Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60 and R40, respectively, the relation between these resistances is

Question

Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60 and R40, respectively, the relation between these resistances is (A) (1/R100) = (1/R40) + (1/R60) (B) R100 = R40 + R60  (C) R100 > R60 > R40  (D) (1/R100) > (1/R60) > (1/R40)

Tardigrade Admin · Accepted Answer

R = (V2/P) or R propto (1/P) ∴ (1/R100) > (1/R60) > (1/R40)

$\frac{1}{R _{100}} = \frac{1}{R _{40}} + \frac{1}{R _{60}}$

$R_{100} = R_{40} + R_{60}$

$R_{100} > R_{60} > R_{40}$

$\frac{1}{R _{100}} > \frac{1}{R _{60}} > \frac{1}{R _{40}}$

$R = \frac{V ^{2}}{P} or R \propto \frac{1}{P}$
$∴ \frac{1}{R _{100}} > \frac{1}{R _{60}} > \frac{1}{R _{40}}$

Q. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100W,60W and 40W bulbs have filament resistances R100​,R60​ and R40​, respectively, the relation between these resistances is

R100​1​=R40​1​+R60​1​

R100​=R40​+R60​

R100​>R60​>R40​

R100​1​>R60​1​>R40​1​

R=PV2​orR∝P1​ ∴R100​1​>R60​1​>R40​1​

$\frac{1}{R _{100}} = \frac{1}{R _{40}} + \frac{1}{R _{60}}$

$R_{100} = R_{40} + R_{60}$

$R_{100} > R_{60} > R_{40}$

$\frac{1}{R _{100}} > \frac{1}{R _{60}} > \frac{1}{R _{40}}$

$R = \frac{V ^{2}}{P} or R \propto \frac{1}{P}$
$∴ \frac{1}{R _{100}} > \frac{1}{R _{60}} > \frac{1}{R _{40}}$