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# Resistance of Silver and Graphite Wires

EANDM-5GHQIJ

An electric potential of $V = 7.52 \text{ V}$ is placed across a thin silver wire with a cylindrical cross-section whose radius is $r_{\rm{Au}} = 19.4 \: \mu\rm{m}$ and whose length is $l = 1.26 \text{ m}$. The resulting current through the wire is $I = 444 \text{ mA}$.

In order to save money, the silver is being replaced with graphite. The potential $V$ across the graphite wire and the current through it must remain the same as before. The length of the wire must also remain the same. The only parameter you may change is the radius of the wire. The resistivity of silver is $\rho_{\rm{Ag}} = 1.59 \times 10^{-8} \: \Omega \cdot \rm{m}$ and the resistivity of graphite is $\rho_{\rm{g}} = 5.60 \times 10^{-5} \: \Omega \cdot \rm{m}$.

What must be the radius $r_{\rm{g}}$ of the graphite wire?

A

$r_{\rm{g}} = 36.4 \: \mu \rm{m}$

B

$r_{\rm{g}} = 6.83 \text{ cm}$

C

$r_{\rm{g}} = 1.15 \text{ mm}$

D

$r_{\rm{g}} = 68.3 \: \mu \rm{m}$