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Q. An alternating current of $rms$ value $10 \,A$ is passed through a $12\, \Omega$ resistor. The maximum potential difference across the resistor is

ManipalManipal 2009Alternating Current

Solution:

$I_{0} =I_{ rms } \times \sqrt{2}$
$=10 \sqrt{2} A$
$\therefore $ Maximum potential difference $=V_{0}$
$=12 \times 10 \sqrt{2}$
$=120 \times 14.14=169.68 \,V$