Question

a wire has a resistance R it is broken into two equal parts and these two parts are joined in parallel the effective resistance is​

Answer 1

we know

R=p×L/A

if lenth will decrease to half then resistance will also, decrease to half,with this formula

so ,new resistance of both rods=R/2

since they r in parallel

1/Req=1/R/2 +1/R/2

1/Req=(2+2)/R

Req=R/4

Answer 2

Given

Resistance of R

broken into two equal parts=$\frac{R}{2}$ &

$\frac{R}{2}$

To Find

effective resistance in parallel

Solution

we know that

$\boxed{\boxed{\sf\red{\frac{1}{R_{equ}}= \frac{1}{R_1}+\frac{1}{R_2} }}}$

$\implies\frac{1}{R_{equ}}= \frac{1}{R_1}+\frac{1}{R_2}$

Putting the values

$\implies\frac{1}{R_{equ}}= \frac{2}{R}+\frac{2}{R}$

Taking lcm R

$\implies\frac{1}{R_{equ}}= \frac{2+2}{R}$

$\implies\frac{1}{R_{equ}}= \frac{4}{R}$

$\implies\:R_{equ}= \frac{R}{4}$

Hence,the effective resistance wil be

$R_{equ}= \frac{R}{4}$