Q43P - Fundamentals Of Physics

Question

An unknown resistor is connected between the terminals of a 3.00 V battery. 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 V battery. At what rate is energy now dissipated?

Step-by-Step Solution

Answer

The rate of energy dissipated now is 0.135 W.

Step 1: The given data

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

Step 2: Significance of the energy dissipation

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.

Step 3: Determining the rate of energy dissipated

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 

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

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