Question

two identical resistors of 12 ohm each are connected to a battery of 3 volt calculate the ratio of the power consumed by the resulting combinations with minimum resistance resistance and maximum resistance

Answer 1

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

here, $R_2=24\Omega,R_1=6\Omega$

or, $\frac{P_1}{P_2}=\frac{24}{6}=\frac{4}{1}$

Answer 2

Ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance is 4:1.

In between the combinations of the resistance, the series combination provides maximum resistance and the parallel combination provides minimum resistance.

R1 = R2 = 12Ω.

V = 3 Volt.

• In case of Parallel combination, R = R1 * R2/ (R1 + R2) = 6Ω. Power consumption across this connection, P1 = V²/R = 3²/6 = 3/2 W.
• In case of Series combination, R = R1 + R2 = 24Ω. Power consumption across this connection, P2 = V²/R = 3²/24 = 3/8 W.

Required ratio of the power consumption,

P1 : P2 = 3/2 : 3/8 = 4 : 1.