Question

A wire of resistance 9ohm is broken in two parts . The length ratio being 1:2 . The two pieces are connected in parallel . The net resistance will be ​

Answer 1

Let the starting resistance is R ,now cut it in half in two peice now resistance is R/2 AND R/2. RIGHT

now come to parallel combination

1/Req = 1/r1 +1/r2

There for. 1/(r/2) +1/(r/2)

It will look like = 2/R+2/R = 4/R

Now R equivalent= R/4 so the answer Will be R/4

Thanks

Answer 2

$\huge\underline\purple{\sf ♡Answer♡:-}$

★$\large{\boxed{\sf R_{net}=2ohm}}$

$\huge\underline\purple{\sf ♡Solution♡:-}$

Given :-

Resistance (R) = 9 ohm

Ratio of Lenght = 1:2

To find :-

$\large{\sf R_{net}=?}$

━━━━━━━━━━━━━━━━━━━━━━━━━

We know that ,

★$\large{\boxed{\sf R \alpha l}}$

R = resistance

l = Lenght

$\large{\sf {\frac{R_{1}}{R_{2}}}={\frac{l_{1}}{l_{2}}}}$

$\large\implies{\sf {\frac{1}{2}}}$

Therefore,

$\large{\sf R_{1}={\frac{1}{3}}×9}$

$\large\implies{\sf 3\:ohm}$

$\large{\sf R_{2}={\frac{2}{3}}×9}$

$\large\implies{\sf 6\:ohm}$

$\large{\boxed{\sf R_{net}={\frac{R_1R_2}{R_1+R_2}}}}$

On putting value :-

$\large\implies{\sf {\frac{3×6}{3+6}}}$

$\large\implies{\sf {\frac{18}{9}}}$

★$\large\red{\boxed{\sf R_{net}=2ohm}}$