An inductor of 10H has 1omh internal resistance. What is the current through the inductor after 10V is applied for 1 second?

 

It will be much larger in size than this: 10H close wound coil on a 36 inch cylindrical form 36 inches long using 378 feet of wire.
It would require 4,012 turns of copper enameled wire. When you try to get below approx 35 Ohms the self capacitance takes over the resonance drops out of sight. so it would take a super conductor at some astronomical cost. The effort is extremely unpractical and foolish. approximately 35 Ohms minimal resistance. The 378 feet wire in a straight line converted to meters would give you the resonance needed so I think a circular inductor will be huge something like what Tesla used Colorado.

 

An inductor of 10H has 1omh internal resistance.

When the voltage is switched on, the inductor current is zero because an inductor resists any change in current (whereas a capacitor resists any change in voltage). Here the inductor current follows a rising exponential curve and reaches the maximum current allowed by the resistance after ~5 time constants. So what is the time constant of this circuit?

 

It will be much larger in size than this: 10H close wound coil on a 36 inch cylindrical form 36 inches long using 378 feet of wire.
It would require 4,012 turns of copper enameled wire. When you try to get below approx 35 Ohms the self capacitance takes over the resonance drops out of sight. so it would take a super conductor at some astronomical cost. The effort is extremely unpractical and foolish. approximately 35 Ohms minimal resistance. The 378 feet wire in a straight line converted to meters would give you the resonance needed so I think a circular inductor will be huge something like what Tesla used Colorado.

It could have an iron core and be much smaller than that?

 

Welcome to AAC!
Since this is Homework, please post you best effort at solving the problem, so that guidance can be given where necessary.

 

An inductor of 10H has 1omh internal resistance. What is the current through the inductor after 10V is applied for 1 second?

We are not going to simply hand you the answer to this problem. We don't do that here on AAC; we'll help you figure out how to solve it yourself, but we're not going to solve it for you. Show us what you've got so far, and we'll try to steer you in the right direction.

For starters, you might try Googling on the phrase, "RL time constant" and see what comes up. This page looks like it might be helpful.

 

It will be much larger in size than this:

How is that pertinent to the TS's question?

 

It will be much larger in size than this:

Much larger in size than what??

It's a math problem. Don't make it into something it's not.

 

Two equations for the current through the inductor can be the V/L=∆I/∆T voltage divided by the inductance equals the change in current divided by the change in time and L/R for the time constant with about 3-5 time constants allowing the inductor to carry the current of V/R.

The resistance is not in series with the inductor and as the current increases the voltage remains across the inductor leaning towards equation 1. However, I'm thinking the maximum current is V/R leaning towards equation 2. Could the current increase linearly as in equation 1 and flat line at V/R?

 

Two equations for the current through the inductor can be the V/L=∆I/∆T voltage divided by the inductance equals the change in current divided by the change in time and L/R for the time constant with about 3-5 time constants allowing the inductor to carry the current of V/R.

The resistance is not in series with the inductor and as the current increases the voltage remains across the inductor leaning towards equation 1. However, I'm thinking the maximum current is V/R leaning towards equation 2. Could the current increase linearly as in equation 1 and flat line at V/R?

Do you have any calculus background?

If not, then perhaps this will be helpful.

If you plot the inductor current versus time when you apply voltage across an inductor having resistance you will see that it is NOT linear. It starts out changing fast (i.e., the fastest rate it is going to have) and then the rate steadily decreases until eventually it stops increasing and it then stays at it's final value.

There is a parameter known as the time constant, which for an inductor in series with a resistor (which is how real inductors are generally modeled) the time constant is L/R. The initial slope of the change in current is such that IF it stayed at that slope from beginning to end the inductor would be at its final current after one time constant. For times much shorter than a time constant, you can often approximate the change as being linear. At 10% of one time constant, the error is about 5% and at 20% of a time constant it is about 10%.

 

The resistance is not in series with the inductor and...

Not true. The internal resistance of the inductor (i.e., the resistance of the copper windings) IS effectively in series with the inductance. This needs to be treated as a series RL circuit.

 

Since an ordinary 10 Henry close wound inductor has too much self capacitance the mathematics results in no solution.
So, What does size have to do with it ? removing enough self capacitance so that the expression having 1 Ohm is possible.
Size is related to wire size and wire size will improve the chances of developing an induction of 10H with resistance of 1 ohm.
Given : An inductor of 10H has 1omh internal resistance.
The question is: What is the current through the inductor after 10V is applied for 1 second?

If you assign a value of 1 Ohm to a 10 Henry inductor and set a pulse generator to single shot square wave 1 Hz 10V DC
I think that would be a measurement of current during that 1 second pulse.
If the pulse was repeated every second the current measurement would be set up different.
It would make sense that an instructor would compare both methods and the setups in order to deduct the result.
When a question of this nature lacks a procedure it probably is a what if the inductor had only 1 ohm internal resistance.
Now that matches some physics course asking for a mathematical deduction. What proceded this is unknown.

 

Just like introductory physics problems are unrealistic because they lack this thing called friction, so too are many introductory problems in many fields. Learn to crawl in an ideal world first.

 

It will be much larger in size than this: 10H close wound coil on a 36 inch cylindrical form 36 inches long using 378 feet of wire.
It would require 4,012 turns of copper enameled wire. When you try to get below approx 35 Ohms the self capacitance takes over the resonance drops out of sight. so it would take a super conductor at some astronomical cost. The effort is extremely unpractical and foolish. approximately 35 Ohms minimal resistance. The 378 feet wire in a straight line converted to meters would give you the resonance needed so I think a circular inductor will be huge something like what Tesla used Colorado.

Hi there,

That is a very interesting viewpoint from which to look at this problem, but that really illustrates the huge difference between theory and practice that often persists in problems like this.
If we look at the possible answers in theory vs practice we end up with only one of these two:
1. The inductor is impossible to build so there is no answer.
2. The current is very approximately 0.9 amps (note this is VERY approximate so as not to give away the real answer, and the real answer will be significantly different than this).

In most of these kinds of problems we would have to answer #2 even though it is difficult to build an inductor such as 10H with only 1 Ohm series resistance.

It may not be impossible to build that inductor either given some environmental and proactical constraints. We can probably find a core material that has a permeability of 10 thousand and so we'd have to build a coil (without a core) of around 1mH, which is much smaller than 10H but still not easy to make given the 1 Ohm requirement. But take the temperature down below freezing and the problem starts to become possible as the resistance decreases. You could use the resistance and temperature coefficient of copper or silver to figure out if this will work. So the coil plus the high perm core material plus the temperature drop may give us the required inductor. The capacitance of course will also come into play making the problem significantly more difficult to answer but they dont even give a requirement on the capacitance so i think we are ok there too.

But of course the problem is more theoretical than as practical as you have implied it should be so i dont think we have anything to worry about if we just go with an answer such as #2 above.

 

Mathematics can be confidential in some areas like exotic core materials.
The Chinese were handed secrets, the scrutiny will be directed to those officials.
Blame game here.