Question

6. (i) The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if

it operates on a 6 V battery (Neglect the change in resistance due to unequal heating of the filament in the

two cases).

(ii) A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length.

Calculate the resistance of the wire in the new Situation.​

Answer 1

Explanation:

1st solution

P= V^2 / R

24 watt = (12 V) ^2 / R

R= 144 / 24

if the effective wattage is P1 when operates on a 6 V battery then,

P1 = (6 v)^2 /R

now put the value of R

P1 = 36 * 24/ 144

P1 = 6W

2nd solution

Suppose the length of R1 =20Ω resistance wire is I, its area of cross-section and its resistivity is ρ Then,

R1= ρl / A = 20Ω

When length becomes twice (2l), then its area of cross-section will become half (A/2). The new resistance

R2 = ρ.2l / A/2

R2 = 4.ρl / A

R2 = 4* 20

R2 = 80Ω

here is ur answer

Answer 2

Explanation:

R1= 20 ohm and

R2= 80 ohm

hope it helps