Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, $100 \mathrm{~W}, 60 \mathrm{~W}$ and 40 W bulbs have filament resistances $\mathrm{R}_{100}, \mathrm{R}_{60}$ and $\mathrm{R}_{40}$, respectively, the relation between these resistances is

(D) $\frac{1}{\mathrm{R}_{100}}>\frac{1}{\mathrm{R}_{60}}>\frac{1}{\mathrm{R}_{40}}$

(A) $\frac{1}{\mathrm{R}_{100}}=\frac{1}{\mathrm{R}_{40}}+\frac{1}{\mathrm{R}_{60}}$

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Current Electricity

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

R_{100}, R_{60}

and

R_{40}

, respectively, the relation between these resistances is

(A) $\frac{1}{R _{100}} = \frac{1}{R _{40}} + \frac{1}{R _{60}}$
(B) $R_{100} = R_{40} + R_{60}$
(C) $R_{100} > R_{60} > R_{40}$
(D) $\frac{1}{R _{100}} > \frac{1}{R _{60}} > \frac{1}{R _{40}}$ ✓ Correct

Answer: (D)

Solution

At room temperature (before heating), the power rating determines the resistance. For a bulb rated at power

P

operating at voltage

V

P = V^{2} / R

, so

R = V^{2} / P

. Higher power means lower resistance. Therefore,

R_{100} < R_{60} < R_{40}

, which gives

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

. The answer is (D).

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Current Electricity

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