The possible percentage error in computing the parallel resistance R of three resistances `1/R=(1)/R_1+ 1/R_2 1/R_3` from the formula , if `R_1, R_2, R_3` are each in error by plus 1.2% is :

The possible error in computing the parallel resistance R of three resistances 1/R=(1)/R_1+ 1/R_2 + 1/R_3 from the formula , if R_1, R_2, R_3 are each in error by plus 1.2% is :

The possible percentage error in computing the parallel resistance R of three resistances R_(1),R_(2),R_(3) from the formula 1/R=1/R_(1)+1/R_(2)+1/R_(3) , if R_(1),R_(2),R_(3) are each in error by 1.2%

Three resistance R_1 , R_2 and R_3 are connected on parallel. Show that their equivalent resistance is 1/R=1/R_1+1/R_2+1/R_3

Write the expressions for the equivalent resistance R of three resistors R_1, R_2 , and R_3 joined in (a) parallel, and (b) series.

Two resistance R1 and R2 connected in parallel, this equivalent resistance is.

In the situation shown, resistance R_(1),R_(2) and R_(3) are in the ratio 3:2:1 , then

Four resistance R_1,R_2,R_3 and R_4 are connected in parallel. The resultant resistance R is

You are given three resistors, each of resistance R_1, R_2 and R_3 such that R_1 < R_2 < R_3 :- How will you arrange them to get a net resistance greater than R_3 ?