Q:I think I understand this more now after looking at my notes a little more. I need to turn them into polar form to make the division easier. Then I can put it back into the original form after.

The magnitude of the circuit impedance is equal to the inductive reactance provided by the inductance: L X ωL, but ω 200 rad/s so that X 200(0.25) 50 ohm while V 100 volts the phasor current for the circuit: I V/Z where V is the voltage phasor V 100 angle 0? and the complex impedance: Z jX j50 Therefore: I 100/(jX) -j2 2 angle -90? The phasor current lags the phasor voltage by 90?.

Q:at which the maximum current flows. Explain briefly.

Have u lost ur mind???,,,,I do know your feeble attempt .but grow up and contact me if you need answers.

Q:i need kno how do to measure inductors in a circuit. i have one reisstor, one capatior, and one inductor. i had to find out what the componts where, and their values. i found the reistor and the capaitor. i can put them in series or in parrellel, and i only have a multi meter, and a ollciscope, and i have a varaible genator, that can give me a variaible frequency. i forgot how to measure the inductor, i kno after a yeart i should know, but we had a few weeks off lol.please only help if u kno what u are doing.

Connect a known source of alternating voltage, a resistor R ohm , an ammeter and the inductor in series. Measure the current. Let VVoltage and I the current. Then the impedance is V/Z. Impedance of R and L in series is Z Square root of (R^2+ Xl^2) where Xl 2 pi x f x L where L inductance in H and f frequency of voltage in Hz. Knowing pi, f, V, I and R, L is calculated.

Q:how does the energy gets stored in the inductor if we are connecting it to an rlc circuit either in parallel or in seriesi also wanted to ask about the phenomena of resonance ocuring in this circiut in details,i know that when the inductor has potential energy the capacitor doesnt have and vice versa,thats why its called electrical resonance,but i want someone to tell me the reason of this phenomena,an athountic answer please.thank u buddy

1kWh3.6*10^6 J SO 3kWh3*3.6*10^610.8*10^6 J Inductors shop the potential in variety of magnetic container. potential saved by using inductor0.5*L*I^2 10.8*10^60.5*L*I^2 21.6*10^6L*I^2 L21.6*10^6/(9*10^4) L2.4*10^2240 H (Unit of inductance is Henry)

Q:And the inductor is in series with a resistor and the battery was connected for a long time

After a long time an inductor acts like a short circuit, meaning all the current flows through the element, and there is no effective resistance. Therefore, the voltage across it would be V 0, as there is no voltage across a short circuit.

Q:a) oscillatesb) makes x-raysc) is uselessd) shorts out one or bothe) stops all current

It will oscillate and form a series filter.

Q:If the current through the inductor drops from 110mA to 50mA in 13useconds? u is micro I think?Any ideas?Thanks

The back emf generated by the collapsing magnetic field is given by . ? (-) L.di/dt L self inductance (H) di/dt rate of change of current (A/s) di/dt (110-50)mA / 13μs 60^-3A / 13^-6s 4.62^3 A/s ? (-) 12.0^-3H x 4.62^3A/s . . ?? (-) 55.40 V (- sign indicating a back emf)

Q:An 8.0 mH inductor and a 2.0 ohm resistor are wired in series an ideal battary. A switch in the circuit is closed at a time 0 sec. at which time current is zero the current reaches half its final value at a time of ofThe answer is 2.8 ms, How?????

Let Vo be the voltage across the battery and VL be the voltage across the inductor then for your circuit: VL Vo e-(tR/L) If you divide both sides by R you get the current relationship instead (Ohms law) VL/R Vo/R e-(tR/L) this equals:- I Io e-(tR/L) or you can simply start with this line instead. for the current to reach half its value I/Io 1/2 I/Io e-(tR/L) 1/2 e-(tR/L) Now, to get rid of the exponential part we take NATURAL logs (ln on your calculator) of both sides giving:- ln(1/2) -(tR/L) this gives t as:- t-L/R ln(1/2) t -[(8x10^-3)/2 x (-0.6931) t 0.00277 t 2.8 ms

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 :)

L 0.001 N?r? / (228r + 254l) where L is the inductance in henrys, r is the coil radius in metres, l is the coil length in metres (0.8r) and N is the number of turns. This formula applies at 'low' frequencies. At frequencies high enough for skin effect to occur a correction of up to about -2% is made. Old school version in Imperal measurement is La?n?/(9a+10l) a and l are in inches/

Q:A homework question for my Electromagnetic Fields and Waves course asks, How would you build a 0.1H inductor? I am extremely stuck! I have researched and tried to figure this out but I am having no luck. My classmates are having trouble as well. Please help!

There okorder