An aluminum wire with a diameter of 1.02 mm and volume of 1.57x10-5 m3, has a battery attached to it with an emf of 30 volts. Find (a) the resistance of the wire and (b) the corresponding current.
(a) the resistance of the wire
"R=\\rho\\frac{l}{A}=\\rho\\frac{V}{A^2}\\\\\n=2.65*10^{-8}*\\frac{1.57*10^{-5}}{(3.14*(1.02*10^{-3})^2\/4)^2}\\\\\n=0.624\\:\\Omega"(b) the corresponding current
"I=\\frac{V}{R}=\\frac{30}{0.624}=48.1\\:\\rm A"
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