(8c27p110) A battery of ? = 2.70 V and internal resistance R = 0.700 ? is drivin

Question

(8c27p110) A battery of ? = 2.70 V and internal resistance R = 0.700 ? is driving a motor. The motor is lifting a 2.0 N mass at constant speed v = 0.50 m/s. Assuming no energy losses, find the current i in the circuit. Enter the lower current. Enter the higher current. Find the potential difference V across the terminals of the motor for the lower current. Find the potential difference V across the terminals of the motor for the higher current. Discuss the fact that there are two solutions to this problem.

Explanation / Answer

Mass is not in Newtons, I'll assume you mean force of 2 Newtons. In one second, the motor lifts with a force of 2 newtons over a distance of 0.5 meter, which is work/energy of Fd = 1 Joule. Since this occurs every second, this is 1 Joule/second or 1 Watt of power. Now we have a circuit with 2.7 volts in series with 0.5 ohms and a R dissipating 1 watt. R total = R +0.5 I total = 2.7/(R +0.5) Voltage across R = VL = 2.7R/(R +0.5) power = 2.7/(R +0.5) x 2.7R/(R +0.5) = 1 2.7

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