Answer to Question #147314 in Electric Circuits for david royn

Question #147314
Copper has a resistivity of Ω m. A wire of diameter 1.5 mm and length 25 m is connected across a potential difference of 50 V.

Calculate:

(a) The resistance of the wire

(b) The current

(c) The power dissipated in the wire
1
Expert's answer
2020-11-30T14:53:57-0500

Resistivity of copper is 1.7×1081.7\times10^{-8}


Wire diameter=1.5mm    radius=7.5×104m=1.5mm \implies radius=7.5\times10^{-4}m

Wire length=25m=25m


Voltage applied across=50V=50V


a) Resistance of the wire(R)=ρla=1.7×10825π×(7.5×104)2(R) =\rho\frac{l}{a}=1.7\times10^{-8}\frac{25}{\pi \times(7.5\times10^{-4}) ^2}


R=0.2405ΩR=0.2405\varOmega


b) Current(I)=VR=500.2405=207.9A(I) =\frac{V}{R}=\frac{50}{0.2405}=207.9A

c) Power dissipated=I2R=(207.9)2×0.2405=10394.989W=I^2R=(207.9) ^2\times 0.2405=10394.989W


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