A heater coil is cut into two parts of equal | vedclass

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

Question

(A) The power $P$ dissipated by a heater coil connected to a constant voltage $V$ is given by $P = \frac{V^2}{R}$,where $R$ is the resistance of the c

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(A) The power $P$ dissipated by a heater coil connected to a constant voltage $V$ is given by $P = \frac{V^2}{R}$,where $R$ is the resistance of the coil.Since $P \propto \frac{1}{R}$ and the resistance $R$ is directly proportional to the length $l$ of the wire $(R = \rho \frac{l}{A})$,we have $R \propto l$.Therefore,$P \propto \frac{1}{l}$.Let the original length be $l$ and the original resistance be $R$. The new length is $l' = \frac{l}{2}$,so the new resistance is $R' = \frac{R}{2}$.The power produced by the original coil is $P = \frac{V^2}{R}$.The power produced by the half coil is $P' = \frac{V^2}{R'} = \frac{V^2}{R/2} = 2 \frac{V^2}{R} = 2P$.Thus,the ratio of the heat produced by the half coil to that by the original coil is $P' : P = 2:1$.

$A$ heater coil is cut into two parts of equal length and one of them is used in the heater. The ratio of the heat produced by this half coil to that by the original coil is

Similar Questions

If current in an electric bulb changes by $1\%$,then the power will change by ............ $\%$

If the resistance of the filament increases with temperature,what will be the power dissipated in a $220\, V- 100\, W$ lamp when connected to a $110\, V$ power supply?

Which of the following is not equal to watt?

An electric bulb rated $50 \, W-200 \, V$ is connected across a $100 \, V$ supply. The power dissipation of the bulb is: (in $ \, W$)

The emf of a car battery is $12 \ V$. If the internal resistance of the battery is $0.4 \ \Omega$,then the maximum power drawn from the battery is . . . . . . $W$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module