In a LR circuit connected with battery what happens if $R=0$ assuming the switch is closed at $t=0$? Will emf be induced in the inductor? I am confused so please help me.
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2 Answers
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R is never zero, because of the internal resistance of the battery. Assuming both resistances to be zero does not lead to a physical solution, because the current cannot be infinite. You must have a however small, but a non-zero resistance in the circuit for a realistic case.
Then what if we use a superconductive capacitor as a battery and connect it to a superconductive coil? Then the charge of the capacitor will generate a current in the coil that will recharge the capacitor to the opposite polarity and after that the cycle will repeat. The energy of the capacitor will transfer to the coil and back periodically at the LC resonance frequency. The circuit will emit radio waves of this frequency and gradually lose power until all stored energy is emitted and the process stops.
answered Sep 28, 2017 at 19:03
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$\begingroup$ Suppose I have a circuit in which the induced emf causes current to flow into inductor.Do I have to assume that equivalent battery due to induced emf will have to have some resistance so as to make it a LR circuit and apply KVL? $\endgroup$ Commented Sep 28, 2017 at 19:13
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$\begingroup$ @user42172 Not necessarily. If the closed circuit that you are describing is superconductive (zero total R), then the current you induce in it will produce a magnetic field equal and the opposite to the inducing field. After that the energy that the circuit consumes from tbe inducing field will be equal to the energy that your circuit emits as radio waves. For example, a magnet positioned over a superconducting dish will levitate, because the current in the dish will create a field repulsing the magnet regardless of its orientation. $\endgroup$ Commented Sep 28, 2017 at 19:47
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In a LR circuit connected with battery what happens if R=0 assuming the switch is closed at t=0? Will emf be induced in the inductor?
Yes, there will be an emf and, in the context of ideal circuit theory, the voltage across the inductor is a step function ($0$ for $t \lt 0$, $V_{DC}$ for $t \gt 0$) while the current through the inductor is the integral of a step, i.e., a ramp.
That is, at $t = 0$, the inductor current is zero and then increases linearly with time without bound:
$$i_L = \frac{V_{DC}}{L}t,\quad t \ge 0$$
For a physical circuit, the inductor current cannot increase without bound and thus will be limited by, e.g., the winding resistance of the inductor and/or the short-circuit current of the voltage source.
answered Sep 28, 2017 at 21:48