Question

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?

Short answer

The power is$0.28W$.

Step-by-step solution

Given Information

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

Explanation

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.

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

:  Calculation.

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