See what a little piece of metal does, Iron-man!

Level 2

You are ready for your next vacation to a land of peacefulness. The only thing between you and your destination is a shameless metal detector that requires you to walk through it without making it go beep !

You took all the articles off but forgot your watch, and as soon as you walked in, the alarms went off and you are required to walk through again, removing your watch this time. You're good to go and you enter your airplane where you sit down and work out the incident at the airport.

You do know that the metal detector is just a simple application of an LCR circuit . The next piece of info you know is that the sound alarm that blared off was most likely a basic one which requires an RMS current of 10 A 10\text{ A} to go off.

Consider the LCR circuit to be made up of a sound source of resistance 5 Ω , 5\, \Omega, an ideal inductor of 2 mH , 2\text{ mH}, and an ideal capacitor of 20 pF . 20\text{ pF}.

All you need to do is to work out the angular frequency of the current that your watch generated, i.e. ω \omega and the RMS value of the alternating voltage for the circuit, i.e. V rms . {V}_\text{rms}.

Submit your answer as ω 10 7 + V rms . \frac {\omega} {{10}^{7}} + {V}_\text{rms}.


The answer is 50.5.

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1 solution

Sunia Tanweer
Mar 7, 2017

Since the alarm went off, the circuit must have been at resonance. So, w r = 1 L C w_r = \frac{1}{\sqrt{LC}} Putting values: w r = 5 1 0 6 H z w_r = 5 * 10^6 Hz

Now, V r m s = R I r m s V_{rms} = R * I_{rms} Putting values: V r m s = 50 V V_{rms} = 50 V

  • So, the result is 50.5.

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