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this circuit[1]

Please help me solve this problem. I have one question

But before that, this is my step by step approach of the problem:

$\mathtt{ i_{max} = 200/20 = 10} A $ so $ 60$ % $ = 6 A $

Now $\mathtt{emf = 200 - 6 . 20 = 80 }V$

the question is >>> this voltage "$80V$"

voltage across in the inductor or what?

because the inductor has a resistance.

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  • $\begingroup$ Please use mathjax to format mathematical expressions. To learn more about mathjax, please read MathJax basic tutorial and quick reference. $\endgroup$ – Yashas Mar 3 '17 at 16:41
  • $\begingroup$ @YashasSamaga Are you looking for a batch or something ( I don't know whether 'this comment's leads to a badge. If not Sorry!) .But I have seen this comment of yours for the 5th question today. The guy has only 6 reputation. If he is answered on Stack he will learn it. But at the first only if his question is sort of treated like this he may leave the site . There isn't much formatting or Equations required here. The guy did his best. Wouldn't it be better if some of his questions are answered and then he be asked to learn Math Jax . Atleast I feel that way.. opportunity .....if not ignore $\endgroup$ – Shashaank Mar 3 '17 at 19:27
  • $\begingroup$ There is no badge for comments (except the Pundit badge which is given for 10 comments with 5 or more score) but the comments I make don't get upvoted that much. Moreover, I already have that badge. I edit the question, add the mathax code then inform the OP that this site supports mathjax. $\endgroup$ – Yashas Mar 3 '17 at 19:29
  • $\begingroup$ @YashasSamaga Ok if you do edit them also , that's great. I use the app ao it doesn't show who edited (neither does it allow to post a picture). I just felt that questions can be first answered. Great if you do ;) I thought that that's why ! $\endgroup$ – Shashaank Mar 3 '17 at 19:34
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I recommend you first consider the inductor to consist of a series connection of an ideal inductor and a 3 Ohm resistor. Now you can see that the maximum current you can develop in the system is $\rm{I_{max}=\frac{200 ~V}{20 ~Ohm} = 10~ A}$

60% of the max current is 6 A. When that current is flowing, the voltage drop across the resistor is 17x6 V. The remaining voltage is across the inductor. Of this, 3x6 is due to the resistivity of the inductor, and the rest is due to its reactance. So the total voltage across the inductor is the sum of the back emf (80 V), and the voltage due to its internal resistance (18 V) for a total of 98 V. and indeed, 17x6 = 102 V - that's where the rest of the voltage is.

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  • $\begingroup$ but 98 V is the total voltage between the inductor ? i mean this inductor have a resistance so> back emf will be divider on inductor and internal resistance or for inductor only or for internal resistance only $\endgroup$ – Ferris adam Mar 3 '17 at 17:18
  • $\begingroup$ i want VL When current equal 6A $\endgroup$ – Ferris adam Mar 3 '17 at 17:19
  • $\begingroup$ No it's not 98VDC. The 80V counter-emf potential is in the magnetic field or magnetic flux. The additional 80V of the inductor is due to the reactance (measured in ohms), not resistance. Reactance and resistance are the components of impedance. Look at the 80 as if it were 80VAC. $\endgroup$ – Misunderstood Mar 3 '17 at 21:41
  • $\begingroup$ @Misunderstood - I don't know if a comment was deleted, but I can't figure out what you are responding to. Neither my answer, nor the comments, mention DC anywhere. The instantaneous voltage across the inductor when the current is 6A is 98 V. $\endgroup$ – Floris Mar 3 '17 at 21:58
  • $\begingroup$ But no one has definitive evidence or explanation for this $\endgroup$ – Ferris adam Mar 6 '17 at 13:38
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The way the question has been written means that you have to make at least one assumption before before you can answer it.

The first thing to say is that if the current has not reached its maximum it must be changing.

What is not clear is whether or not the current started at a maximum and then the circuit was broken (look carefully at the diagram and there is a break in the circuit to the left of the positive terminal of the battery), or the initial current was zero and the current is building to to a maximum value.

So you either have a "discharging" of a LR circuit or a "charging" of a LR circuit both of which will give you exponential type relationships for the current and voltages.

To find these relationships you assume that the imperfect inductor is an ideal inductor with no resistance in series with a $3\; \Omega$ resistor which in turn is in series with a $17 \; \Omega$ resistor and a $200\;\rm V$ battery.

The voltage that you require is the voltage across the ideal inductor with no resistance in series with a $3\; \Omega$ resistor or the voltage of the battery minus the voltage across the $17\; \Omega$ resistor.

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  • $\begingroup$ this RL circuit with DC Source but.. this inductor have a resistor this resistor is in series with 17 okay so if we calculate the maximum current when i closed the switch it will be equal 10 A and this is 100% emf = vb emf = 200 i want know how i can calculate voltage i mean emf when the current is 6 A and the inductor have resistance ?! $\endgroup$ – Ferris adam Mar 3 '17 at 17:00
  • $\begingroup$ <img src="i.imgur.com/jtU6P2B.png" width="100" height="100"> $\endgroup$ – Ferris adam Mar 3 '17 at 17:07
  • $\begingroup$ @Ferrisadam What I have tried to explain to you is that when the current is changing there is a voltage across an ideal inductor. The way forward is to find a value for the maximum current which will not be changing so you can ignore the ideal inductor, find 60% of that current and use the very last relationship that I gave in my answer to find the required voltage. $\endgroup$ – Farcher Mar 3 '17 at 17:16
  • $\begingroup$ okay but i calculate the back emf is equal = 200-120=80 v now back emf will be divider on inductor and internal resistance or for inductor only or for internal resistance only $\endgroup$ – Ferris adam Mar 3 '17 at 17:22
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    $\begingroup$ 240−17I60% what do you mean by this value ? $\endgroup$ – Ferris adam Mar 3 '17 at 17:29

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