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A 100 muF capacitor in series with a 40 omega resister connected to a 110 V, 60 Hz supply What is the maximum current in the circuit ? |
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Answer» SOLUTION :`X_c = sqrt(R^2 + (1/(OMEGAC))^2` `= sqrt ((40)^2 + (1/(2PI xx 60 xx 100 xx 10^(-6)))^20mega` = 48 W |
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