Establish a relationship to determine the equivalent resistance R of a combination of three resistors having resistances `R_(1), R_(2)` and `R_(3)` connected In series. Calculate the equivalent resistance of the combination of three resistors of `2 Omega, 3 Omega` and `6Omega` joined in parallel.

Explanation

An applied potential V produces current I in the resistors `R_(1),R_(2)` and `R_(3)`, causing a potential drop `V_(1) V_(2)` and `V_(3)` respectively through each resistor.

Total Potential `, V = V_(1) +V_(2) +V_(3)`
By Ohm.s Law `V_(1) =IR_(1)`
`V_(2)= IR_(2)`
`V_(3)=IR_(3)`
Thus `V=IR_(1)+IR_(2)+IR_(3)`
`=I(R_(1)+R_(2)+R_(3))`
IF R is the equivalent resistance `V=IR `
Hence `IR = I(R_(1)+R_(2)+R_(3))`
`R=R_(1)+R_(2)+R_(3)`
This proves that overall resistance increases when resistors are connected in series. Three resistors `2 Omega, 3 Omega` and `Omega`, are joined in parallel combination.
Equivalent resistance
`(1)/(R_(p))=(1)/(2)+(1)/(3)+(1)/(6)`
`= (2+2+1)/(6) =(6)/(6)`
`R_(p)= 1Omega`