For the series combination of three resistors establish the relation :
` R=R_(1)+R_(2)+R_(3)`
where the symbols have their usual meanings.
Calculate the equivalent resistance of the combination of three resistors of `6Omega , 9 Omega` and `18 Omega` joined in parallel.

Explanation

Same current (I) flows through different resistances, when these are joined I series , as shown in the figure.

Let R be the combined resistance
Then, V=IR
`V_(1)=IR_(1),V_(2)=IR_(2),V_(3)=IR_(3)`
`because V=V_(1)+V_(2)+V_(3)" "therefore " "IR=IR_(1)+IR_(2)+IR_(3)`
`implies IR=I(R_(1)+R_(2)+R_(3))" "therefore" "R=R_(1)+R_(2)+R_(3)`
Now , `R_(1)=6 Omega" "R_(2)=9Omega" "R_(3)=18Omega`
In parallel combination
`(1)/(R)=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))" "implies " "(1)/(R)=(1)/(6)+(1)/(9)+(1)/(18)=(3+2+1)/(18)=(6)/(18)=(1)/(3)`
`therefore (1)/(R)=(1)/(3)" "implies " "R=3Omega`