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?

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 VoltsPower = 0.0625 WattsEmf = 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;

Substituting the parameters into the formula, we have;

Cross-multiplying, we have:

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):

Substituting the parameters into the formula, we have;

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: link

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