In a simple electric circuit, Ohm's law states that V=IRV=IR, where V is the voltage in volts, I is the current in amperes, and R is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.04 ohms per second. When the resistance is 100 ohms and the current is 0.03 amperes, at what rate is the current changing?
In a simple electric circuit, Ohm's law states that V=IRV=IR, where V is the voltage in volts, I is the current in amperes, and R is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.04 ohms per second. When the resistance is 100 ohms and the current is 0.03 amperes, at what rate is the current changing?
Chapter3: Polynomial Functions
Section3.5: Mathematical Modeling And Variation
Problem 7ECP: The kinetic energy E of an object varies jointly with the object’s mass m and the square of the...
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In a simple electric circuit, Ohm's law states that V=IRV=IR, where V is the voltage in volts, I is the current in amperes, and R is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.04 ohms per second. When the resistance is 100 ohms and the current is 0.03 amperes, at what rate is the current changing?
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