# Calculating induced voltage

I am studying basic electricity in Delmars Standard Textbook of Electricity. I am a complete beginner with electronic concepts.

In one unit I am presented with the following:

Assume an inductor is connected to a 120 VAC line. Assume the inductor has an induced voltage of 116 V. Subtract the induced voltage from the applied voltage leaves you with 4 V to push current through the wire resistance of the coil.

This makes sense. However, in the next paragraph when it comes to how the book presents calculating induced voltage I start to get confused. The book goes on to explain another scenario where we calculate the amount of voltage required to achieve a particular current flow and are given the applied voltage and resistance. This allows us to calculate the required voltage under such a scenario. Again, this makes sense.

An inductor is connected to a 120 V circuit, the coil has a resistance of 6 Ω, and a current of 0.8 A flows through the circuit. Thus, the voltage necessary to push the 0.8 A through the 6 Ω of resistance is 4.8 V.

Where things fall apart on my understanding is the following statement presented in the book:

Only 4.8 V are needed to push the current through the wire resistance of the inductor, the remaining applied voltage is used to overcome the coil's induced voltage of 119.904 V. The induced voltage is calculated with the following formula -

√(1202 - 4.82) = 119.904 V

I have no idea where this formula comes from.

The book goes on to say after subtracting the 4.8 V from 120 V of applied voltage, the remaining voltage is used to overcome the coils induced voltage of 119.904 V. This would suggest that 115.2 V of applied voltage can overcome 119.904 V of induced voltage. Is this possible?

Put another way, if we take 120 V (applied voltage) - 119.904 V (induced voltage) we are left with 0.096 V which is less than the 4.8 V that is required to push 0.8 A of current. Am I thinking about this incorrectly?

Does anyone know where the square root formula presentef above comes from?

Here’s the passage I’m referring to. Basically what confuses me is at the top the pythagorean theorem is not used in calculating how much voltage is available but at the bottom it is used. When applying the same math used at the top of the passage things don’t make sense in the context of the example at the bottom of the passage.

Your "subtract the induced 116V from 120 volts and get 4V pushing current through the resistance part" may make sense for you. That's because you do not take into the account the phase difference of the sinusoidal AC voltages that you subtract. The RMS value is got right by subtracting the squares of the voltages and by taking a squareroot. That can be proven with a lengthy calculation which needs tricky trigonometric formulas and an integral according the mathematical meaning of RMS voltages.

About 150 years ago one genius named Oliver Heaviside invented a way to simplify the practical calculation of circuits. In 1900 his invention was fully exploited. It was taken in to use in electrical engineering and also in the schools where the engineers learned the theory of their profession. Today the practical method of solving AC circuits which contain linear parts is called "Phasor Calculus". The squareroot of the sum or difference of squares is used there from the very beginning.

• So this book hasn’t covered things like the impact of being 90 degrees out of phase yet. So the RMS value derived by the pythagorean theorem is the way the induced voltage of an inductor is calculated? Mar 8, 2023 at 22:25
• I need the full book in front of my eyes to judge it. Nothing less is enough. As well you may have bypassed something. So I cannot claim anything of the book. But if there's 116V over the inductive part of the inductor it leaves 30.7V over the resistive part. That's the sqrt of the difference of the squares of 120V and 116V. Mar 8, 2023 at 22:41
• Continued: If the book claims that the current is 0.8A, the wire resistance is 6 ohms and it means the induced voltage i.e. the imagined voltage over the inductive part of the inductor is 119.904 volts that's right - as long as we take properly into the account the 90 degrees phase shift of the induced voltage and the current through the inductor. Mar 8, 2023 at 23:12
• I added the passage from the book, might provide a bit more clarity to what I'm looking at. The disconnect is why is the Pythagorean theorem only needed in the second example and not the first? Mar 9, 2023 at 17:53
• Heaviside did invent the p operator, but credit Steinmetz for applying that concept with phasors for circuit analysis. Btw, Heaviside was much criticized at the time for his lack of rigor; turned out that he was right though Mar 10, 2023 at 13:33

What is missing in your understanding is that the voltage drop across the resistance is 90 degrees out of phase with respect to the induced voltage.

The voltage/current relationship for an ideal inductor is $$\v=L*di/dt\$$. The induced voltage is proportional not to the inductor current, but to the rate of change of the current. The derivative of $$\sin(x)\$$ is $$\cos(x) = sin(x+90)\$$. So if a sinusoidal voltage is applied across an inductance, the resulting current is also sinusoidal, lagging the voltage by 90 degrees.

Now a physical inductor has resistance as well, so the equation becomes $$\v = L*di/dt + Ri\$$. Even this simple-looking first-order differential equation can be tricky to solve (search 'RL circuit' on Wikipedia for an example). One old method to avoid that and arrive at the steady-state solution is by using phasors, which is what they did in the second example without telling you. Phasors are vectors in the complex number plane; they are useful for solving sinusoidal systems. A phasor diagram is shown below

This shows the induced voltage phasor $$\\overline{V_L}\$$ leading the current phasor $$\\overline{I}\$$ by 90 degrees. The (vector) sum of the induced voltage and the resistive voltage drop $$\\overline{I}R\$$ is the applied voltage $$\\overline{V}\$$. This is where the Pythagorean relationship comes from: $$\V^2 = V_L^2 + (IR)^2\$$

Knowing the applied voltage, resistance, and current you can calculate the induced voltage. Note that the magnitude of $$\\overline{V_L}\$$ is proportional to the magnitude of the current and the inductance. So you can also determine the inductance from $$\V_L = 2\pi fL*I\$$, and also the phase lag $$\\theta = tan^{-1}(V_L/IR) = tan^{-1}(2\pi fL/R)\$$. In this example, $$\L=398\$$ mH and $$\\theta = 87.7^{\circ}\$$ (assuming the frequency is 60 Hz, given that the line voltage is 120V)

The first example, whether it is confusingly worded or just plain wrong, I would just ignore it. That's not how inductors behave. What actually happens is that the current settles at a magnitude where the induced and the resistive voltage drops balance the applied voltage, as shown in the phasor diagram

• Ok, that makes sense. Looking at the attached screenshot, which value is the resistive voltage drop? The 4.8v? I know the induced voltage is the 119.904 and the applied is the 120v. Secondly, why is the Pythagorean theorem only needed in the second example and not the first in the screenshot? Mar 9, 2023 at 17:58
• @Amasephy - I would ignore the first example. I went and expanded my answer, hope that makes it more clear Mar 10, 2023 at 13:18