A bulb is rated 220;V and 100;W. What is the | vedclass

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

Question

(D) The power rating of the bulb is given by $P = 100\;W$ and the voltage rating is $V = 220\;V$.We know the relationship between power,voltage,and re

Vedclass · Accepted Answer

$484\;\Omega$

Vedclass · Answer

(D) The power rating of the bulb is given by $P = 100\;W$ and the voltage rating is $V = 220\;V$.We know the relationship between power,voltage,and resistance is given by the formula $P = \frac{V^2}{R}$.Rearranging the formula to solve for resistance $R$,we get $R = \frac{V^2}{P}$.Substituting the given values: $R = \frac{(220)^2}{100}$.$R = \frac{48400}{100}$.$R = 484\;\Omega$.Therefore,the resistance of the bulb is $484\;\Omega$.

$A$ bulb is rated $220\;V$ and $100\;W$. What is the resistance of this bulb?

Similar Questions

$A$ series circuit has only one conducting path for the electrons that move through it; a parallel circuit has multiple conducting paths.

The amount of charge passing through a cell in $4 \, s$ is $12 \, C$. Find the current supplied by the cell. (in $, A$)

Power of a lamp is $60\, W$. Find the energy in joules consumed by it in $1\, s$. (in $, J$)

What is the resistance of an ideal voltmeter?

Three $2\,\Omega$ resistors,$A$,$B$,and $C$,are connected as shown in the figure. Each of them dissipates energy and can withstand a maximum power of $18\,W$ without melting. Find the maximum current (in $A$) that can flow through the circuit such that none of the resistors melt.

Vedclass Test Series

Exam Paper Generator

Online Exam Module