An electric iron of resistance 20, Omega take | vedclass

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

Question

(D) The amount of heat $(H)$ produced is given by Joule's law of heating as $H = I^2Rt$.Given:Resistance,$R = 20\, \Omega$Current,$I = 5\, A$Time,$t =

Vedclass · Accepted Answer

$1.5 	imes 10^4\, J$

Vedclass · Answer

(D) The amount of heat $(H)$ produced is given by Joule's law of heating as $H = I^2Rt$.Given:Resistance,$R = 20\, \Omega$Current,$I = 5\, A$Time,$t = 30\, s$Using the formula:$H = (5)^2 	imes 20 	imes 30$$H = 25 	imes 20 	imes 30$$H = 500 	imes 30$$H = 15000\, J$$H = 1.5 	imes 10^4\, J$Therefore,the amount of heat developed in the electric iron is $1.5 	imes 10^4\, J$.

An electric iron of resistance $20\, \Omega$ takes a current of $5\, A$. Calculate the heat developed in $30\, s$.

Similar Questions

What is the highest and lowest total resistance that can be secured by combinations of four coils of resistance $4 \,\Omega, 8 \,\Omega, 12 \,\Omega$ and $24 \,\Omega$?

How much energy is given to each coulomb of charge passing through a $6\, V$ battery (in $, J$)?

Which of the following terms does not represent electrical power in a circuit?

Calculate the number of electrons constituting one coulomb of charge.

What determines the rate at which energy is delivered by a current?

Vedclass Test Series

Exam Paper Generator

Online Exam Module