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
The resistance of the wire is R=ρ⋅lS=49⋅10−8 Ohm⋅m⋅20 mπ(0.004 m)2/4=0.78 Ohm.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}.R=ρ⋅Sl=49⋅10−8Ohm⋅m⋅π(0.004m)2/420m=0.78Ohm.
a. The current is I=UR=6 V0.78 Ohm=7.69A.I = \dfrac{U}{R} = \dfrac{6\,\mathrm{V}}{0.78\,\mathrm{Ohm}} = 7.69\mathrm{A}.I=RU=0.78Ohm6V=7.69A.
b. The power is P=UI=6 V⋅7.69 A=46.14 W.P=UI = 6\,\mathrm{V}\cdot 7.69\,\mathrm{A} = 46.14\,\mathrm{W}.P=UI=6V⋅7.69A=46.14W.
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