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Why is self induced emf much greater than the source voltage in a DC supply when the switch is opened?

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Notice that the induced e.m.f. is directly proportional to the rate of change of magnetic flux in a conductor. $\epsilon_0\propto \frac{d\phi}{dt}$ where the flux,$\phi$ is proportional to the current and hence the voltage of the DC source.

But the important point to note is that the induced e.m.f. depends not only on the magnitude of the voltage causing the induction but how quickly the surge in that voltage occurs. When you switch on a DC circuit the voltage reaches very quickly(a rate which is of-course more than suppose a DC voltage of 10V/1s which would result in an induced e.m.f equal to the max DC voltage but in reality it is much more than that as the time is much less than 1 second for the change to occur). Thus the induced e.m.f. is much higher than the max DC voltage.

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The induced emf $\mathcal E_{\rm L} = - L \frac {dI}{dt}$ so it depends on the rate of change of current in the circuit.
The time constant, a measure of the rate at which current changes in a circuit) of a series inductor $L$ and resistor $R$ is $\tau = \frac{L}{R}$ which means that for constant inductance $L$ the time constant depends on the resistance in the circuit.

When the switch is opened the resiatnce of the circuit rises dramatically as you now have in the circuit air between the two contacts of the switch.
This dramatic increase in resistance means that the time constant of the circuit is now very, very small.
In turn this means that the rate of current change in the circuit is very large which means that the induced emf is very large and therefore largesr than the voltage of the dc supply.
Indeed it can be so large that the air actually becomes ionised (a conductor) and an arc/spark appears across the contacts of the switch.

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Those two voltages play very different roles here and knowing one you cannot predict the other.

The voltage generated when the switch is opened is dependent on the inductance of the circuit and on the rate of the current change: $emf=L\frac {di}{dt}$. The rate of the current change, in turn, depends on the magnitude of the initial current and the characteristics of the switch.

The DC voltage in the circuit does affect the initial magnitude of the current (i.e., the magnitude before the switch is opened), but even a very small DC voltage can produce a large current, if the resistance is low, and, vise versa, a large DC voltage can produce a small current, if the resistance is high.

So, we may have a circuit with a $1V$ battery, $1ohm$ resistance, some inductance L and some switch, which will produce a greater emf than a circuit with the same inductance and the same switch, but with a $10V$ battery and $10Kohm$ resistance, because the initial current in the second circuit would be $1000$ times lower than in the first circuit.

Of course, if the second circuit also had a smaller inductance, the difference would be even more dramatic.

In summary, for the same DC, the range of possible self-induced emf voltages is almost unlimited.

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