# RL circuit

Discussion in 'Homework Help' started by TsAmE, Oct 28, 2010.

1. ### TsAmE Thread Starter Member

Apr 19, 2010
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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

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2. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Last edited: Oct 28, 2010

Feb 17, 2009
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4. ### TsAmE Thread Starter Member

Apr 19, 2010
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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.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.

6. ### TsAmE Thread Starter Member

Apr 19, 2010
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Howcome? Shouldnt it not have the emf across it since there is no current?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Becaues at T = 0 coil act like open circuit.

Last edited: Oct 29, 2010
8. ### JoeJester AAC Fanatic!

Apr 26, 2005
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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.

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
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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.

10. ### TsAmE Thread Starter Member

Apr 19, 2010
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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.

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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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.

12. ### TsAmE Thread Starter Member

Apr 19, 2010
72
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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?

13. ### The Electrician AAC Fanatic!

Oct 9, 2007
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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).