How can three resistors of resistances 2 Omega,3 Omega and 6 Omega be connected to give a total resistance of (a) 4 Omega (b) 1 Omega?

How can three resistors of resistances 2 Omega,3 Omega and 6 Omega be

1.	How can three resistors of resistances 2 Omega,3 Omega and 6 Omega be connected to give a total resistance of (a) 4 Omega (b) 1 Omega?
Answer» SOLUTION :a) If we connect RESISTANCES of `3 Omega and 6 Omega` in parallel and then RESISTANCE of `2 Omega` is connected in series of the combination, the total resistance of combination is `4 Omega` as shown in fig.12.3(a) The combined resistance R. on the parallel grouping of `3 Omega and 6 Omega` resistances is given as: `1/R.=1/3+1/6=(2+1)/6=3/6=1/2` `implies R=2 Omega` and total resistance of ENTIRE combination `R=2 Omega +R.=2+2=4 Omega` (b) IF all the three resistances are joined in parallel as shown in fig.12.3(b) we have `1/R=1/2+1/3+1/6=(3+2+1)/6=6/6=1/1 implies R=1 Omega`