Question

There are two electric bulbs marked 60 W, 240 V and 40 W, 240 V respectively. Which one
them has higher resitance ? Calculate.​

Answer 1

For first bulb P=60W V=220V

P,=VI=V(V/R)'2 =V2/R ,,R=V2/P=240X240/60=960 Ohm. For bulb 2 =240x240/40=1440 Ohm Bulb 2 has higher resistance

Answer 2

Answer:

The bulb having a power of 40W has higher resistance.

Explanation:

The power of an electric bulb is given as,

$P=\frac{V^2}{R}$           (1)

Where,

P=power of the bulb

V=volatge across an electric bulb

R=resistance of an electric bulb

From the question we have,

P₁=60W

V₁=240v

P₂=40W

V₂=240v

P₁=60W

By using equation (1) we get;

$60=\frac{(240)^2}{R}$

$R_1=\frac{240\times 240}{60}=960\Omega$          (2)

P₂=40W

By using equation (1) we get;

$40=\frac{(240)^2}{R}$

$R_2=\frac{240\times 240}{40}=1440\Omega$      (3)

From the equation (2) and equation (3) we get;

$R_2 > R_1$           (4)

Hence, the bulb having a power of 40W has higher resistance.