Question

16. Two electric bulbs marked 100W 220V and 200W 200V have tungsten

filament of same length. Which of the two bulbs will have thicker
filament?
​

Answer 1

The second bulb will have thicker filament

Step-by-step explanation:

Given;

First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V

Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V

Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m

Resistance of the first bulb:

`P = IV = (V)/(R) .V = (V^2)/(R) \\\\R = (V^2)/(P) \\\\R_1 = (V_1^2)/(P_1) = ((220)^2)/(100) = 484 \ ohms`

Resistance of the second bulb:

`R_2 = (V_2^2)/(P_2) = ((200)^2)/(200) = 200 \ ohms`

Resistivity of the tungsten filament is given by the following equation;

`\rho = (RA)/(L)`

where;

L is the length of the filament

R is resistance of each filament

A is area of each filament

`A = \pi r^2`

where;

r is the thickness of each filament

`\rho = (R (\pi r^2))/(L) \\\\(\rho L)/(\pi) = Rr^2 \\\\Recall ,\ (\rho L)/(\pi) \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\((r_1)/(r_2) )^2 = (R_2)/(R_1) \\\\(r_1)/(r_2) = \sqrt{(R_2)/(R_1) } \\\\(r_1)/(r_2) = \sqrt{(200)/(484) } \\\\(r_1)/(r_2) = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1`

Therefore, the second bulb will have thicker filament