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When a resistor with resistance r is 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.) what power does the resistor consume if it is connected to a 12.6-v car battery? assume that r remains constant when the power consumption changes?

Respuesta :

AL2006
The key to unlock this complicated mystery is to know the value of the resistor.  We can calculate it from the description of Scene 1:

Power dissipated by a resistance = (voltage)² / (resistance)

Resistance = (voltage)² / (power)

In scene 1:

Resistance = (1.5 v)² / (0.0625 w)

Resistance = 36 Ohms

That helps a bunch !  It's all we need to solve the second part.

Power = (voltage)² / (resistance)

Power = (12.6 v)² / (36 Ω)

Power = 4.41 watts
=====================================

Another (easier) way:

Power = (voltage)² / (resistance)

SO ... if the resistance doesn't change, then the power
is proportional to the square of the voltage.

(power₂ / power₁) = (voltage₂ / voltage₁)²

power₂ = (power₁) ·  (voltage₂ / voltage₁)²

power₂ = (0.0625 w) · (12.6v / 1.5v)²

power₂ = (0.0625 w) · (8.4)²

power₂ = (0.0625 w) · (70.56)

power₂ = 4.41 w
Lanuel

The power consumed by the resistor if it is connected to a 12.6-V car battery is 4.41 Watts.

Given the following data:

  • Voltage = 1.5 Volts
  • Power = 0.0625 Watts
  • Emf = 12.6 Volts

To find the power consumed by the resistor if it is connected to a 12.6-V car battery:

First of all, we would determine the value of this resistor by using the following formula;

[tex]Power = \frac{Voltage^2}{Resistance}[/tex]

Substituting the parameters into the formula, we have;

[tex]0.0625 = \frac{1.5^2}{Resistance}[/tex]

Cross-multiplying, we have:

[tex]0.0625(Resistance) = 1.5^2\\\\Resistance = \frac{1.5^2}{0.0625}\\\\Resistance = \frac{2.25}{0.0625}[/tex]

Resistance = 36 Ohms

Now, we can find the power consumed by the resistor if it is connected to a 12.6-V car battery (resistance remained constant):

[tex]Power = \frac{Voltage^2}{Resistance}[/tex]

Substituting the parameters into the formula, we have;

[tex]Power = \frac{12.6^2}{36}\\\\Power = \frac{158.76}{36}[/tex]

Power = 4.41 Watts.

Therefore, the power consumed by the resistor if it is connected to a 12.6-V car battery is 4.41 Watts.

Read more: https://brainly.com/question/23534762

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