Question

Answer the 25th question. Urgently please.
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Answer 1

Answer:

$\frac{40A}{\rho l} Amp$

Step-by-step explanation:

We know that:

$R=\rho \frac{l}{A}$

Where:

$R=Resistance\\\rho = Resitivity\\A=Area\: of\: cross\:section\\l= Length\: of\: the\: wire$

∴According to the question:

$R=\rho\frac{l}{A}$

We also know that:

$V=IR \;\;\;\;\;(Ohm's\:Law)$

⇒$I=\frac{V}{R}$

Where:

$V=Potential \;Difference\\I=Current\\R=Resistance$

According to the question:

$V=40V$

∴$I=\frac{40}{\frac{\rho l}{A} } \\\\I=\frac{40A}{\rho l} \;\;\;\;\;\;\;\;\;\;\;\;Answer$

Answer 2

Answer:

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