Question #183726

A 20 m constantan wire with 4 mm thickness connected to a 6 V battery. Given 

resistivity of constantan is 49  10-8  m. Calculate

a. Current in the circuit

b. Power dissipated by the wire


1
Expert's answer
2021-04-22T07:27:47-0400

The resistance of the wire is R=ρlS=49108Ohmm20mπ(0.004m)2/4=0.78Ohm.R = \rho \cdot \dfrac{l}{S} = 49\cdot10^{-8}\,\mathrm{Ohm\cdot m}\cdot \dfrac{20\,\mathrm{m}}{\pi (0.004\,\mathrm{m})^2/4} = 0.78\,\mathrm{Ohm}.


a. The current is I=UR=6V0.78Ohm=7.69A.I = \dfrac{U}{R} = \dfrac{6\,\mathrm{V}}{0.78\,\mathrm{Ohm}} = 7.69\mathrm{A}.


b. The power is P=UI=6V7.69A=46.14W.P=UI = 6\,\mathrm{V}\cdot 7.69\,\mathrm{A} = 46.14\,\mathrm{W}.


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