In the circuit shown the battery and the inductor have negligible internal resistance and there is no current in the circuit.

Determine reading on each of the meters the instant after the switch is closed.

Attempt:

I = V/R = 25/15 = 1.67A

V1 = IR = 1.67 x 15 = 25.05V

V2 = 25 - 25.05 = -0.05V

but the correct answer was:

V1 = 0V

V2 = 25V

I am not sure why

 

The question refers to the instant the switch is closed.

Read http://www.allaboutcircuits.com/vol_1/chpt_16/4.html

 

Read this
http://www.crovel.svedra.com/index.php?option=com_content&task=view&id=84&Itemid=49

 

I read it, but I am still a bit confused. I tried using the formula I = I0 e^(-t/ tau), but obviously at t = 0 (the instant the switch is closedd) I = I0, which didnt really help. I dont understand why all the voltage is across the inductor, and some isnt across the resistor.

 

Actually the current will be given by

\(i(t)=\frac{E}{R}(1-e^{-\frac{t}{\tau}})\)

So the current at t=0 is zero. Meaning no voltage drop across the resistor. The inductor has it all.

 

Howcome? Shouldnt it not have the emf across it since there is no current?

 

Becaues at T = 0 coil act like open circuit.

Why is there any voltage even present across the inductor? We always accept a voltage across a resistor without argument because we know Ohm’s law (V = I × R) all too well. But an inductor has (almost) no resistance it is basically just a length of solid conducting copper wire (wound on a certain core). So how does it manage to “hold-off” any voltage across it?
In fact, we are comfortable about the fact that a capacitor can hold voltage across it. But for the inductor, we are not very clear!
A mysterious electric field somewhere inside the inductor! Where did that come from?
It turns out, that according to Lenz and/or Faraday, the current takes time to build up in an inductor only because of ‘induced voltage.’ This voltage, by definition, opposes any external effort to change the existing flux (or current) in an inductor. So if the current is fixed, yes, there is no voltage present across the inductor, it then behaves just as a piece of conducting wire. But the moment we try to change the current, we get an induced voltage across it. By definition, the voltage measured across an inductor at any moment (whether the switch is open or closed) is the ‘induced voltage.’
So let us now try to figure out exactly how the induced voltage behaves when the switch is closed. Looking at the inductor charging phase, the inductor current is initially zero. Thereafter, by closing the switch, we are attempting to cause a sudden change in the current. The induced voltage now steps in to try to keep the current down to its initial value(zero).
So we apply ‘Kirchhoff’s voltage law’ to the closed loop in question. Therefore, at the moment the switch closes, the induced voltage must be exactly equal to the applied voltage, since the voltage drop across the series resistance R is initially zero (by Ohm’s law).
As time progresses, we can think intuitively in terms of the applied voltage “winning.” This causes the current to rise up progressively. But that also causes the voltage drop across R to increase, and so the induced voltage must fall by the same amount (to remain faithful to Kirchhoff’s voltage law).
That tells us exactly what the induced voltage (voltage across inductor) is during the entire switch-closed phase.
Why does the applied voltage “win”? For a moment, let’s suppose it didn’t. That would mean the applied voltage and the induced voltage have managed to completely counter-balance each other — and the current would then remain at zero. However, that cannot be, because zero rate of change in current implies no induced voltage either! In other words, the very existence of induced voltage depends on the fact that current changes, and it must change.
We also observe rather thankfully, that all the laws of nature bear each other out. There is no contradiction whichever way we look at the situation. For example, even though the current in the inductor is subsequently higher, its rate of change is less, and therefore, so is the induced voltage (on the basis of Faraday’s/Lenz’s law). And this “allows” for the additional drop appearing across the resistor, as per Kirchhoff’s voltage law!

 

You can work the formula close to t=0 and see the potential across the coil.

Suppose t = one millionth of tau. The voltage across the coil would be within a few hundred microvolts of the source voltage. If that doesn't clear things up, recompute for one-billionth of tau. You could work that down to tau to see the initial voltage.

 

I just can't resist giving a smart-a** answer.

The OP's post didn't ask for the voltage across the resistor and across the inductor. It asked for the reading on the meters the "instant" after the switch is closed. I take "instant" to be an infinitesimal time period.

Since meters can't respond in an infinitesimal time, the answer is that both meters read the same thing they read the "instant" before the switch closed, presumably zero.

 

I just can't resist giving a smart-a** answer.

The OP's post didn't ask for the voltage across the resistor and across the inductor. It asked for the reading on the meters the "instant" after the switch is closed. I take "instant" to be an infinitesimal time period.

Since meters can't respond in an infinitesimal time, the answer is that both meters read the same thing they read the "instant" before the switch closed, presumably zero.

The answers in my notes said V1 = 0V and V2 = 25V.

Also why does the inductor act like an open circuit when there is no current flowing? I mean it is still a wire with no gap.

 

The answers in my notes said V1 = 0V and V2 = 25V.

Also why does the inductor act like an open circuit when there is no current flowing? I mean it is still a wire with no gap.

Because the voltage across an inductor is not proportional to the current through the inductor; it's proportional to the rate of change of the current. The voltage is equal to L*di/dt.

The instant after the switch closes the current is still zero, but the rate of change of the current is not zero.

On the other hand, the voltage across a resistor is proportional to the current, but not the rate of change.

 

Im confused. How can the rate of change of current not be 0 if the current is 0? I mean since the current is 0 and isnt changing, then there shouldnt be a rate of change?

 

Im confused. How can the rate of change of current not be 0 if the current is 0? I mean since the current is 0 and isnt changing, then there shouldnt be a rate of change?

Have you taken calculus yet? It's perfectly possible for a variable quantity to be zero yet have a non-zero rate of change.

Consider a function y = f(x) = x

If you plot the linear function y = x, you will see an increasing ramp, a straight line rising with a non-zero slope. Calculus tells us that the slope is 1.

At the origin (x=0) the function has a value of zero, but the rate of change is not zero.

In your problem at t = 0+ (the instant after the switch closes), the current in the inductor is still zero, but it is increasing (has a positive rate of change).