A 500 Wheating unit is designed to operate with an applied potential difference of115 V. (a) By what percentage will its heat output drop if the applied potential difference drops to110 V? Assume no change in resistance. (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than that calculated in (a)?

The power of the heating unit, P = 500 W

Initial potential difference,${\mathrm{V}}_1=115\mathrm{V}$

The new potential difference,${\mathrm{V}}_2=110\mathrm{V}$

The force is proportional to the square of the potential. From this calculate the percentage change in performance. As temperature decreases, resistance also decreases, but power is inversely proportional to resistance, so as resistance decreases, power increases. Thus, the actual drop in will be smaller if you take into account the temperature dependence of the resistance.

Formula:

The power generated due to a potential difference in a battery,

$\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}$ ….. (1)

Here, V is the potential difference and R is the resistance.

From equation (1), you can get that power is directly proportional to the square of the potential difference. Thus, it gives

$∆\mathrm{P}\infty ∆{\mathrm{V}}^2$

Now, the differential equation of the power using the above condition can be given as:

$∆\mathrm{P}=2\mathrm{V}∆\mathrm{V}$

Where, $∆\mathrm{P}$is the change in power and $∆\mathrm{V}$is the change in the potential.

Thus, you can calculate percentage change in power as follows:

$$\begin{gathered} ∆\mathrm{P}/\mathrm{P}=2∆\mathrm{V}/\mathrm{V} \\ =\frac{2\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)}{{\mathrm{V}}_1} \end{gathered}$$

Substitute known valued in the above equation.

$$\begin{gathered} ∆\mathrm{P}/\mathrm{P}=\frac{2\left(110-115\right)}{110} \\ =-0.086 \\ =-8.6\% \end{gathered}$$

Hence, the value of the percentage is 8.6 %.

At lower temperature, R also decreases, but P is inversely proportional to resistance considering equation (1).

Thus, as resistance decreases, power increases. i.e.

$\mathrm{P}\infty {\mathrm{R}}^{-1}$

Hence, the actual drop in P will be smaller when the temperature dependence of the resistance taken in to consideration.