Question

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.

Answer 1

Answer:

Parallel, R2, R1, series

Explanation:

From Ohm's law, V=IR hence making current the subject of the formula then $I=\frac {V}{R}$

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 $\frac {1}{R1}+\frac {1}{R2}$ 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

Answer 2

According to the amount of current, the arrangement will be "Parallel, R2, R1, Series".

Parallel and Series connection

By using Ohm's law

→ Voltage (V) = Current (I) × Resistance (R)

or,

→ I = $\frac{V}{R}$

Since R1 > R2

Let, Current (V) = 8 then the resistance will be:

R2 = $\frac{8}{2}$ = 4 A

and,

R1 = $\frac{8}{2}$ = 4 A

When the resistance is in parallel connection,

= $\frac{1}{R1} + \frac{1}{R2}$

= $\frac{1}{4} +\frac{1}{2}$

By taking L.C.M, we get

= $\frac{1+2}{4}$ = $\frac{3}{4}$ and,

The current be:

→ I = $\frac{8}{\frac{3}{4} }$ = 10.667 A

When the resistance is in series connection,

= R1 + R2

= 4 + 2

= 6 and,

The current be:

→ I = $\frac{8}{6}$ = 1.33 A

Thus the response above is appropriate.

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