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A battery with an emf of 12.0 V shows a terminal voltage of 11.6 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel. What is the battery's internal resistance?

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Kazeemsodikisola

Answer:

Explanation:

Check attachment for solution

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temdan2001

Answer: 0.6 ohms

Explanation:

For each light bulb,

P = V^2 / R

R = V^2 / P

R = 12.0^2 / 4 = 36 ohms

Where p = power

V = voltage

R = resistance

If connected in parallel, total resistance = 36 / 2 = 18 ohms

Internal resistance = voltage drop / current

Please find the attached file for the solution.

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