An electric bulb marked 220 V , 100 W will get fused if it is made to consume 150 W or more. What maximum voltage can the bulb withstand?

Correct answer: D. 269.4 V The resistance of the bulb is given by R = (V^2)/(P) . Substituting the rated values, R = (220^2)/(100) = 484 . The maximum power the bulb can withstand before fusing is P_ = 150 W . The maximum voltage V_ is given by P_ = (V_ ^2)/(R) . V_ = P_ R = 150 × 484 = 72600 269.4 V .

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Thermal and Chemical Effects of Electric Current

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Thermal and Chemical Effects of Electric Current

An electric bulb marked 220 V , 100 W will get fused if it is made...

An electric bulb marked

220 \text{ V}

100 \text{ W}

will get fused if it is made to consume

150 \text{ W}

or more. What maximum voltage can the bulb withstand?

Options

A $146.7 \text{ V}$
B $495.0 \text{ V}$
C $330.0 \text{ V}$
D $269.4 \text{ V}$

Correct answer

D. $269.4 \text{ V}$

Step-by-step solution

The resistance of the bulb is given by

R = \dfrac{V^2}{P}

.

Substituting the rated values,

R = \dfrac{220^2}{100} = 484 \Omega

.

The maximum power the bulb can withstand before fusing is

P_\max = 150 \text{ W}

.

The maximum voltage

V_\max

is given by

P_\max = \dfrac{V_\max^2}{R}

V_\max = \sqrt{P_\max R} = \sqrt{150 \times 484} = \sqrt{72600} \approx 269.4 \text{ V}

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