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?An inductor of 10H has 1omh internal resistance.
It could have an iron core and be much smaller than that?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.
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.An inductor of 10H has 1omh internal resistance. What is the current through the inductor after 10V is applied for 1 second?
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 will be much larger in size than this:
Do you have any calculus background?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?
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.The resistance is not in series with the inductor and...
Hi there,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.
by Steve Arar
by Jake Hertz
by Steve Arar