You’ve made the finals of the Science Olympics! As one of your tasks, you’re given 1.0 g of aluminum and asked to make a wire, using all the aluminum, that will dissipate 7.5 W when connected to a 1.5 V battery. What length and diameter will you choose for your wire?

We need to find the Length and the Diameter need to choose for the wire.

We want to find the current $I_{2\Omega }$through the given resistor $R=2\Omega$. To get the current we will apply the loop rule where the loop rule is a statement that the electrostatic force conservative. suppose we go around a loop, measuring potential differences across circuit elements as we go and the algebraic sum of these differences is zero when we return to the starting point.

role="math" localid="1650540234081" $\sum V=0...\left(1\right)$

We draw a loop clockwise. let's apply equation (28.2) for the left loop, where the voltage across the resistor $4\Omega$is negative $-\left(4\Omega \right)I_{4\Omega ,R}$because the traveling direction is in the direction of the current. The localid="1649095765807" $emf12V$is positive because the direction of traveling is from negative to positive terminal in the battery and the voltage across the resistor $2\Omega$is positive $+\left(2\Omega \right)I_{2\Omega }$ because the traveling direction is in the opposite direction of the current.

$\underset{}{\sum V=0}$

localid="1650540641993" role="math" $$\begin{gathered} 12v-\left(4\Omega \right)I_{4\Omega ,R}+\left(2\Omega \right)I_{2\Omega }=0 \\ \left(4\Omega \right)I_{4\Omega ,R}-\left(2\Omega \right)I_{2\Omega }=12V...\left(2\right) \end{gathered}$$

Draw a loop clockwise for the right loop, Where the voltage across the resistor $4\Omega$is negative $-\left(4\Omega \right)I_{4\Omega ,L}$because the traveling direction is in the direction of the current. The emf $15V$is positive because the direction of traveling is from negative to positive terminal in the battery and the voltage across the resistor $2\Omega$is negative $-\left(2\Omega \right)I_{2\Omega }$because the traveling direction is in the direction of current.

role="math" localid="1650540710151" $$\begin{gathered} \sum V=0 \\ 15v-\left(4\Omega \right)I_{4\Omega ,L}+\left(2\Omega \right)I_{2\Omega }=0 \\ \left(4\Omega \right)I_{4\Omega ,L}-\left(2\Omega \right)I_{2\Omega }=15V...\left(3\right) \end{gathered}$$

junction law states that the large through the circuit is conserved , so at any junction in the circuit , the current enter the junction equals the leaving current from the junction.

$I_{4\Omega ,L}=I_{4\Omega ,R}+I_{2\Omega }$

Now, using the expression of $I_{4\Omega ,L}$into equation (3) to get the new equation

$$\begin{gathered} \left(4\Omega \right)I_{4\Omega ,L}+\left(2\Omega \right)I_{2\Omega }=15V \\ \left(4\Omega \right)(I_{4\Omega ,R}+I_{2\Omega })+\left(2\Omega \right)I_{2\Omega }=15V \\ \left(4\Omega \right)I_{4\Omega ,R}+\left(6\Omega \right)I_{2\Omega }=15V \end{gathered}$$

Now, subtract equations (2) and (4) to get $I_{2\Omega }$

$$\begin{gathered} \left(4\Omega \right)I_{4\Omega ,R}+\left(6\Omega \right)I_{2\Omega }=15V \\ - \\ \left(4\Omega \right)I_{4\Omega ,R}-\left(2\Omega \right)I_{2\Omega }=12V \\ =========== \\ \left(8\Omega \right)I_{2\Omega }=3V \end{gathered}$$

The current through the resistor $2\Omega$is $I_{2\Omega }==0.375A$

When the current flows through a resistor, it dissipates energy, the rate of the dissipated energy is the power. This rate where the energy is transferred from the current to the resistor is

$P_R=I^{2}R$

So, using equation (5) to get the dissipated power through the resistor $2\Omega$by

role="math" localid="1650542003589" $$\begin{gathered} P_{2\Omega }={I^2}_{2\Omega } \\ ={(0.375A)}^2\left(2\Omega \right) \\ {P}_{2\Omega }=0.28W \end{gathered}$$