Incandescent bulbs are designed by keeping in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The power rating of a bulb is given by $P = V^2 / R$,where $V$ is the rated voltage (usually $220 \ V$ or $240 \ V$) and $R$ is the resistance of

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The power rating of a bulb is given by $P = V^2 / R$,where $V$ is the rated voltage (usually $220 \ V$ or $240 \ V$) and $R$ is the resistance of the filament at its operating temperature.Since $V$ is constant for all bulbs,we have $P \propto 1 / R$,which implies $R \propto 1 / P$.Therefore,a bulb with a higher power rating will have a lower resistance at its operating temperature.Although the resistance of the filament increases with temperature,the relative order of resistances remains the same at room temperature as it is at the operating temperature.Given $P_{100} > P_{60} > P_{40}$,it follows that $R_{100} < R_{60} < R_{40}$.

Similar Questions

$A$ heater draws a current of $2\,A$ when connected to a $250\,V$ source. The rate of energy dissipation is ............. $W$.

$A$ time-varying current $i$ is passed through a resistance $R$ as shown in the figure. The total heat generated in the resistance is

Assertion: Long distance power transmission is done at high voltage.
Reason: At high voltage supply,power losses are less.

$A$ household supply is $220 \, V$ with a fuse of $9 \, A$. What is the maximum number of $60 \, W$ bulbs that can be connected in parallel and operated simultaneously?

The energy dissipated by a resistor is $10 \, mJ$ in $1 \, s$ when an electric current of $2 \, mA$ flows through it. The resistance is $....... \, \Omega$ (Round off to the Nearest Integer).

Vedclass Test Series

Exam Paper Generator

Online Exam Module