Question

When a resistor with resistance Ris connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that Rremains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

Short answer

1. The power consumed by the resistor when connected to 12.6 V car battery is 4.41W
2. The voltage of the battery is 13.4V .

Step-by-step solution

  (a) Determination of the power consumed by the resistor when connected to 12.6 V car battery.

The resistance is constant in both cases, and the expression of power is

$p=\frac{V^2}{R}$

Rearrange for R,

$p=\frac{V^2}{R}$

So, for equal resistance,

$$\begin{gathered} \frac{V_1^2}{P_1}=\frac{V_1^2}{P_2}....\left(1\right) \\ \mathrm{Solve}\mathrm{for}\mathrm{P}2, \\ {P}_2=\frac{V_2^2{P}_1}{V_1^{25}} \end{gathered}$$

Substitute all the values in the above equation,

$$\begin{gathered} P_2=(0.0625W){\left[\frac{\left(12.5V\right)}{\left(1.50V\right)}\right]}^2 \\ =4.41W \end{gathered}$$

(b) Determination of the voltage of the battery.

From equation (i), Solve for V₂ as,

$$\begin{gathered} \frac{V_1^2}{P_1}=\frac{V_2^2}{P_2} \\ {V}_2=V_1\sqrt{\frac{P_2}{P_1}} \end{gathered}$$

Substitute all the values in the above equation,

$$\begin{gathered} V_2=(1.50V)\sqrt{\frac{5.00W}{0.0625W}} \\ =13.4V \\ \mathrm{Thus},\mathrm{the}\mathrm{voltage}\mathrm{is}13.4\mathrm{V}. \end{gathered}$$