A 6 `Omega` resistance wire is doubled on itself. Calculate the new resistance of the wire.
(b) Three 2 `Omega` resistors A, B and C are connected in such a way that the total resistance of the combination is 3 `Omega` Show the arrangement of the three resistors and justify your answer.

Explanation

Let length and cross-section area of given will be L and A respectively and `rho` be its resistivity. Then
Resistance of wire R = (`rho`L)/(A)=6`Omega` …(i)
When the wire is doubled on itself, its new length L. =`L/2`= and new cross-section area A. = 2A. So its new resistance will be
`R.=(rhoL.)/(A.)=(rho(L/2))/((2A))=1/4(rhoL)/(A)=R/4=6/4=1.5 Omega`
(b) The arrangement of 3 resistors of 2 `Omega` each, so as to have a net resistance of 3 `Omega`, is shown here. Here `R_1` is in series of the parallel combination of `R_2` and `R_3`. If combined resistance of `R_2` and `R_3` be R.,then `(1)/(R.)=(1)/(R_2)+(1)/(R_3)=1/2+1/2=1/1`
`implies R.=1 Omega`
`Omega` Net resistance of ciricuit `R=R_1+R.=2+1=3 Omega`