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Electronics Interview Question:
The current through a resistor of 50 ohms in an AC circuit at t = 0.008 s is 65% of the peak value. The smallest possible frequency of the generator delivering the current is
Submitted by: AdministratorIf it is then this is easy, you take the arcsine of 65% that gives you what angle you are at in the sine cycle, take the ratio of that to 360, and multiply that by 8mS to get the time of a full cycle, and then take one over that to get frequency. This will be the lowest frequency. Note that the 50 ohms has nothing to do with it, other than implying it is an RF circuit where 50 ohms is common. You say "smallest possible frequency" which also is not very clear I will assume you mean lowest frequency. There is no highest frequency you can meet this requirement with an arbitrarily high frequency.
Submitted by: Administrator
Submitted by: Administrator
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