You are to connect resistors R1 and R2, with R1 > R2, to a battery, first individually, then in series, and then in parallel. Rank those arrangements according to the amount of current through the battery, greatest first.

Parallel, R2, R1, series

Explanation:

From Ohm's law, V=IR hence making current the subject of the formula then

Since R1>R2 eg 4>2 then current for individual connection of R2 will be greater than that for R1 for example, assume V is 8 then if R2 we will have 8/2=4 A but for R1 we shall have 8/4=2 A.

When in parallel, the equivalent resistance will be given by for example here it will be 1/4+1/2=3/4. Still taking V of 8 then I= 8/(3/4)=10.667 A

When in series connection, equivalent resistance is given by adding R1 and R2 hence using the same figures we shall have 4+2=6 hence I=8/6=1.33A

We can conclude that the arrangement of current from greatest will be parallel, R2, R1, series

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