An electric heater consists of a nichrome coi | vedclass

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

Question

(A) The power consumed by an electric heater is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the coi

Vedclass · Accepted Answer

more than $1 \,kW$

Vedclass · Answer

(A) The power consumed by an electric heater is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the coil.Given that the voltage $V = 220 \,V$ remains constant.From the formula,we see that $P \propto \frac{1}{R}$.When a part of the coil is cut off,the length of the wire decreases. Since resistance $R = \rho \frac{L}{A}$,where $L$ is the length,decreasing the length $L$ results in a decrease in resistance $R$.Since the resistance $R$ decreases,the power $P$ consumed by the heater must increase.Therefore,the new power consumed will be more than $1 \,kW$.

An electric heater consists of a nichrome coil and runs under $220 \,V$,consuming $1 \,kW$ power. Part of its coil burned out and it was reconnected after cutting off the burnt portion. The power it will consume now is

Similar Questions

Two wires have resistances of $2 \ \Omega$ and $4 \ \Omega$ and are connected to the same voltage. The ratio of heat dissipated in the resistances is:

An electric bulb of $100\, W$ is connected to a supply of electricity of $220\, V$. The resistance of the filament is ............. $\Omega$.

According to Joule's law,if the potential difference across a conductor having a material of specific resistance remains constant,then the heat produced in the conductor is directly proportional to

$A$ $100\, W$ bulb working on $200\, V$ and a $200\, W$ bulb working on $100\, V$ have:

Two electric bulbs have power ratings $P_1$ and $P_2$ at voltage $V$. They are connected in parallel to a voltage source $V$. What is the total power consumed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module