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Here are the voltage and current shown for the primary winding of a transformer or an ideal inductor when the voltage is sinusodial.

My question is why the current is increasing when the voltage is decreasing as shown in the black box ? Its obvious that current should be increasing with increasing voltage and current should be decreasing with decreasing voltage but here the current is increasing for decreasing voltage.

And the current decreases during the negative voltage cycles only.

What is the the reason behind all this ?

enter image description here

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  • \$\begingroup\$ Inrush current is another thing than what you talk about in your question. \$\endgroup\$ – linuxfan says Reinstate Monica Apr 27 '20 at 9:21
  • \$\begingroup\$ It is the effect of reactance: en.wikipedia.org/wiki/Electrical_reactance \$\endgroup\$ – linuxfan says Reinstate Monica Apr 27 '20 at 9:25
  • \$\begingroup\$ The current in an inductor is proportional to the integral of the applied voltage. So when the voltage reaches zero, the maximum current is the result of the accumulated integral of the just past cycle of voltage. The energy that flowed into the inductor during that cycle is stored up (integrated) and manifests itself as a peak current. \$\endgroup\$ – G36 Apr 27 '20 at 12:56
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    \$\begingroup\$ This is similar to what happens if you apply a sine wave of current to a capacitor i.stack.imgur.com/cFxy8.png The capacitor integrates the current and when the previous cycle of current comes to an end as the current reaches zero, the voltage across the capacitor reaches a maximum (at T = 10ms). \$\endgroup\$ – G36 Apr 27 '20 at 12:56
  • \$\begingroup\$ Yes @G36. Its like if we apply fixed dc voltage to inductor the current will increase at a constant rate. now if we have any mechanism to reduce the dc voltage, rate of increase of current will be reduced but the magnitude of the current will go higher than its previous value although the dc voltage is reduced now \$\endgroup\$ – Alex Apr 27 '20 at 13:19
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Here are the voltage and current shown for the primary winding of a transformer or an ideal inductor when the voltage is sinusodial.

My question is why the current is increasing when the voltage is decreasing as shown in the black box ? Its obvious that current should be increasing with increasing voltage and current should be decreasing with decreasing voltage but here the current is increasing for decreasing voltage.

It's obvious that into a resistor current increases with increasing voltage, and decreases with decreasing voltage, because that's how a resistor works, the current is governed by the voltage.

However, an inductor is not a resistor. The current in an inductor is governed not by the voltage, but by the time integral of the voltage

And the current decreases during the negative voltage cycles only.

What is the the reason behind all this ?

Because it's only in the negative half cycle that the time integral of the voltage decreases.

I suspect you're having problems with the concept of a time integral. Let's find something more familiar that's also a time integral.

Your credit card debt is the time integral of the rate at which you buy stuff online. If you keep buying stuff, your debt will keep increasing. It doesn't matter whether the rate at which you buy stuff is increasing, or decreasing (hey, I bought less stuff than last week!). As long as the rate is positive, like the voltage is positive, the debt or the inductor current will keep increasing. To get the debt or the current to decrease, the spend rate or the voltage must actually go negative, so sell some stuff and pay off some debt.

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  • \$\begingroup\$ When voltage decreases the current starts to drop flux starts to drop and the energy stored in the magnetic field starts to drop. This energy is transferred from the inductor back to the voltage source where it originated. can this be the reason ? energy is supplied to the source when the voltage of the voltage source start decreasing but the overall voltage in the circuit drives the current in the positive because now we have two source , one the voltage source and other the energy stored in the inductor? \$\endgroup\$ – Alex Apr 27 '20 at 9:52
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    \$\begingroup\$ When the voltage decreases, the current DOES NOT start to drop. Open your eyes and see what's happening. You will make no progress until you see what is, rather than what you think you see. \$\endgroup\$ – Neil_UK Apr 27 '20 at 10:28
  • \$\begingroup\$ if we apply fixed dc voltage to inductor the current will increase at a constant rate. now if we have any mechanism to reduce the dc voltage, rate of increase of current will be reduced but the magnitude of the current will go higher than its previous value although the dc voltage is reduced now ? right ? \$\endgroup\$ – Alex Apr 27 '20 at 10:48
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    \$\begingroup\$ @Alex That's right! \$\endgroup\$ – Neil_UK Apr 27 '20 at 10:53
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My question is why the current is increasing when the voltage is decreasing

Current rises at a rate determined by voltage. If the voltage is positive and small in amplitude, current rises at a small rate. If the voltage is positive and large in amplitude, current rises at a large rate. Current rises and continues to rise when a positive voltage is applied across the inductor terminals even if that voltage is falling towards zero. The current seeks to find a value that represents the "area" of the voltage waveform and, only if that voltage fell to 0 volts will the current remain constant as per: -

$$V = L\dfrac{di}{dt}$$

With the voltage now at 0 volts, there is no area under the voltage waveform and hence, there is no need for the current to increase or decrease with time i.e. it remains at a constant value.

Its obvious that current should be increasing with increasing voltage and current should be decreasing with decreasing voltage

No, that's not how it works.

Consider the case that the voltage is negative but rising towards zero volts (i.e. an increasing voltage) - the current will continue to fall negatively while the voltage is rising towards zero volts. Only when the voltage reaches 0 volts will the current stop changing. That is because the slope of current is determined by the applied voltage or: -

$$V = L\dfrac{di}{dt}$$

Simple mechanical analogy

You have an object that can travel only forwards or backwards.

  • If the object's velocity was +1 m/s then after 10 seconds the distance traveled will be +10 metres.
  • If the forward speed slowed, the object's distance from the starting point would still increase even though the object was slowing down.
  • If the speed fell to 0 m/s then the object would still be at a positive distance from the starting point and, that distance will remain fixed until the object started moving again.
  • If the object then moved (backwards) at -1 m/s, then the positive distance would start to fall towards zero.
  • If the object continued to move backwards, the distance from the starting point would eventually become negative.

In this analogy: -

  • Velocity behaves like inductor voltage
  • Distance behaves like inductor current
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    \$\begingroup\$ That's what I showed you in this answer. Look at the final picture - I added this after we discussed it in comments. \$\endgroup\$ – Andy aka Apr 27 '20 at 8:44
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    \$\begingroup\$ @Alex No, we can relate the voltage with the rate of change of current in an inductor. \$\endgroup\$ – Neil_UK Apr 27 '20 at 9:07
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    \$\begingroup\$ NO NO NO - when voltage decreases, current DOES NOT start to drop. The rate at which current was rising starts to drop but that does not mean the current drops - current continues to rise but at a slower rate of increase with time. That rate of increase of current only stops when the voltage becomes zero. Then current remains at some fixed value. At that point the flux is maximum and the energy is at a maximum because, current is at a maximum. \$\endgroup\$ – Andy aka Apr 27 '20 at 10:03
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    \$\begingroup\$ No, the magnitude of the current won't reduce until the voltage becomes a negative value. \$\endgroup\$ – Andy aka Apr 27 '20 at 10:25
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    \$\begingroup\$ if we apply \$\color{blue}{variable}\$ dc voltage to inductor the current will increase at a constant rate - No, if you apply a \$\color{red}{fixed}\$ dc voltage, the current will increase at a constant rate. \$\endgroup\$ – Andy aka Apr 27 '20 at 10:33

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