Discussion in 'Homework Help' started by amilton542, Jan 8, 2011.

1. ### amilton542 Thread Starter Active Member

Nov 13, 2010
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hey guys im baffled by a few things,
when an inductor and a capacitor are at a state of resonance using one another to discharge when the supply is removed the current flow increases through the inductor by the capacitor discharging and it will react by creating a volt drop, but when the current flow decreases through the inductor it generates a voltage, how does it generate this voltage by the magnetic field??
Also when i use the hosepipe theory and wrap the hosepipe into a coil the coil will resist the water, i got a multimeter and cut 12" of 2.5mm solid copper wire layed it flat and measured the resistance, i then wrapped it round a thin pole to create my turns, i then removed it and measured the resistance and i still had the same value. i know an inductor will have resistance because of the thin wire and length of wire to create the turns, but i always thought if a length of cable was wrapped into a coil you would increase the resistance but it doesnt change.its easy to read about inductors but its hard to picture whats happening

Last edited: Jan 8, 2011

Apr 5, 2008
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3. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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The voltage on the coil MUST be changing to see any resistance effects due to the inductance. A DC ohm meter will not measure inductive reactance as a resistance reading.

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4. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Roughly:
Inductors resist a change in current with opposing voltage.
Capacitors resist a change in voltage with opposing current.

The inductor is "trying" to keep the same level of current through it, the magnetic field collapsing generates a current. This current charges the capacitor, creating a voltage. When the inductor can no longer charge the capacitor, the capacitor sees the inductor as a load, and discharges through it, before current can flow quickly, (remember, inductors resist changes in current), the capacitor's energy needs to generate the magnetic field in the inductor, at which point current flows faster until C is discharged, and the entire process starts over again.

"Q" is the quality of the LC tank, roughly, the number of times the tank will "ring" after external power is removed. Parasitic resistance in the L, C and wires make it impossible to get an infinite Q.

The damping factor α is $\alpha = \frac{R}{2L}$

Q is related to the damping factor by $Q = \frac{\omega_0}{2\alpha}$

Where ω0 (resonant frequency) is $\omega_0 = sqrt{\frac{1}{LC}}$

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