Q:A constant voltage of 12.00 V has been observed over a certain time interval across a 2.60 H inductor. The current through the inductor, measured as 3.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 7.00 A at the end of the time interval. How long was this time interval?

Given v12.00V, L2.60H, Δi4.00A The voltage across an inductor is given by: v L*di/dt v L*Δi/Δt Δt (L/v)Δi 0.87s

Q:Can anyone help me with this problem? I thought that the answers are zero for the inductor and the resistor. Thanks!!!!!!!!!!!!!An AC source operating at 59 Hz with a maximum voltage of 170 V is connected in series with a resistor (R 1.2 kilo-ohms) and an inductor (L 3.8 H). ΔVR, max 110 V and ΔVL, max 129 V. The max current is 0.092A. When the current is zero What are the potential differences across the resistor, inductor and AC source?

For a correct working of this Google, Inductive Reactance, and plug in your values.

Q:An inductor has an inductance of 0.075 H. The voltage across this inductor is 64 V and has a frequency of 660 Hz. What is the current in the inductor?

Ohm's law still applies only this time you are working with inductive reactance instead of resistance. XL 2 pi f L V I XL I V / XL f 660 Hz; L 0.075 H; V 64 V Plug and chug!!

Q:A 12.6-V battery is in series with a resistance of 0.330 Ω and an inductor. I got part a right, but I can't seem to figure out the rest. help please. a) After a long time, what is the current in the circuit?I(tinfinity) 38.1 A(b) What is the current after one time constant?I(tτ) AHow is the time constant defined for an RL series circuit? How does the current change as a function of time in such a circuit (c) What's the voltage drop across the inductor at this time?ΔVL VUse Kirchoff's Loop rule to describe the relative sizes of the potential differences across the battery, inductor, and resistor.(d) Find the inductance if the time constant is 0.110 s.L H

b) Current is given by i (E/R)*(1 - e^(-t/tL)) where tL is the inductive time constant. If t tL then i (E/R)*(1 - e^(-1)) You can crank that out. In an RL series circuit, tL L/R c) If the current is i, the voltage across it is i*R. The voltage across the circuit is E. So the voltage across the inductor is E - i*R. d) tL L/R 0.110 s Solve for L. The L H threw me at first -- I think it's telling you that the units are henries.

Q:Can anyone tell the different types of inductors.

I'd like to add, that no-one uses inductors if at all possible to design them out lossy, costly, imprecise (too much resistance for a given size) Filters use R and C's (and opamps) or DSP in digital In power supplies, frequencies are moving into the 1MHz area so that they can use flat coils made into the pcb You can still buy inductors, basically two types tiny minature inductors with relatively high resistance for tuning and old fashioned filters or power inductors for switching regulators Powetr factor control or control of harmonics in power circuits or ballast (as in lighting)

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.

An easy way to remember the answer is to apply DC (frequency 0) and see what happens: In inductor is simply a coil of wire. If DC is applied, it will be nearly a short circuit, and very high current will flow. As the frequency increases, the current will get lower. A capacitor is simply 2 parallel plates that don't touch, separated by a dielectric. If DC is applied, you will have an open circuit and no current will flow. As the frequency increases, the current will also increase.

Q:A variable inductor with negligible resistance is connected to an ac voltage source. By what factor does the current in the inductor decrease if the inductance is increased by a factor of 6.0 and the driving frequency is increased by a factor of 8.0?

X wL where X is the inductor's Reactance and w is the frequency in rads/sec the reactance will increase by 6 x 8 48 so the current will decrease by the same factor

Q:Calculate the inductance of an air-core coil with the following specifications: length 20 cm, #of turns 200 and diameter of core 2 cm.Could you show your work if you know how to do it? Thanks :)

See: okorder

Q:Help please!

resistors and inductors in series add up R12R1+R2 resistors and inductors in parallel dont: 1/R121/R1 + 1/R2 or R121/[(1/R1) + (1/R2)] or R12R1*R2/(R1+R2) remember this formula, it will be time saver on tests (even if it only for two elements) for capacitors it is the opposite: parallel capacitors add up C12C1+C2 series capacitors dont: C12C1*C2/(C1+C2) a) Inductors - output is open circuit (nothing connected to it). in this case D and B are in, series DBD+B. this is in parallel with A, so ADB AD*B/(AD+B) this ADB is then in series with C so we get ABCD ADB+C using given numbers we get Ldb4+26H La1H Ladb6*1/(6+1)6/7H0.857H Lc3H Labcd3+0.8573.857H now the output is shorted and B and C are in parallel: LbcLb*Lc/(Lb+Lc)2*3/(2+3)6/5H1.2H this is then in series with A so LabcLa+Lbc1H+1.2H2.2H finally this is in parallel with D so LabcdLabc*Ld/(Labc+Ld)2.2*4/(2.2+4) b) capacitors, first output is open: you get Cb in series with Cd. this is then in parallel with Ca. finally Cabd is in sereis with Cc. numerical results will match second part of a. then output X-Y is shorted and you get Cb and Cc in parallel CbcCb+Cc. This is in series with Ca CabcCa*Cbc/(Ca+Cbc) finally this is in parallel with Cd. CabcdCd+Cabc and results must match first part of a.

Q:An inductor is connected to a sinusoidal voltage with an amplitude of 120v. A peak current 3.0A appears in the inductor.a) What is the maximum current if the frequency of the applied voltage is doubled?b) What is the inductive reactance at each of the two frequencies?I started it but can't calculate the frequency or know what formal to use to get max Currentthanks Jack

The impedance (or reactance) of an inductor is: Z jωL And this tells you the ratio of the voltage to current: Z V/I So if you re-arrange this for I, you get: I V / Z V / (jωL) Now you can see that if you double the frequency (which is the same as doubling ω) that the current will be halved. Part b should be straightforward using the equations above, you don't need to find the frequency (and actually, you can't in this case).