You are given a transformer pictured below. what is the current through the secondary coil, if the number of turns in the primar
y coil is 86, the number of turns in the secondary coil is 85, the load resistance is 19 ohms, and the voltage input to the primary coil is 36 volts? you can assume that r1 has negligible resistance and that the transformer is ideal. give your answer in milliamps (non-negative & 3 significant digits).
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Explanation:
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rho-hdfe-rda
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Answer:
10000 Bq / 625 Pq = 16
Radioactivity has decreased by a factor of 16
2^4 = 16
So the sample has gone thru 4 half-lives
24 da / 4 = 6 da
6 da is the half-life
