Question

if a long wire is connected to the terminals of a 12 - v battery, 9.4×10^19 electrons pass a cross section of the wire each second. part a determine the current in the wire. express your answer with the appropriate units. part b determine the resistance in the wire. express your answer with the appropriate units.

Explanation:

Step1: Calculate the current using electron - charge relationship

The charge of one electron is $e = 1.6\times10^{- 19}\ C$. The number of electrons $n = 9.4\times10^{19}$ pass a cross - section per second. The current $I$ is given by $I=\frac{Q}{t}$, and $Q = ne$. So $I=\frac{ne}{t}$. Since $t = 1\ s$, $I=n e$. Substitute $n = 9.4\times10^{19}$ and $e=1.6\times10^{-19}\ C$ into the formula: $I=(9.4\times10^{19})\times(1.6\times10^{-19})\ A$. Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $I=(9.4\times1.6)\times10^{19-19}\ A=15.04\ A$.

Step2: Calculate the resistance using Ohm's law

Ohm's law is $V = IR$, where $V = 12\ V$ and $I = 15.04\ A$. We can solve for $R$ by rearranging the formula to $R=\frac{V}{I}$. Substitute $V = 12\ V$ and $I = 15.04\ A$ into the formula: $R=\frac{12}{15.04}\ \Omega\approx0.798\ \Omega$.

Answer:

Part A:
$I = 15.04\ A$
Part B:
$R=0.798\ \Omega$