An unknown resistor is connected between the terminals of a 3.00 Vbattery. Energy is dissipated in the resistor at the rate of 0.540 W.The same resistor is then connected between the terminals of a 1.50 Vbattery. At what rate is energy now dissipated?

1. The initial potential difference applied$V_1=3.00V$
2. The newpotential difference applied$V_2=1.50V$
3. The initial rate of dissipation is,$P_1=0.540\mathrm{W}$

The power or rate of energy transfer, in an electrical device across which a potential difference is equal to the product of current and potential difference. It can also be defined as energy transferred per unit time.

The rate of energy dissipation of an electrical device is calculated from the potential difference across it and its resistance using the corresponding formula.

Formulae:

The rate at which the energy is dissipated from the given state, $P=\frac{V^2}{R}$ …(i)

Here, P is the power, R is resistance, V is the potential difference.

The rate of energy dissipated in both initial and the case of new applied potential difference can be given using equation (i) as:

$\begin{array}{l}{P}_1=\frac{V_1^2}{R}\\ P_2=\frac{V_2^2}{R}\end{array}$

As the battery is connected across the same resistance, its value R is same in both the cases. Thus, from the above equations, we get

$$\begin{gathered} \frac{V_1^2}{P_1}=\frac{V_2^2}{P_2} \\ {P}_2=P_1\times {\left(\frac{V_2}{V_1}\right)}^2 \\ =0.540W\times {\left(\frac{1.50V}{3.00V}\right)}^2 \\ =0.540\times \frac{1}{4} \\ =0.135W \end{gathered}$$

Hence, the value of the rate of energy dissipated is 0.135 W.