Chapter 4: Q34E (page 843)

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?

Expert verified

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

Step by step solution

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,

$\frac{V_{1}^{2}}{P_{1}} = \frac{V_{1}^{2}}{P_{2}} . . . . (1) Solve for P 2, P_{2} = \frac{V_{2}^{2} P_{1}}{V_{1}^{25}}$

Substitute all the values in the above equation,

$P_{2} = (0.0625 W) {[\frac{(12.5 V)}{(1.50 V)}]}^{2} = 4.41 W$

From equation (i), Solve for V₂ as,

$\frac{V_{1}^{2}}{P_{1}} = \frac{V_{2}^{2}}{P_{2}} V_{2} = V_{1} \sqrt{\frac{P_{2}}{P_{1}}}$

Substitute all the values in the above equation,

$V_{2} = (1.50 V) \sqrt{\frac{5.00 W}{0.0625 W}} = 13.4 V Thus, the voltage is 13.4 V .$

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