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

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

Question

(A) The heat produced by a heater is given by the formula $H = \frac{V^2 t}{R}$,where $V$ is the voltage,$t$ is time,and $R$ is the resistance.Since t

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

(A) The heat produced by a heater is given by the formula $H = \frac{V^2 t}{R}$,where $V$ is the voltage,$t$ is time,and $R$ is the resistance.Since the voltage $V$ and time $t$ are constant,the heat produced is inversely proportional to the resistance: $H \propto \frac{1}{R}$.When the coil is cut into two equal parts,the resistance of each part becomes $R' = \frac{R}{2}$.Let $H_{Full}$ be the heat produced by the full coil with resistance $R$,and $H_{Half}$ be the heat produced by the half coil with resistance $R/2$.Then,$\frac{H_{Half}}{H_{Full}} = \frac{R}{R_{Half}} = \frac{R}{R/2} = \frac{2}{1}$.Therefore,the ratio of heat produced is $2 : 1$.

$A$ heater coil is cut into two equal parts and only one part is used in the heater. What is the ratio of the heat produced by the half coil to that produced by the original full coil?

Similar Questions

The resistance of a hot tungsten filament is about $10$ times its cold resistance. What will be the resistance of a $100\, W$ and $200\, V$ lamp when not in use? (in $\Omega$)

In the circuit shown,the point '$B$' is earthed. The potential at the point '$A$' is ............. $V$.

The emf of a cell of internal resistance $2 \ \Omega$ is measured using a voltmeter of resistance $998 \ \Omega$. The error in the emf measured is (in $\%$)

In the circuit shown,the capacitance of the capacitor is $2\,\mu F$. Find the current flowing through the $2\,\Omega$ resistor in $A$.

If each resistance in the figure is $9 \Omega$,then the reading of the ammeter $(A)$ is (in $A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module