A wire has a resistance of `6 Omega`. It is cut into two parts and both half values are connected in parallel. The new resistance is ....

Options

A. `3Omega`
B. `6Omega`
C. `12Omega`
D. `1.5Omega`

Answer

D

Explanation

Resistance of original wire is
`R = rho (l)/(A)`
`rho`, being the specific resistance of wire. When the wire is cut in few equal halves then resistance becomes
`R'=rho(l)/(2)=(R )/(2)`
Thus, the net resistance of parallel combination of per halves is given by `R_("net")=(R'xxR')/(R'+R')`
`= (R')/(2)=(R )/(2xx2)=(6)/(4)=1.5 Omega`