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Expression of power dissipation in a resistor of resistance R is,

$P=l^{2}R$

Solve for current using Ohm’s Law,

$$\begin{gathered} l=\frac{V}{R} \\ =\frac{8.0V}{17\mathrm{Î\copyright }} \\ =0.47A \end{gathered}$$

Use the value of I in the power expression,

$$\begin{gathered} {\mathrm{P}}_{5\mathrm{Î\copyright }}={\mathrm{l}}^2\mathrm{R} \\ ={(0.47\mathrm{A})}^2\left(5.0\mathrm{Î\copyright }\right) \\ =1.1\mathrm{W} \\ \mathrm{Similarly}, \\ {\mathrm{P}}_{9\mathrm{Î\copyright }}={\mathrm{l}}^2\mathrm{R} \\ ={(0.47\mathrm{A})}^2\left(9.0\mathrm{Î\copyright }\right) \\ =2.0\mathrm{W} \\ \mathrm{Thus},\mathrm{the}\mathrm{total}\mathrm{power}\mathrm{is}, \\ \mathrm{P}={\mathrm{P}}_{5\mathrm{Î\copyright }}+{\mathrm{P}}_{9\mathrm{Î\copyright }} \\ =1.1\mathrm{W}+2.0\mathrm{W} \\ =3.1\mathrm{W} \end{gathered}$$

The power output is given by,

$$\begin{gathered} {\mathrm{P}}_{16\mathrm{V}}=\mathrm{Î\mu \pm ô}-{\mathrm{l}}^2\mathrm{r} \\ \mathrm{Substitute}\mathrm{Î\mu }=16\mathrm{V},\mathrm{l}=0.47\mathrm{A}\mathrm{and}\mathrm{r}=1.6\mathrm{Î\copyright } \\ {\mathrm{P}}_{16\mathrm{V}}=(16\mathrm{V})(0.47\mathrm{A})-{(0.47\mathrm{A})}^2(1.6\mathrm{Î\copyright }) \\ =7.2\mathrm{W} \end{gathered}$$

Thus, the power output of the 16.0 V battery is 7.2 W

Rate of conversion of electrical energy is the power output and the expression for the 8.0 V battery is given as,

$$\begin{gathered} {\mathrm{P}}_{8\mathrm{V}}=\mathrm{Î\mu \pm ô}-{\mathrm{l}}^2\mathrm{r} \\ \mathrm{Substitute}\mathrm{Î\mu }=8.0\mathrm{V},\mathrm{l}=0.47\mathrm{A}\mathrm{and}\mathrm{r}=1.4\mathrm{Î\copyright } \\ {\mathrm{P}}_{8\mathrm{V}}=(8\mathrm{V})(0.47\mathrm{A})-{(0.47\mathrm{A})}^2(1.4\mathrm{Î\copyright }) \\ =4.1\mathrm{W} \end{gathered}$$

Thus, the power output of the 8.0 V battery is 4.1 W.

Infer from part (a), (b) and (c)

It is shown that the power output for 16.0 V in part (b) is equal to the power dissipated for 8.0 V battery in part (c) and the combined power loss across 5.0Ω and 9.0 Ω resistor in part (a).