Operating voltageResistance of the device -R- Currents flowing in the deviceI - V-R Heat generated per second- HeatH - V times I times t Power given by the deviceP - V times I or P - H-t Reason the change in power230V 57-5 Omega 4A 920J 920WDifference in resistance230V 11-5 Omega 2A 460J 460W Difference in resistance230V 230 Omega 1A 230J 230W Difference in resistance115V 57-5 Omega 2A 230J 230W Difference in resistance460V 57-5 Omega 8A 3680J 3680W Difference in resistanceThe table shows details of an electric heating device designed to work in 230 V- Complete the table by calculating the change in the heat and power on changing the voltage and resistance of the device- Analyze the table and answer the following question-What change happens to power on increasing the resistance without changing the voltage-

Question

Toppr Admin · Accepted Answer

On increasing the resistance without changing the voltage- power decreases-P-dfrac-V-2-R-Since V is constant-P -propto -dfrac-1-R-

Operating voltage	Resistance of the device (R)	Currents flowing in the device $$I = V/R$$	Heat generated per second, Heat $$H = V \times I \times t$$	Power given by the device $$P = V \times I$$ or $$P = H/t$$	Reason for the change in power
230V	57.5 $$\Omega$$	4A	920J	920W	Difference in resistance
230V	11.5 $$\Omega$$	2A	460J	460W	Difference in resistance
230V	230 $$\Omega$$	1A	230J	230W	Difference in resistance
115V	57.5 $$\Omega$$	2A	230J	230W	Difference in resistance
460V	57.5 $$\Omega$$	8A	3680J	3680W	Difference in resistance

On increasing the resistance without changing the voltage, power decreases.$$P=\dfrac{V^2}{R}$$Since V is constant,$$P \propto \dfrac{1}{R}$$

On increasing the resistance without changing the voltage, power decreases.
$$P=\dfrac{V^2}{R}$$
Since V is constant,
$$P \propto \dfrac{1}{R}$$