Q:I understand that the switch gets incinerated. What exactly happens when you open the switch? I understand that the inductor opposes the change in current, but what happens when that current is forcibly taken away instantly? Please explain.Thanks.

Inductors store energy when they are first energized so they release that energy when the circuit is turned off. The result is that the amperage stays high for many milliseconds after the switch starts to open. NOTE: when you are dealing with electrical systems like this, there is no such thing as instantly. This increased amount of energy means the arc has more energy to dissipate before it extinguishes. That does not, necessarily, incinerate the switch. Switches are (or should be) designed to survive the arcing that they will encounter in service. The design life of a swtich can, of course, be controlled by the design (which also influences the cost). Almost every switch arcs. The only ones which don't operate at, like, less than 20 millivolts. An arc is a high temperature plasma. The plasma temperature is on the order of 10000 C, more than hot enough to melt any substance we know of. So temperature is part of it but it is really the arc energy that is most important. The material science (metallurgy) of arcing contacts is a rather specialized field. The details of what happens depend on many things including opening vs closing, AC vs DC, frequency, voltage, current, the circuit (resistive vs inductive), opening speed, atmosphere (or lack thereof), the number of cycles the switch has seen, the temperature of environment, among others. Electrical engineering deals with the electrical behavior during steady state and during transients (when the circuit is turned on or turned off). If you want to see the equations that describe some of the physics involved, check out Wikipedia for inductance or induction.

Q:A series circuit contains a resistor with R 24 ohms, an inductor with L 2 H, a capacitor with C 0.005 F, and a generator producing a voltage of E(t) 12 sin(10t). The initial charge is Q 0.001 C and the initial current is 0. Find the charge at time t.Q(t) ??

First thing to realize is that this circuit is in resonance. The characteristic frequency is equal to 1/sqrt(LC) 1/sqrt(2*.005) 10 This means the driving source won't provide net new charge to the capacitor and we can look at this in two pieces so for the output across the capacitor we have v-out 1/jωC/(R+jωL+1/(jωC))*v-in 1/(1-ω^2LC+jωRC) When we plug in values we get v-out1/(1-10^2*(2*0.005)+j10*24*.005) v-out1/(1-1+1.2j)*v-in -j/1.2*12sin(10t) 10sin(10t-π) so the change in Q across the capacitor from the driven source will be Q(t)-driven 0.005*10sin(10t-π) 0.0510sin(10t-π) C now we have to look at the initial conditions with a voltage of 0.2 V stored in the capacitor. We'll short out the driven source and solve for the decay in voltage through the resistor and inductor Because there is in essence a DC voltage stored on the capacitor and discharged through the resistor, the inductor has no role. The voltage discharge is 0.2exp(-t/RC)0.2exp(-8.33t). The charge is just this multiplied by C. or 0.2*0.005exp(-8.33t)0.001exp(-8.33t) the total will be Q(t)0.0510sin(10t-π)+0.001exp(-8.33t)

Q:for the current to reach 67% of its maximum value

The final value will be 24/146. The equation is [24/146]*[1-exp(Rt/L)]. So if answer is T, 0.67 1-exp(-RT/L). Find T

Q:A series LR circuit of a 2.0H inductor with negligible internal resistance, a 100 ohm resistor, a switch, and a 9.0V power source. After the switch is closed, What is the maximum power delivered by the power supply?

P E?/R 9?/100 0.81 watts inductor is not a factor as after a small amount of time, it has no effect on the circuit. .

Q:Current through an inductor is turned on at time t0, as shown in the figure. Vscos(200*pi*t). Calculate the energy delivered to the inductor at t21 ms.Here's the figure:

I(1/L)*integral(V dt). Note that you need to consider the initial conditions, when finding the solution. Once you get the solution, just plutg in t21ms.

Q:A 1.15 k-ohms resistor and a 560mH inductor are connected in series to a 1450Hz generator with an rms voltage of 12.6 V. A)What is the rms current in the circuit? B)What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part A?i would realli appreciate a walk through of this problem and what answer i should expect. thanks

A) the reactance of the inductor is XL 2π*f*L 2π*1450*0.560 5102Ω So the impedance Z sqrt(1150^2 + 5102^2) 5230Ω Therefore the rms current V/Z 12.6V/5230Ω 2.41x10^-3A 2.41 mA B) If the current is reduced to 1/2 then then impedance must double So Z sqrt(R^2 + (XlL- XC)^2) 10460 So 1150^2 + (5102 - XC)^2 10460^2 or 5102 - XC +-sqrt(10460^2 - 1150^2) +-10397 or XC 5102 + 10397 15499 Now XC 1/2πf*C or C 1/(2π*1450*15499) 7.08x10^-9F 7.08nF

Q:A project of mine requires that I make a set of custom coils, some of which are tapped. I've never scratch-built inductors before so this is new to me. So how to you build a tapped coil?

Get an O ring shape iron core, passing many turns of wire.Passing wire through the center hole for tapping. [2] Get a E shape iron core,make coil fit the center E. Put the open E side attach to wire being tapped.

Q:I need this very badly for my Electronics Project, but unfortunately I still can't find it! What are the defects and testing of Inductors??? Please help! X(

TESTING INDUCTORS An inductor is a device consisting of one or more windings of wire with or without a magnetic core. Frequent causes of inductor coil failures are shorted turns, open turns, and changes in inductor value. Small power supply transformers are similar in construction to inductors and can be tested with the capacitor-inductor analyzer shown in figure 4-30. Inductors may be tested in the circuit, but the circuit impedance will have some effect on the readings. It is recommended that you remove the inductor or transformer from the circuit before you perform any tests.

Q:Also ,when there are two capacitors and two inductors in a circuit,how do I calculate the resonant frequency ?

Inductance of a wire is ??L/8π ?? is the magnetic constant 1.2566e-6 H/m (or T·m/A) L is length in meters Inductance of a hollow cylinder is (??L/2π)(ln((2L/a)–1) a is the radius. thickness is negligible. 2) depends on how they are connected. Combine the two caps as series or parallel, ditto inductors.

Q:A 1E-6 F capacitor has a charge of 10.0 microC. It is connected to a 1.0 H inductor at t 0. What is the potential energy of the capacitor and the potential energy of the inductor at t 1 second?

The angular velocity (omega) here sqrt(1/LC) sqrt(1/(1x10^-6*1) 1000 rad/s Q0 10^x10^-6C Now the charge on a capacitor in an LC oscillator q Q0*cos(omega*t) 10.x10^-6*cos(1000*1) 5.62x10^-6C So the energy q^2/2C (5.62x10^-6)^2/(2*1x10^-6) 1.58x10^-5J Now the current i -omega*Q0*sin(omega*t) -1000*10x10^-6*sin(1000) -0.00827A Energy in an inductor 1/2*l*i^2 1/2*1.0*(0.00827)^2 3.42x10^-5J