Chapter 18, Problem 7Q

To determine

To explain:

On increasing the resistance of a small immersion heater, will the heating process speed-up or slow-down

Answer to Problem 7Q

Solution:

It will slow-down the heating process.

Formula used:

      • The Ohm’s law tells that the ratio of voltage drop (V)and the current (I)passing through a resistance (R)is ideally constant which is equal to the value of that resistance itself, i.e.,

               $R=\frac{V}{I}$

      • The power consumption by a resistor is given as,

               $P=I^{2}R=V\times I=\frac{V^2}{R}$

Explanation of Solution

For the immersion rod, the supply voltage will remain constant. So, with the help of Ohm’s law $R=\frac{V}{I}$, we can say that when the resistance is increased, keeping the voltage to be constant, the current will decrease. Now, the equation $P=V\times I=\frac{V^2}{R}$ tells that as the voltage is constant and resistance is increased, therefore the power-consumption in the resistance will decrease.

Now, because the power-consumption in the resistance of the immersion heater is decreasing, it will slow-down the heating process.