## College Physics (7th Edition)

a). $1A,0.5A,0.5A$ b). $20V,10V,10V$ c). $30W$
a) $R_{eq}=\frac{R_{1}(R_{2}+R_{3})}{R_{1}+R_{2}+R_{3}}=13.33-Ohm$ Thus, $i=\frac{20}{40}\times3=1.5A$ Potential across $R_{2}+R_{3}$ and $R_{1}$ are equal, so, $40(1.5-i_{1})=20i_{1}$, i.e. $i_{1}=1 A$ Thus, $i_{2}=i_{3}=0.5A$ b) $V_{1}=20\times1=20V$, $V_{2}=20\times0.5=10V$, $V_{3}=20\times0.5=10V$, c) Power = $i^{2}R=1.5^{2}\times 13.33=30W$