I was told by my professor that the current in a purely LR Circuit with at initial state (JUST after closing the switch) is zero. Can you please help me understand why it that so?
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asked Aug 17, 2014 at 16:23
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$\begingroup$ After closing what switch? If you start with an inductor and a resistor connected only at one terminal, and then connect the other one, the current will obviously start out at zero... and stay at zero forever! $\endgroup$– leftaroundaboutCommented Aug 18, 2014 at 0:40
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4 Answers
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Intuitively, inductors create back EMFs that "resist" changes in current. Anytime there is an inductor in a circuit, it will resist such changes. Mathematically, inductors force the current in a circuit to be continuous. Let's consider an LR circuit where the inductor and resistor are both in series. Regardless of the initial voltages or current, we can calculate how such quantities will involve in time. The voltage across the inductor is given by
$$ V_L = L\frac{dI}{dt}, $$
where $I$ is the current running through the entire circuit. Similarly, Ohm's law gives us the voltage across the resistor:
$$ V_R = IR. $$
According to Kirchoff's laws, conservation of energy dictates that the total voltage in a closed loop (without any external sources of EMF) is zero. Thus,
$$ V_{tot} = V_L+V_R = L\frac{dI}{dt}+IR = 0. $$
We now have a mathematical reason why the current must be continuous. The differential equation requires that the time derivative of $I$ exist, implying that $I(t)$ itself must be at least a $C^1$ function (continuous and differentiable). If we wish, we can stop here and conclude that if we know $I(t)$ for some time $t$, then $I(t+\Delta t) \approx I(t)$ for some small change in time $\Delta t$. Thus, if the initial current is zero, some small time after the switch is closed it will still be approximately zero.
Small note: The voltage across an inductor is equal and opposite to the induced EMF in the circuit. Thus, it is a matter of semantics if the voltage is across the inductor or if an opposite voltage is induced in the loop.
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$\begingroup$ $C^1$ does not mean "continuous and differentiable". $C^1$ means that the derivative is continuous (and hence, it is differentiable and continuous). Also, you say $I$ must be $C^1$ but you probably wanted to mean $I$ is continuous and differentiable. I'll take the counterexample of $Q$ in an $RC$ circuit: it sure is continuous and differentiable, but its derivative ($I(t)$) is not continuous. $\endgroup$– niobiumCommented Oct 31 at 14:46
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The voltage drop across an inductor is proportional to the change in current, or V = L*dI/dt. When the switch is closed, the circuit is completed and a current starts to increase dramatically. This then causes a time increasing flux of magnetic field in the inductor. According to Lenz's Law, an opposite current will be induced in the inductor to oppose the time changing flux. Now we have two currents in the circuit. You can think of them as moving in opposite directions and canceling out. This only applies the exact instant you close the switch, as the current due to the voltage source will overcome this opposition and eventually approach a steady state solution.
It is also good to work out the equation from Kirchoff's law for the simple circuit which I'll start here. Try solving for I(t):
V - L*dI/dt - I*R = 0
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$\begingroup$ Possibly this is the right answer (I can't tell because I don't really understand what the OP is asking), but at any rate it's imprecise wording. You don't induce currents, you induce EMFs (voltage). What Lenz's law states is that this voltage is such that the current it causes through a resistor or capacitor is such as to inhibit the change in magnetic flux. Cancelling currents in opposite directions... sure you can somehow interpret stuff like that, but I doubt it's very useful here. $\endgroup$ Commented Aug 18, 2014 at 0:47
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$\begingroup$ So would it be better to say that at the instant the switch is turned on, the induced EMF is equal to and opposite in sign as the battery? The Kirchoff loop would then be Vbattery - Vemf = 0. Now there is no voltage drop across the resistor, and no current in the circuit. $\endgroup$– hopCommented Aug 18, 2014 at 12:06
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In problems of this type, it's required to determine the initial conditions. Often you might have to work them out from analyzing a modified circuit (for example, with a switch in a different position). But it's also possible the initial conditions are simply given as part of the problem statement, and that seems to be what was done here.
If your professor said, "I have a ball on top of a 1 m incline, with initial velocity 0, and it begins rolling...", you wouldn't need to ask how the ball got to the top of the hill to determine how far it rolls after that. Similarly, if a problem takes as a premise that the current through an inductor is initially some value, there's no need to know how it got to be that way in order to solve the problem for $t\gt{}0$.
answered Aug 18, 2014 at 1:38
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$\begingroup$ Your example was good, my question is different...it was told that the current IS zero! So how can u prove it? It is NOT to solve a problem so please don't compare with that $\endgroup$ Commented Aug 21, 2014 at 16:44
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$\begingroup$ My point is, it's not subject to proof. Your professor can just say that it's so and it won't affect the rest of the analysis. How would you prove a ball is 1 meter above the ground? If your professor said, "Imagine you have a laser pointed at a screen..." would you want to know how to prove you have a laser? $\endgroup$ Commented Aug 21, 2014 at 17:39
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At $t=0$ in an LR circuit, the current is zero because at $t=0$ the inductor opposes the current. It becomes itself like an opposing battery.
