An electric bulb is designed to draw power $P_0$ at voltage $V_0$. If the voltage is $V$,it draws a power $P$. Then:

  • A
    $P = \left( \frac{V_0}{V} \right)^2 P_0$
  • B
    $P = \left( \frac{V}{V_0} \right)^2 P_0$
  • C
    $P = \left( \frac{V}{V_0} \right) P_0$
  • D
    $P = \left( \frac{V_0}{V} \right) P_0$

Explore More

Similar Questions

The resistance of a device component decreases as the current through it increases and it is described by the relation,$R = \frac{0.2 I}{I-4}$,where $I$ is the current. Determine the minimum power delivered. (Assume $I > 4$) (in $W$)

$A$ room heater is rated $400 \text{ W}$, $220 \text{ V}$. If the supply voltage drops to $200 \text{ V}$, what will be the power consumed (approximately) (in $\text{ W}$)?

An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true?
$(A)$ The temperature distribution over the filament is uniform
$(B)$ The resistance over small sections of the filament decreases with time
$(C)$ The filament emits more light at higher band of frequencies before it breaks up
$(D)$ The filament consumes less electrical power towards the end of the life of the bulb

You are given three bulbs of $25\, W$,$40\, W$,and $60\, W$. Which of them has the lowest resistance?

$A$ bulb of $100 \text{ W}$ rating is connected with $220 \text{ V}$ supply. The resistance of the bulb is . . . . . . .

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo