Question

one electric bulb is rated 40 W and 240 V and other 25 W and 240 V. Which bulb has higher resistance and how many times?

Answer 1

use power as V.V/R and get resistance corresponding to powers and then gat ratio.

Answer 2

Answer: The higher resistance is $r_{2}$ and $r_{2}$ is 1.6 times of $r_{1}$

Explanation:

Given that,

Power of first bulb $P_{1} = 40 W$

Voltage of first bulb $V_{1} = 240 V$

Power of second bulb $P_{2}=25 W$

Voltage of second bulb $V_{2} = 240 V$

We know that,

The power is the product of the voltage and current.

$P = VI$

Now, the resistance of first bulb

$R_{1} = \dfrac{V^{2}}{P}$

$R_{1} =\dfrac{240\times 240}{40}$

$R_{1} = 1440 ohm$

The resistance of the second power

$R_{2} = \dfrac{240\times240}{25}$

$R_{2} = 2304 ohm$

Therefore,

$R_{2}>R_{1}$

$\dfrac{R_{2}}{R_{1}} = \dfrac{2304}{1440}$

$\dfrac{R_{2}}{R_{1}} = 1.6$

$R_{2} = 1.6 R_{1}$

Hence, the higher resistance is $r_{2}$ and $r_{2}$ is 1.6 times higher then $r_{1}$