I am doing some problems in linear and nonlinear circuit analysis, so I have a question...why is the voltage on an inductor represented as \$V=L\frac{di}{dt}\$ instead of \$V=-L\frac{di}{dt}\$, I know \$-L\frac{di}{dt}\$ is back EMF, but why is voltage written without the minus? Can someone explain?
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\$\begingroup\$ Can you provide a link to the source? Add a diagram too if possible. \$\endgroup\$– AJNCommented Dec 3, 2020 at 19:19
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\$\begingroup\$ If current is decreasing then \$\frac{di}{dt}\$ is negative. As per Andy’s answer, no minus sign needed in the formula as it is inherently in the \$\frac{di}{dt}\$. \$\endgroup\$– relayman357Commented Dec 3, 2020 at 20:01
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\$\begingroup\$ Also, welcome to EE StackExchange. Making your post easily readable will help get good responses. Use MathJax to format equations nicely. \$\endgroup\$– relayman357Commented Dec 3, 2020 at 20:13
1 Answer
When you apply a voltage across an inductor, the formula that describes the current flow is this: -
$$V_{APPLIED} = L\cdot\dfrac{di}{dt}$$
So, with 1 volt applied to a 1 henry inductor, the result is a current rising at 1 amp per second. No negative signs were harmed in this formula!
When you induce a voltage (with a changing flux) you put a negative sign in front of the voltage formula to remind us that any current drawn from the terminals will be in a direction that tends to reduce the flux that initially induced the voltage.
Given the comments below and the misconceptions about back-emf I have decided to add a few more words about back-emf. What some folk levitate towards is the belief that the back-emf causes the inductor current to ramp up. This is naïve. Consider a pure inductor with 1 volt applied across its terminals. A back-emf of 1 volt is produced and this appears to suggest that current can never flow. However, appearances can be deceptive because we know that current does flow. So, back-emf is a red herring - it exists but it doesn't stop current flowing as per the relationship described earlier in my answer.
But, this seems to be at odds with the back-emf exactly equalling the applied emf. But, consider what is happening; the back-emf and the applied voltage are both constant values and, they are across a zero ohm impedance hence, the current cannot be defined other than saying: -
$$\boxed{\text{The current is: }\hspace{1cm}\dfrac{0}{0}}$$
Why did I say that the impedance is zero?
Answer: it's an inductor and, the back-emf is induced in series with that inductor hence, with the applied voltage and back-emf voltage being equal, there is 0 volts across the "true" inductance. Zero volts means zero spectral content across that inductor and, it therefore has to be represented by 0 ohms.
In other words, you cannot use the value of applied voltage and back-emf to calculate the current through the inductor. The current rising or falling delivers the back-emf and that's it.
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\$\begingroup\$ That is not correct.You forgot Lenz's law and I don't think it is a correct way to explain this. \$\endgroup\$– Se1fieCommented Dec 3, 2020 at 18:54
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\$\begingroup\$ Ok, but, current increases slow, that slow increasing is because of negative voltage induced by changing magnetic flux...I would be very grateful if You can explain a little bit deeper... \$\endgroup\$ Commented Dec 3, 2020 at 18:54
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\$\begingroup\$ @Se1fie in what respect is it incorrect? \$\endgroup\$– Andy akaCommented Dec 3, 2020 at 18:55
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\$\begingroup\$ Andy the change in current flowing through an inductor creates a back EMF not the other way around. \$\endgroup\$– Se1fieCommented Dec 3, 2020 at 18:57
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2\$\begingroup\$ @MatijaStankovic the back-emf is irrelevant - it's a red herring in these circumstances - the back-emf is in opposition to the applied voltage and, it has exactly the same value for a pure inductor so, on that basis might you be inclined to believe that no current can flow? \$\endgroup\$– Andy akaCommented Dec 3, 2020 at 19:17