A wire of $9 Ω$ resistance having 30 cm length is tripled on itself. What is its new resistance?

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Solution

$R_{1} = \frac{ρ l_{1}}{a_{1}}$
$l_{1} = \frac{l}{3}, a_{1} = 3 a$
Hence, net $R_{1} = ρ \frac{l}{9} a$
Thus, $R_{1} = \frac{R_{2}}{9}$ = $\frac{9}{9}$ = 1 $Ω$

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A wire of 9Ω resistance having 30 cm length is tripled on itself. What is its new resistance?

R1=ρl1a1 l1=l3,a1=3a Hence, net R1=ρl9a Thus, R1=R29 = 99 = 1 Ω

A wire of $9 Ω$ resistance having 30 cm length is tripled on itself. What is its new resistance?

$R_{1} = \frac{ρ l_{1}}{a_{1}}$
$l_{1} = \frac{l}{3}, a_{1} = 3 a$
Hence, net $R_{1} = ρ \frac{l}{9} a$
Thus, $R_{1} = \frac{R_{2}}{9}$ = $\frac{9}{9}$ = 1 $Ω$