Question

Two identical resistors each of resistance 12 ohm are connected 1) in series 2) in parallel, in line to a battery of 6volts. Calculate the ratio of power consumed in the combination of resistors in the two case

Answer 1

ratio of power is 4:1

Answer 2

Answer: The ratio of the power consume is 1:8

Explanation:

Given that,

Two identical resistor each of resistance 12 ohm.

Let us consider $R_{1}$ and $R_{2}$ are two identical resistor.

Firstly, both resistors connected in the parallel

Then, the resistance is

$\dfrac{1}{R} = \dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$

$\dfrac{1}{R} = \dfrac{1}{12\Omega}+\dfrac{1}{12\Omega}$

$\dfrac{1}{R} = \dfrac{1}{6\Omega}$

$R= 6\Omega$

Now, the power consumed

$P = \dfrac{V^{2}}{R}$

$P = \dfrac{36 Volt}{6\Omega}$

$P = 6\ watt$

Now, both resistors connected in the series

$R' = R_{1} + R_{2}$

$R' = 12\Omega +12\Omega$

$R' = 24\Omega$

The power consume

$P' = \dfrac{36\ volt}{24\Omega}$

$P' = 1.5\ watt$

Now, the ratio of power consumed in the combination of resistors in series and parallel

$\dfrac{P'}{P} = \dfrac{1.5}{12}$

$\dfrac{P'}{P} = \dfrac{1}{8}$

Hence, the ratio of the power consume is 1:8