Q:If the current through the inductor (L 34.8 mH) at 448 Hz has an amplitude of 55 mA, what is the amplitude of the AC voltage?What is the rms value of the AC voltage?

Inductive Reactance X? 2πfL 98 ohms E I*X 55 mA x 98 ohms 5.39 volts The problem is, what do you mean by amplitude ? Voltages are measured in peak, peak-peak, or RMS. If none are stated, RMS is assumed. So lacking a meaning for amplitude, I have to assume they are in RMS current and voltage. So the RMS answer is 5.39 volts RMS. .

Q:I understand that power is that rate at which work is done and that because of this the power is equal to d/dt (1/2Li^2). I also understand that the power is also equal to Li di/dt since Ldi/dt is v and v*i is power. I understand that since the power is equal to both of these equations that they are equal to each other. The part that I don't get is mathematically how to get from one to the other.

An ideal inductor consumes no power. However an inductor stores energy E integral of (v)(i)dt, and since v Ldi/dt E integral of (Ldi/dt)(i)dt integral of L(i)di (1/2)L(i)^2 This energy is stored in the magnetic field.

Q:A resistor, capacitor, and inductor are connected in series across an AC generator. Which of the following statements is true? a. All the power is lost in the capacitor.b. All the power is lost in the resistor.c. Power is lost in all three elements.d. All the power is lost in the inductor.

If these are the only choices, C is the most correct answer. This is due to an Electrical phenomenon called Impedance. Impedance in AC Circuits because of interaction between Resistors, Capacitors and Inductors is similar to resistance in DC Circuits. Whenever you have all three (resistance, capacitance, inductance) in an AC Circuit you will have currents due to Impedance and therefore all elements cause power to be lost.

Q:Hello! I am a college student who wants to get extra credit in my Electronics class (I need it desperately).I heard that in some circuits you can replace a capacitor with an inductor, so I wanted to do a experiement:-Build a standard common emmitter circuit with a capacitor-Build another common emmitter circuit with a equivalent inductor-Compare and contrast the responseI did a google search on how to replace a inductor with a capacitor in a circuit, but to no avail. I haven't found anything.Do any of you brilliant engineers know about this circuit technique? Can you direct me to more information on how to do this?Thanks!

I believe that the substitution is real, but not literal - for example in op amp filters a feedback inductor is the same as a feed in capacitor, and so on. Perhaps at a certain frequency a capacitor will look the same as an inductor in a circuit, but it seems impossible to for a capacitor to replace an inductor across all frequencies. Then again, I'm still an undergrad, and perhaps in something like a dc oscillator, it could be done.

Q:Help! I am having trouble trying to compute the inductor current changes in a parallel R (1Kohms), and L (1mH). They are being driven by a current source of 1A. How do I verify that the current flows in the inductor changes by 63% of its initial value in one time constant.Also, How to,1. Analytical derivation of voltage and current across L2. How the specific values of R and L affect the transient behavior of voltage and current across the inductor3. Compare the analytical solution with the numeric results (I used LT SPICE to successfully obtain the voltage and current waveform across R and L)any help on any of the questions, particularly, the 63% question and Q1 and Q2 would be greatly appreciated. Thanks.

For this circuit, the current of the source is equal to the current through the resistor plus the inductor current. Therefore Ise/R+(1/L)integral(e dt), where e is the voltage across the current source. I'm not into solving equations like this any more, but the solution will be something like eI R e^-t/T where TL/R. e^-1.368 and 1-.36863.2%.

Q:here's the text which contains it,Referring to figure, it can be seen that no power is disspated in a pure inductor. In the first quarter of cycle, both V and I are positive so the power is positive, which means that energy is supplied to the inductor. In the second quarter, V is positive but I is negative. Now power is negative which implies that the energy is returned by the inductor(The figure looks something like this) ::

Yes, you're correct. Your picture shows the correct phase relationship between voltage and current in an inductor, and in that case your description is correct.

Q:If you double the current through an inductor, what happens to the inductance of the inductor?It is doubled.It remains the same.It is halved.

Inductance of an inductor does not depend on either voltage or current. It is a constant quantity just as resistance! The only way to decrease or increase inductance is using parallel and series combinations of inductance respectively

Q:A 265 mH inductor whose windings have a resistance of 6.70 Ω is connected across a 10.5 V battery with an internal resistance of 3.35 Ω. How much energy (in J) is stored in the inductor?Thanks for the help!

PI*R^2 (electric power wiki)

Q:A student wants to figure out if an unlabeled item is a capacitor or inductor. He applies a voltage with diferent frequencies and find that as the frequency goes up, the current through the item goes down. Is this a capcitor or an inductor? Why? Thanks for the help! Please explain in detail.

inductor. Inductive reactance is proportional to frequency, so it goes up as frequency goes up. Since Current is inversely proportional to reactance, as frequency goes up, current goes down. A cap would be the opposite. .

Q:I have built a basic telegraph to test the inductor on my scope to verify what I have been taught. I wrapped my coil around a drill bit from bottom to top in a clockwise direction with the negative lead on the bottom, positive on top. I connect the battery positive directly to the top of the coil ( positive) and the negative battery to the key switch, from the switch to the lower end of my coil. I expected as the switch was closed and current was increasing that the coil will attempt to keep it the same ( zero) by reducing the voltage by an equal amount while it is building its magnetic field. However, I am seeing a positive spike rather than a negative and I do not know why. I have connected my scope negative to the bottom of the coil ( negative) and positive to the top of the coil ( positive). Like i said, I expected a negative spike in voltage, but i am seeing a positive spike which reduces to the positive battery voltage. Please Explain ?

It is a bit difficult to visualize the experimental setup, but what I would expect that you see is this: when you close the key, the voltage across the coil should rise rapidly to the battery voltage, as the current increases, limited only by the coil and battery resistance. The scope will measure the sum of the voltage due to the coil inductance and the coil resistance, the voltage due to inductance will quickly deline to zero, and the voltage will approach the battery voltage asymptotically. Life gets much more interesting when you open the key: the current flowing through the inductor attempts to flow through the opening key contacts and will produce a spark, and you will see a large voltage spike on your oscilloscope. Figure out what the polarity should be, and see if that is what you get.