Chapter 21 - AC Circuits - Learning Path Questions and Exercises - Exercises - Page 748: 7

a) $I_{rms}=4.47A$ and $I_{peak}=6.32A$ b) $V_{rms}=112V$ and $V_{peak}=158V$

a) $P_{avg}=I_{rms}^{2}R$ or, $500=I_{rms}^{2}\times25$ or, $I_{rms}=4.47A$ So, $I_{peak}=\sqrt 2\times 4.47=6.32A$ b) $V_{rms}=4.47\times 25=112V$ and $V_{peak}=\sqrt 2\times112=158V$