A 20.70 V battery is used to supply current to a 25 kN resistor. Assume the voltage drop across any wires used for connections is negligible. (a) What is the current (in mA) through the resistor? 8.28e-7 X mA (b) What is the power dissipated by the resistor (in mW)? 0.0 X mW (c) What is the power input from the battery (in mW), assuming all the electrical power is dissipated by the resistor? mW (d) What happens to the energy dissipated by the resistor? O The energy dissipated by the resistor is converted to elastic potential energy. O The energy dissipated by the resistor is converted to electric potential energy. O The energy dissipated by the resistor is converted to heat. The energy dissipated by the resistor is converted to gravitational potential energy.

A 20.70 V battery is used to supply current to a 25 kN resistor. Assume the voltage drop across any wires used for connections is negligible. (a) What is the current (in mA) through the resistor? 8.28e-7 X mA (b) What is the power dissipated by the resistor (in mW)? 0.0 X mW (c) What is the power input from the battery (in mW), assuming all the electrical power is dissipated by the resistor? mW (d) What happens to the energy dissipated by the resistor? O The energy dissipated by the resistor is converted to elastic potential energy. O The energy dissipated by the resistor is converted to electric potential energy. O The energy dissipated by the resistor is converted to heat. The energy dissipated by the resistor is converted to gravitational potential energy.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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