Question

Two bulbs have the following ratings (a) 40 w at 22o V, and (b) 40 wat 110 V. The filament of which bulb has a higher resistance? What is the ratio of their resistance?

Answer 1

We know power of a bulb P= V^2/ R

For bulb A, Power=40W at V= 220W

R1=V²/P = (220) ² / 40 = 1210 ohms

For Bulb B, Power= 40W and V=110W

R2= V² /P = (110) ² / 40 = 302.5

Now R1:R2 = 1210:302.5 = 1/4

Hene the ratio R1/R2= ¼

Answer 2

The resistance of bulb (a) has higher resistance.

The ratio of their resistance is 4.

Explanation:

To calculate the resistance of the bulb use the relation between power of the bulb, resistance and voltage as

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

Here, R resistance of the bulb and V is the voltage.

Given for bulb (a), P = 40 W and V = 220 V.

Substitute these given values, we get

$40W=\frac{(220)^{2} }{R}$

$R=\frac{(220)^{2} }{40W}=1210 ohm$

Similarly for bulb (b), P =40 W and V = 110 V.

$40 W=\frac{(110)^{2} }{R}$

R = 302.5 ohm

Thus, it is clear from the result the for bulb (a)has a higher resistance.

The ratio of their resistance,

$=\frac{1210}{302.5}$

 = 4

#Learn More:

Topic : Power and resistance relation.

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