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Registered Member #15
Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068
Welcome to the Electrical Quizzes thread.
RULES:
1. Anyone may post a quiz, however, only one (1) quiz may be active at a time.
2. All posts must stay on topic - any offtopic posts or one line comments will be deleted.
3. Keep it fair for everyone. Please do not cheat by searching for the quiz question on the internet. Do your own work and if you have already seen a quiz that is posted here and know the answer, please give others a chance. I really can't enforce this, so you are basically on the honor system here.
4. Quizzes should be related to electrical circuits or electrical theory.
5. You may use the internet to research equations and theory however to help you solve a problem.
Thanks and enjoy!
Can you figure out whats in the box???
The meters are old fashioned analog meters and the black box does not dissipate any heat.
Registered Member #15
Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068
Part Scavenger wrote ...
It's empty? The voltage drop across the 1 ohm would be .5V, so the voltage drop across the meter would be .5V.
EDIT=> Another thought. It could be a capacitor. Once it fills up, it'd never dissipate any heat and the volt drop would still be .5 across the meter.
Not correct. Remember that virtually no current flows through these type of analog voltage meters, therefore, the 0.5A of current has to flow through the black box.
Registered Member #79
Joined: Thu Feb 09 2006, 11:35AM
Location: Arkansas
Posts: 673
Pooh. I knew it couldn't be that easy.
EDIT=> It can't be a supercondutor can it? It has an impedance of 1ohm. What would use up .25W without dissipating heat? But come to think of it, you might be on the right track with the 1/2 time thing. If it's analog, they wouldn't change fast enough to see it.
Registered Member #15
Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068
wrote ...
If ti is dissipating almost no heat, my guess would be a low voltage oscillator that shorts the output 1/2 of the time...
Congrats! You got it. The answer is a switch operating at 50% duty cycle which is basically what you said. Switched fast enough, the analog voltage meter will not respond quick enough and measure just the averaged voltage which would be 0.5V.
Registered Member #113
Joined: Fri Feb 10 2006, 01:40AM
Location:
Posts: 49
Dan, is the duty cycle 50% on both square waves? Because if it is, it seems rather easy. Whenever the clock input is high and the D is high as well, the output would change. Just figure out how often 49 KHz and 100 KHz have a beat, right?
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Electrical Quizzes - QUIZ 8 POSTED
