Answer to Question #147314 in Electric Circuits for david royn

Question

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

Answer 1

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

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

Wire length$=25m$

Voltage applied across$=50V$

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

$R=0.2405\varOmega$

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

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