Answer
(a) $8.8 \mathrm{~\Omega}$
(b) 3.18 A
(c) 3.18 A
(d) $V_{1.6} = 5.09 \,\text{V}, V_{2.4} = 7.63 \,\text{V}, V_{4.8} = 15.26 \,\text{V} $
(e) $P_{1.6} = 16.18 \,\text{W}, P_{2.4} =24.27 \,\text{W}, P_{4.8} = 48.54 \,\text{W}$
(f) Greatest resistance
Work Step by Step
\subsection*{Solution}
(a) The equivalent resistance for the three reistsors in series is given by
\begin{align*}
R_{eq}& = 1.6 \mathrm{~\Omega} + 2.4 \mathrm{~\Omega} +4.8 \mathrm{~\Omega}\\
& = \boxed{8.8 \mathrm{~\Omega}}
\end{align*}
(b) The current is the same for the resistors in series and from Ohm's law we get the current by
$$I_{1.6} = I _{2.4} = I_{4.8 } = I$$
And from Ohm's law, we could calculate the current for the combination $I$ by
\begin{align*}
I = I_{1.6} = I _{2.4} = I_{4.8 }= \dfrac{V}{R_{eq}} = \dfrac{28.0 \,\text{V}}{8.8 \mathrm{~\Omega}} = \boxed{3.18 \,\text{A}}
\end{align*}
(c) The battery has the same current as it is in series with the resistors
\begin{align*}
I = \boxed{3.18 \,\text{A} }
\end{align*}
(d) Each resistor in series has its own voltage with the same current, so for each resistor the voltage is
\begin{gather*}
V_{1.6} = IR_{1.6} =(3.18 \,\text{A})(1.6 \mathrm{~\Omega}) = \boxed{5.09 \,\text{V}}\\
V_{2.4} = IR_{2.4} =(3.18 \,\text{A})(2.4 \mathrm{~\Omega}) = \boxed{7.63 \,\text{V}}\\
V_{4.8} = IR_{4.8 } =(3.18 \,\text{A})(4.8 \mathrm{~\Omega}) = \boxed{15.26 \,\text{V}}
\end{gather*}
(e) The dissipated power in each resistor is given by
\begin{gather*}
P_{1.6} = I V_{1.6} = (3.18 \,\text{A}) (5.09 \,\text{V}) = \boxed{16.18 \,\text{W}}\\
P_{2.4} = IV_{2.4} = (3.18 \,\text{A}) (7.63 \,\text{V}) = \boxed{24.27 \,\text{W}}\\
P_{4.8} = IV_{4.8} = (3.18 \,\text{A}) (15.26 \,\text{V}) = \boxed{48.54 \,\text{W}}
\end{gather*}
(f) As in part (e), the most dissipated power is due to the $\textbf{greatest resistance}$
