Answer:
Parallel, R2, R1, series
Explanation:
From Ohm's law, V=IR hence making current the subject of the formula then [tex]I=\frac {V}{R}[/tex]
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 [tex]\frac {1}{R1}+\frac {1}{R2}[/tex] 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
According to the amount of current, the arrangement will be "Parallel, R2, R1, Series".
By using Ohm's law
→ Voltage (V) = Current (I) × Resistance (R)
or,
→ I = [tex]\frac{V}{R}[/tex]
Since R1 > R2
Let, Current (V) = 8 then the resistance will be:
R2 = [tex]\frac{8}{2}[/tex] = 4 A
and,
R1 = [tex]\frac{8}{2}[/tex] = 4 A
When the resistance is in parallel connection,
= [tex]\frac{1}{R1} + \frac{1}{R2}[/tex]
= [tex]\frac{1}{4} +\frac{1}{2}[/tex]
By taking L.C.M, we get
= [tex]\frac{1+2}{4}[/tex] = [tex]\frac{3}{4}[/tex] and,
The current be:
→ I = [tex]\frac{8}{\frac{3}{4} }[/tex] = 10.667 A
When the resistance is in series connection,
= R1 + R2
= 4 + 2
= 6 and,
The current be:
→ I = [tex]\frac{8}{6}[/tex] = 1.33 A
Thus the response above is appropriate.
Find out more information about resistor here:
https://brainly.com/question/24858512
