Two resistances `R_(1)andR_(2)` provides series to parallel equivalents as n/1, then the correct relationship is

Two resistance R_(1) and R_(2) are connected in (i) series and (ii) parallel. What is the equivalent resistance with limit of possible percentage error in each case of R_(1)=5.0+-0.2Omega and R_(2)=10.0+-0.1Omega

Two resistors of resistance R_1 and R_2 having R_1>R_2 are connected in parallel. For equivalent resistance R, the correct statement is

Two resistors are connected (a) in series (b) in parallel The equivalent resistance in the two cases are 9 ohm and 2ohm respectively. Then the resistances of the component resistor are

Two resistors of resistance R_(1) and R_(2) having R_(1) gt R_(2) are connected in parallel. For equivalent resistance R, the correct statement is

Two resistance R_1 and R_2 when connected in series produce equivalent resistance equal to x. and when the same two resistance are connected in parallel then equivalent resistance becomes y. Calculate the product of magnitudes of R_1 and R_2 in terms of x and y.

Two resistance R_(1) and R_(2) may be connected either in series or in parallel across a battery of zero internal resistance. It is required that the joule heating for the parallel combination be five times that for series combination. If R_(1) is 10Omega find R_(2) .

Two resistance R_(1) and R_(2) may be connected either in series or in parallel across a battery of zero internal resistance. It is required that the joule heating for the parallel combination be five times that for series combination if R_(1) is 100 Omega find R_(2) .

When two resistors of resistances R_1 and R_2 are connected in parallel, the net resistance is 3 Omega . When connected in series, its value is 16 Omega . Calculate the values of R_1 and R_2 .