In an emergency, you need to run a radio that uses \(30.0 \mathrm{~W}\) of power when attached to a \(10.0-\mathrm{V}\) power supply. The only power supply you have access to provides \(25.0 \mathrm{kV}\) but you do have a large number of \(25.0-\Omega\) resistors. If you want the power to the radio to be as close as possible to \(30.0 \mathrm{~W}\), how many resistors should you use, and how should they be connected (in series or in parallel)?

In order to calculate the required total resistance, we have to use the formula for power: $$ P = \frac{V^2}{R_{total}} $$ Here, \(P\) represents the power, \(V\) represents the voltage, and \(R_{total}\) is the total resistance. Since we want the power to be as close as possible to 30.0 W, we can plug in the values and solve for the total resistance: $$ R_{total} = \frac{V^2}{P}$$ $$ R_{total} = \frac{(10.0 \mathrm{V})^2}{30.0 \mathrm{W}}$$ $$ R_{total} = 3.33 \Omega $$

Now we have an idea of the total resistance needed, which is \(3.33 \Omega\). Given that we only have access to \(25.0-\Omega\) resistors, we need to find out the number of resistors required and their connection to get as close as possible to the desired total resistance. Let's consider the two configurations: series and parallel. 1. Series: If we connect resistors in series, their resistances simply add up: $$ R_{series} = R_1 + R_2 + \cdots + R_n $$ To get close to the required resistance of \(3.33 \Omega\), we would need fewer resistors. However, connecting \(25.0-\Omega\) resistors in series will result in a higher total resistance than the desired value. Therefore, this configuration will not work. 2. Parallel: If we connect resistors in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of their individual resistances: $$ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} $$ Now, let's plug in the numbers and solve for the number of resistors \(n\) required: $$ \frac{1}{R_{total}} = n \cdot \frac{1}{25.0 \Omega} $$ $$ n = \frac{1}{R_{total}} \cdot 25.0 \Omega $$ Using the required total resistance of \(3.33 \Omega\): $$ n = \frac{1}{3.33 \Omega} \cdot 25.0 \Omega $$ $$ n \approx 7.50 $$ Since we need a whole number of resistors, we can round this value up to 8. Thus, we need 8 resistors connected in parallel.

To achieve a power dissipation as close as possible to 30.0 W using a 10.0-V power supply and 25.0-\(\Omega\) resistors, we need to use 8 resistors connected in parallel.