Question

out of 60 watt and 40 watt lamps which one of has high electrical resistance when it used and why?

Answer 1

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Answer 2

The bulb with 40 W has higher electrical resistance.

Explanation:

The power of a bulb is found using the formula

$\text{Power} =\text {voltage} \times \text { current }=(I)^{2} \times \text { resistance } \rightarrow(1)$

Let us consider $P_1 = 60 \ W$ and $P_2 = 40 \ W$, the voltage supplied will be constant denoted as V.

The current produced in 60 W bulb be $I_1$ and current produced in 40 W bulb be $I_2$.

Then from (1)

$I_{1}=\frac{P_{1}}{V}=\frac{60}{V} \rightarrow(2)$

And

$I_{2}=\frac{P_{2}}{V}=\frac{40}{V} \rightarrow(3)$

Let the resistance for 60 W bulb be denoted as $R_1$ and for 40 W bulb is denoted as $R_2$.

From Ohm’s law, we know that,

$I=\frac{V}{R}$

So for $R_1$

$I_{1}=\frac{V}{R_{1}}$

$\Rightarrow R_{1}=\frac{V}{I_{1}}$

Substitute the value of $I_1$ from eqn (1),

$\Rightarrow R_{1}=\frac{V}{\frac{60}{V}}=\frac{V^{2}}{60}$

Similarly, $R_2$ can be calculated,

$\Rightarrow R_{2}=\frac{V}{I_{2}}$

Substitute the value of $I_2$ from eqn (2),

$\Rightarrow R_{2}=\frac{V}{\frac{40}{V}}=\frac{V^{2}}{40}$

Now divide $R_1$ with $R_2$,

$\frac{R_{1}}{R_{2}}=\frac{V^{2}}{60} \times \frac{40}{V^{2}}=\frac{40}{60}=\frac{2}{3}$

$R_{1}=\frac{2}{3} \times R_{2}$

So, the resistance for 60 W bulb is 2/3rd times less than the resistance for 40 W bulb.

Thus, the 40 W bulb has higher resistance.