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I have done a simulation of a simple circuit using Multisim software to simulate the behaviour of the coil in DC circuit. The behaviour of the coil in DC circuit is shown by the below picture:

enter image description here

enter image description here

However using the Multisim software, when I open the switch S1 I obtain a negative voltage through the inductor not positive voltage as indicated in the first picture above:

enter image description here

Same things, when I close the switch, I obtain a positive voltage (inductor charging) not a negative voltage:

enter image description here

Is there any explanation for this issue?

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    \$\begingroup\$ Your first image doesn't show what direction of current and what direction of voltage are considered positive, and what directions are considered negative. \$\endgroup\$
    – The Photon
    Sep 27, 2020 at 13:47
  • \$\begingroup\$ @ThePhoton: It's OK now. I have added the direction of the current. \$\endgroup\$
    – geek225
    Sep 27, 2020 at 13:53
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    \$\begingroup\$ It's still not showing the direction of the voltage \$\endgroup\$
    – The Photon
    Sep 27, 2020 at 13:54
  • \$\begingroup\$ electronics.stackexchange.com/questions/470171/… \$\endgroup\$
    – G36
    Sep 27, 2020 at 14:09
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    \$\begingroup\$ But how would a negative voltage at this point opposite anything? \$\endgroup\$
    – G36
    Sep 27, 2020 at 16:34

2 Answers 2

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Just to show:

enter image description here

The inductance is here 10mH and the input voltage V1 is 10V for easy calculations.

The timed switch SW1 starts to conduct at t=10ms and stops at t=50ms. The inductor has zero resistance, so its current grows to 1A.

At t=50ms the inductor current bulldozes its way through R2 because it's the only available route. The inductor voltage jumps to 10V, it jumps as high as needed for continuing the current. The polarity of the node 3 voltage goes negative because the inductor pulls current through R2 upwards (=from GND towards node 3).

From t=10ms to t=50ms inductor current is fed by voltage divider which as unloaded outputs 5V. Inductor current pulls V(3) onto it knees, the inductor current grows and voltage V(3) decays with time constant L/R. Resistance R=5 ohms (=Thevenin equivalent resistance of the voltage divider).

When the switch opens at t=50ms the inductor current and V(3) decay again with time constant L/R, but this time R=10 ohms, the decay happens with double speed.

I guess your problem is the attempt to approach the practical behaviour of an inductor via a circuit theoretical equation Emf=-L(di/dt). The minus is inserted to make the unmeasurable imagined quantity "electromotive force" to be compliant with the measurable quantities such as voltages between circuit nodes. Check this old discussion of inductor's practical operation: How does the inductor ''really'' induce voltage?

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  • \$\begingroup\$ Thanks for explanation. I understand the point that the inductor will resist to the increase/decrease of current by producing a voltage between its leads in opposing polarity to the change (at t=50ms negative sign voltage through the inductor). However, at t=0s the switch is open so there is no current flowing (0 Amps) and when we close the switch at t=10ms the current will increase so the inductor should resit to this increase of current , but we get a positive voltage same if I replace the inductor by a resistor( positive sign of the voltage) . Is there any explanation ? \$\endgroup\$
    – geek225
    Sep 27, 2020 at 17:20
  • \$\begingroup\$ You should take that inductor has a working law (otherwise it's not an inductor). The law says that the voltage over an inductor is in any case = L(di/dt) where di/dt is the growth rate (amperes per second) of the inductor current. In any circuit at any time the voltage over the inductor and the current through the inductor take values which satisfy this law and all other circuit laws. For this reason often exact solutions can be got by solving differential equations. But simple cases can be solved directly, say Constant DC voltage U over inductor causes linearly growing current, di/dt=U/L. \$\endgroup\$
    – user136077
    Sep 27, 2020 at 17:34
  • \$\begingroup\$ Voltage over an inductor is equal to minus L(di/dt) \$\endgroup\$
    – geek225
    Sep 27, 2020 at 17:39
  • \$\begingroup\$ @geek not true, see my simulation; voltage of node 3 is at t=10ms plus 5V and the inductor current grows. The internal emf in inductor at the same time is minus 5V . Emf is unnecessary extra thing which actually explains nothing of the physics, \$\endgroup\$
    – user136077
    Sep 27, 2020 at 17:46
  • \$\begingroup\$ So I have to consider the voltage over inductor in a circuit by applying this formula VL= Ldi/dt ? \$\endgroup\$
    – geek225
    Sep 27, 2020 at 17:50
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When the switch goes open-circuit, the inductor tries to keep the same current flowing AND in the same direction. The only way it can achieve this is by making its upper terminal go negative. So, at the instant in time that the switch opens, the current is forced through the 10 kohm resistor by a large negative voltage at its upper terminal.

Your upper multisim experiment is incorrect because you are using the formula for back emf (i.e. it has a negative sign) and you have no resistance in your circuit to develop the voltage across when the inductor has charged to its peak current.

In other words, the two "experiments" in your question are not the same. Just think about what your upper multisim "experiment" is telling you; when the switch is closed, how can the inductor voltage be anything else other than the applied supply voltage? Your analysis is faulty.

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    \$\begingroup\$ Andy aka: Thanks for your answer. But I do not understand why formula : VL= -Ldi/dt is incorrect. Can you explain to me more ? Thanks. \$\endgroup\$
    – geek225
    Sep 27, 2020 at 14:02
  • \$\begingroup\$ When you are applying a voltage to an inductor and looking at the rate of change of current, the negative sign is inappropriate i.e. it is positive. When you are considering the back emf inside the inductor, you use a negative sign. Just think about it a minute: If you apply 1 volt across an inductor, you would expect current to rise positively and not rise negatively. \$\endgroup\$
    – Andy aka
    Sep 27, 2020 at 14:04
  • \$\begingroup\$ Andy aka: So which circuit I have to use to simulate the beahviour of the inductor in DC circuit ? Thanks for your support. \$\endgroup\$
    – geek225
    Sep 27, 2020 at 14:07
  • \$\begingroup\$ Your lower circuit is correct in that when you open the switch the inductor's top terminal produces a negative voltage. Your upper circuit is faulty. \$\endgroup\$
    – Andy aka
    Sep 27, 2020 at 14:10
  • \$\begingroup\$ Andy aka : Same thing when I close the switch S1 (I have add a resistor to simulate the charging phase for the inductor). The voltage through the inductor is positive not negative (VL= -Ldi/dt) \$\endgroup\$
    – geek225
    Sep 27, 2020 at 14:19

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