Answer
$R_{T}$ $=$ $20$ $ohms$
$I_{T}$ $=$ $1.2$ $ohms$
$P_{T}$ $=$ $28.8$ $watts$
$v_{1}$ $=$ $12$ $volts$
$v_{2}$ $=$ $2.4$ $volts$
$P_{1}$ $=$ $14.4$ $watts$
Work Step by Step
a. equivalent resistance $R_{T}$
$R_{T}$ $=$ $R_{0}$ + $R_{1}$ + $R_{2}$
$R_{T}$ $=$ $8$ + $10$ + $2$
$R_{T}$ $=$ $20$ $ohms$
b. current $i$
$I_{T}$ $=$ $\frac{V}{R_{T}}$
$I_{T}$ $=$ $\frac{24}{20}$
$I_{T}$ $=$ $1.2$ $ohms$
c. power delivered by the source
$P_{T}$ $=$ $I_{T}\times{V_{T}}$
$P_{T}$ $=$ $1.2\times{24}$
$P_{T}$ $=$ $28.8$ $watts$
d. the voltages $v_{1}$ and $v_{2}$
$v_{1}$ $=$ $I_{T}\times{R_{1}}$
$v_{1}$ $=$ $1.2\times{10}$
$v_{1}$ $=$ $12$ $volts$
$v_{2}$ $=$ $I_{T}\times{R_{2}}$
$v_{2}$ $=$ $1.2\times{2}$
$v_{2}$ $=$ $2.4$ $volts$
e. minimum power rating required for $R_{1}$
$P_{1}$ $=$ $I_{T}\times{v_{1}}$
$P_{1}$ $=$ $1.2\times{12}$
$P_{1}$ $=$ $14.4$ $watts$
