A coil develops heat of 800 , cal/sec when 20 | vedclass

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

Question

(C) The rate of heat production is given by $P = \frac{H}{t} = 800 \, cal/sec$. Converting this to $SI$ units (Joules per second or Watts): $P = 800 \

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(C) The rate of heat production is given by $P = \frac{H}{t} = 800 \, cal/sec$. Converting this to $SI$ units (Joules per second or Watts): $P = 800 	imes 4.2 \, J/sec = 3360 \, W$. The formula for electrical power in terms of voltage $V$ and resistance $R$ is $P = \frac{V^2}{R}$. Substituting the given values: $3360 = \frac{20^2}{R}$. $3360 = \frac{400}{R}$. $R = \frac{400}{3360} = \frac{40}{336} \approx 0.119 \, \Omega$. Rounding to two decimal places,we get $R \approx 0.12 \, \Omega$.

$A$ coil develops heat of $800 \, cal/sec$ when $20 \, V$ is applied across its ends. The resistance of the coil is .............. $\Omega$ $(1 \, cal = 4.2 \, J)$.

Similar Questions

How many $60 \,W$ bulbs may be safely run on $220 \,V$ using a $5 \,A$ fuse?

What amount of heat will be generated in a coil of resistance $R$ due to a charge $q$ passing through it if the current in the coil decreases to zero uniformly during a time interval $\Delta t$?

$A$ $10\, V$ storage battery of negligible internal resistance is connected across a $50\,\Omega$ resistor. How much heat energy is produced in the resistor in $1$ hour? (in $J$)

$A$ heater of $220\, V$ heats a volume of water in $5\, \text{min}$ time. $A$ heater of $110\, V$ heats the same volume of water in ............... $\text{min}$.

In India,electricity is supplied for domestic use at $220\,V$. It is supplied at $110\,V$ in the $USA$. If the resistance of a $60\,W$ bulb for use in India is $R$,then the resistance of a $60\,W$ bulb for use in the $USA$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module