Question

Two identical resistors of 12$\Omega$ each are connected to a battery of 3V.
Calculate the ratio of the power consumed by the resulting combination with minimum resistance and maximum resistance! ​

Answer 1

Let a Conductor A and Conductor B of 12 ohm each.

In series combination,

R = R₁ + R₂

R = ( 12 + 12 ) ohm

R = 24 Ω

i = V/R

i = 3/24 = 1/8 A

P = Vi

Power = 3 × 1/8 W or 3/8 W

In parallel combination,

1/R = 1/R₁ + 1/R₂

1/R = 1/12  1/12

1/R = 2/12 ohm

1/R = 1/6 or

R = 6 Ω

i = V/R

i = 3/6 or 1/2 A

Power = 3 × 1/2 W or 3/2 W

Since question demands minimum to maximum ratio,

Here, parallel resistance is minimum while series one is maximum.

Therefore, Ratio = ( 3/2 ) ÷ ( 3/8 )

= 3/2 × 8/3

= 4 / 1

Ratio = 4 : 1

Answer 2

answer : 4 : 1

explanation :

Two identical resistors of 12 ohm each are connected to a battery 3 volt.

minimum resistance is possible only when both resistors are connected in parallel combination.

so, Req = 12 × 12/(12 + 12) = 6 ohms.

Maximum resistance is possible only when both the resistors are connected in series combination.

so, Req = 12 + 12 = 24 ohms.

we know, power consumed by wire is given by, P = V²/R

hence, Power is inversely proportional to resistance.

so,

$\frac{P_1}{P_2}=\frac{R_2}{R_1} \:$

$or, \frac{P_1}{P_2}=\frac{24}{6}=\frac{4}{1} \: </p><p>$