Voltage in an inductor - DC circuit

Question

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:

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:

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

Is there any explanation for this issue?

Answer 1

Just to show:

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?

Answer 2

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.