Question

PLEASE DO ANSWER THIS QUICKLY ....
Two identical resistors of 12 ohm each are connected to a battery of 3 V. Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.

Answer 1

Answer:

Step-by-step explanation: As we know , minimum resistance is provided in a parallel series . Thus , these resistances will give the minimum resistant only when both the resistors are connected in parallel =1÷R = 1 ÷ R1 +1 ÷ R2

= 1 ÷ R = 1 ÷ 12 + 1 ÷ 12

= 1 ÷ R = 1 + 1 ÷ 12 = 2 ÷ 12 = 1 ÷ 6

= 1÷ R = 1 ÷6

= R = 6 ohm

Now by ohms law , R = V÷I

= I = V÷R =3÷6= 1 ÷ 2A

Also , maximum resistance is provided in series circuit . Thus , the two resistors have to be connected in series to give maximum resistance.

= R = R1 + R2 = 12 + 12 = 24 ohm

Now by ohms law , R = V÷I

= I = V÷R

= 3÷24 = 1÷8A

Now , ratio rato of the power consumed by the resulting combinations with minimum and maximum resistance = P1÷P2

Where P1 = VI =3 × 1÷2 = 3÷2

Also , P2 =VI = 3 × 1 ÷ 8 = 3 ÷8

That means ratio = 3÷2÷3÷8

3÷2×8÷3 = 4:1

Thus the ratio is 4 :1 ...

Hope it helps ...

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