Question

The length of a 60 W, 240 Ω light bulb filament is 60 cm Remembering that the current in the filament is proportional to the electric field, what is the current in the filament following the doubling of its length?

Answer 1

Answer:

Finally current will be

i = 0.35 A

Explanation:

As we know that power of the bulb is given by the formula

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

now we have

$P = 60 W$

R = 240 ohm

so we have

$60 = \frac{V^2}{240}$

$V = 120 Volts$

now the current in the bulb is given as

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

$i = \frac{120}{240} = 0.5 A$

now when length of the filament is double

so the resistance of the wire also gets double

so we have

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

$60 = \frac{V^2}{480}$

$V = 169.7 volts$

now the current in the bulb is given as

$V = i R$

$169.7 = i(480)$

$i = 0.35 A$