Question

The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current in the wire? Express your answer using three significant figures.

Answer 1

Answer:

$I=0.047A$

Explanation:

Let's use Ohm's law:

$V=IR$  

or

$I=\frac{V}{R}$   (1)

Where:

$V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance$

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

$R=\rho*\frac{l}{A}$    (2)

Where:

$R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}$

Keep in mind that the electrical resistivity of the gold is a known constant which is $\rho_g_o_l_d=2.35*10^{-8}$ and the cross sectional area of the conductor is calculated as:

$A=\pi *(r^{2})=\pi *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}$

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

$R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} } =14.96056465\Omega$

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

$I=\frac{0.7}{14.96056465}\approx0.047A$