LECT18_20721A

LECT18_20721A - Quick Quiz 20.5 In this RL circuit, driven...

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Unformatted text preview: Quick Quiz 20.5 In this RL circuit, driven by an AC source of EMF, the inductor is an air-core solenoid. The switch is closed and after some time, the light bulb glows steadily. An iron rod is inserted into the solenoid, thereby increasing the strength of the B-field inside the solenoid. As the rod is being inserted, what happens to the brightness of the light bulb? Answer: L is increasing. So | V L | = L( I/ t) is increasing. = V R + V L V R is decreasing, the power dissipated in the lightbulb decreases, and its brightness decreases. Energy stored in an inductor EMF induced by the inductor prevents the battery from establishing the maximum current immediately. Battery must do extra work to produce current. Extra work: becomes potential energy stored in the B-field of the inductor PE = 1/2 L I 2 An RL circuit switch is closed at t=0, and the current begins to increase. R = 0.5 . L = 3x10 3 H. = 9V. How much potential energy is stored in the inductor at t=0, and t= ? At t=0: I=0. PE = 1/2 L I 2 = 0 At t= : I = /R: PE = 1/2 L I 2 = 1/2 L ( /R) 2 = 0.49J Application: traffic signals Several coils of wire create a solenoid Air-core L = ( ( N 2 / l ) )A car-core: L increases by / Device which measures inductance registers the change, triggers green signal. L Application: traffic signals Several coils of wire create a solenoid Air-core L = ( ( N 2 / l ) )A car-core: L increases by / Device which measures inductance registers the change, triggers green signal....
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LECT18_20721A - Quick Quiz 20.5 In this RL circuit, driven...

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