltage. However whenever we calculate the voltage drop across an inductor anywhere else, for example in an RLC circuit, we say that the voltage drop across the inductor is equal to inductive reactance multiplied by current passing through it, i.e. $$X_L\times I$$

However, these 2 cannot be the same. We derive an expression for current in a purely Inductive circuit by fulfilling the following condition (AC voltage is mentioned as $$V(t)$$) $$V(t)=L\frac{di}{dt}\\ V(t)\ dt=L\ di$$ Assuming $$V(t)=10\sin(\omega t)$$ and integrating both sides $$i = -10\cos(\omega t)/\omega L\\ i = 10\sin(\omega t + 90)/\omega L$$

Now if we multiply current with inductive reactance, we get $$i\omega L = 10\sin(\omega t+90)$$.

Multiplying current with inductive reactance should supposedly give us the voltage across the inductor, right? But instead of $$10\sin(\omega t)$$ we are getting a curve which is phase shifted by 90 degrees! Where am I going wrong and how does multiplying current with capacitive reactance give you voltage across inductor anyway (intuitively)?

If we assume that multiplying current with inductive reactance gives us voltage drop across the inductor then the net voltage in that purely inductive circuit won't be zero.

Also what exactly does $$10\sin(\omega t+90)$$ tell us: is it the voltage across the inductor, net potential in the circuit, what exactly does it denote?

I know that the current and voltage are out of phase by 90 degrees but how does that relate to these?

Edit: My main question here is, $$X_L\times I$$ is this expression equivalent to voltage dropped across an Inductor in

1. A purely Inductive circuit
2. Any other circuit which isn't purely Inductive

Multiplying current with inductive reactance should supposedly give us the voltage across the inductor, right?

Only in complex phasor representation. In real representation, which you have used, this relation is not valid, due to the phase shift between voltage and current.

• Can you explain why this is true Nov 30, 2021 at 18:55
• Sorry, that would take a lot effort. Find some electrical engineering textbook where they explain complex representation (phasors) as used in AC circuits. Nov 30, 2021 at 21:03

Current and voltage waveforms (assuming a sine wave) are phase shifted by 90 degrees for capacitors and inductors.

There are 4 quadrants for a sine wave:

• Q1 0-90 deg,
• Q2 90-180 deg,
• Q3 180-270 deg,
• Q4 270-360 deg.

In Q1-Q2 the sine wave is positive, and in Q3-Q4 the sine wave is negative.

The capacitor voltage-current relationship is defined as

$$v_C(t) = \frac{1}{C} \int_0^t i(\tau) d\tau$$

Think of integrator as an accumulator: in every discrete step add new value to the sum of all previous values:

$$v_C(k) = \frac{1}{C} i(k) + v_C(k-1)$$

Now imagine a sine-wave current source m connected to a capacitor:

• In Q1 voltage increases because current is positive.
• In Q2 the current decreases but is still positive so voltage keeps increasing.
• In Q3 current is negative so voltage starts to decrease. Obviously the voltage is at maximum between Q2 and Q3.
• In Q4 current increases but is still negative which means voltage keeps decreasing.
• In Q1 current becomes positive so voltage starts to increase. The voltage is at minimum between Q4 and Q1.

Now draw the voltage waveform and compare to the current waveform and you will see where the 90 degrees phase shift comes from. The same applies to inductors, it is only less intuitive (at least to me!).

When you write $$V_C = I_C X_C$$ you take into account both amplitude and phase shift! This is called a phasor notation. Here is how it is actually expanded:

$$V_C = I\angle\phi \cdot \frac{1}{\omega C}\angle-\frac{\pi}{2} = \frac{I}{\omega C} \angle \phi-\frac{\pi}{2}$$

The above equation actually tells you that the voltage waveform is shifted by $$\frac{\pi}{2}$$ to the right compared to current waveform (assuming time increases from left to right), and this equals one quarter of the full cycle.

The phasors are translated to sine waves as follows:

$$i_C(t) = \sqrt{2}I \sin(\omega t + \phi), \quad v_C(t) = \sqrt{2} \frac{I}{\omega C} \sin(\omega t + \phi - \frac{\pi}{2})$$

and this gives you information about instantaneous voltage and current values.

It is really easy to use phasor notation. It is the same as with purely resistive circuits, the only difference is that you have to include phases. The three impedances are defined as:

$$X_R = R \angle 0, \quad X_L = \omega L \angle \frac{\pi}{2}, \quad X_C = \frac{1}{\omega C} \angle -\frac{\pi}{2}$$

The equation $$V_L = I \cdot X_L$$ is always valid as long as $$I$$ is the current through inductor. This is true in all (linear) circuits, not just purely inductive circuits.