Chapter 24 - Alternating-Current Circuits - Problems and Conceptual Exercises - Page 869: 50

(a) $4.3A$ (b) $8.7A$

(a) We know that $I_{rms}=\frac{V_{rms}}{R}$ We plug in the known values to obtain: $I_{rms}=\frac{65}{15}=4.3A$ (b) As $R_{eq}=(\frac{1}{R}+\frac{1}{R})^{-1}=\frac{R}{2}=\frac{15}{2}=7.5\Omega$ Now $I_{rms}=\frac{V_{rms}}{R_{eq}}$ We plug in the known values to obtain: $I_{rms}=\frac{V_{rms}}{R_{eq}}=\frac{65}{7.5}=8.7A$