3 years ago

Write and solve a MA or efficiency problem

Physics

1 answer:

GuDViN [60]3 years ago

8 0

Input energy is: 200 joule
Output energy is: 100 joule

100/200*100=%50 efficiency

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The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ . Rearranging the above equation,

$A=\frac{\rho l}{R}\\ \frac{\pi d^2}{4}=\frac{\rho l}{R}\\ l=\frac{\pi d^2 R}{4 \rho}$

Here $d=0.15, R=15, \rho=1.68(10^{-8})$ .

Substituting numerical values,

$l=\frac{\pi 0.15^2(10^{-6}) (15)}{4 (1.68)(10^{-8})} \\ l=15.78$

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