Answer
(a) $0.8 \mathrm{~\Omega}$
(b) $I_{1.6} = 17.5 \,\text{A}, I_{2.4} =11.7 \,\text{A}, I_{4.8} =5.8 \,\text{A}$
(c) $I_{t} =35 \,\text{A}$
(d)$\varepsilon = V_{1.6} = V_{2.4} = V_{4.8} = 28 \,\text{V}$
(e) $P_{1.6} =490 \,\text{W}, P_{2.4} = 326 \,\text{W}, P_{4.8} =163 \,\text{W}$
(f) least resistance
Work Step by Step
\subsection*{Solution}
(a) The resistors are connected in parallel, so the equivalent resistance for them is
\begin{align*}
\dfrac{1}{R_{\text{eq}}} &= \dfrac{1}{ 1.6 \mathrm{~\Omega} } + \dfrac{1}{ 2.4 \mathrm{~\Omega}} + \dfrac{1}{ 4.8 \mathrm{~\Omega}}= \dfrac{5}{4 \mathrm{~\Omega} }
\end{align*}
The value of $R_{\text{eq}}$ is $0.8 \mathrm{~\Omega}$
(b) The current is not the same for the three resistors. By using Ohm's law we get the current through each resistor by
\begin{gather*}
I_{1.6} = \dfrac{28 \,\text{V}}{1.6 \mathrm{~\Omega}} = \boxed{17.5 \,\text{A}}\\
I_{2.4} = \dfrac{28 \,\text{V}}{2.4 \mathrm{~\Omega}} = \boxed{11.7 \,\text{A}}\\
I_{4.8} = \dfrac{28 \,\text{V}}{4.8 \mathrm{~\Omega}} = \boxed{ 5.8 \,\text{A}}
\end{gather*}
(c) The total current is the sum of the current in each resistor
\begin{align*}
I_{t} = I_{1.6} + I_{2.4} + I_{4.8} = \boxed{35 \,\text{A}}
\end{align*}
(d) The potential difference will be the same for all resistors and the battery as they are connected in parallel
$$\varepsilon = V_{1.6} = V_{2.4} = V_{4.8} = \boxed{28 \,\text{V}}$$
(e) The dissipated power equals the voltage times the current and for each resistor is given by
\begin{gather*}
P_{1.6} = (V_{1.6})(I_{1.6}) = (28 \,\text{V})(17.50\,\text{A}) = \boxed{490 \,\text{W}}\\
P_{2.4} = (V_{2.4})(I_{2.4}) = (28 \,\text{V})(11.67 \,\text{A}) = \boxed{326 \,\text{W}}\\
P_{4.8} = (V_{4.8})(I_{4.8}) = (28 \,\text{V})(5.83 \,\text{A}) = \boxed{163 \,\text{W}}
\end{gather*}
(f) The results in part (e) shows that the most dissipated power is for the $\textbf{least resistance}$
