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Praxis Physics (5266) Practice Tests & Test Prep by Exam Edge - Free Test


Our free Praxis Physics (5266) Practice Test was created by experienced educators who designed them to align with the official Educational Testing Service content guidelines. They were built to accurately mirror the real exam's structure, coverage of topics, difficulty, and types of questions.

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Praxis Physics - Free Test Sample Questions

An emf 1.5 V cell with an internal resistance of 0.1Ω is connected across an R resistor. If the current in the circuit is 2A, what is the energy dissipation rate inside the cell?

Correct Answer:
0.4 w


to determine the energy dissipation rate inside the cell, we need to consider the cell's internal resistance and the current flowing through it. the cell in the problem has an internal resistance (r) of 0.1 ohms and it is supplying a current (i) of 2 amperes.

the power dissipated due to the internal resistance of the cell can be calculated using the formula for power dissipation in resistors: \( p = i^2 \times r \). here, \( p \) represents the power (or energy dissipation rate), \( i \) is the current, and \( r \) is the resistance.

plugging the given values into the formula gives: \[ p = (2 \, \text{a})^2 \times 0.1 \, \omega = 4 \times 0.1 = 0.4 \, \text{watts} \]

therefore, the energy dissipation rate inside the cell, which is essentially the power lost as heat due to the internal resistance, is 0.4 watts. this indicates that while the cell is operating to provide current to the circuit, it is simultaneously losing energy at a rate of 0.4 watts due to its own internal resistance. this is an important consideration in battery-powered systems where energy efficiency is crucial.