(8c27p110) A battery of = 2.10 V and internal resistance R = 0.600 is driving a

Question

(8c27p110) A battery of = 2.10 V and internal resistance R = 0.600 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.

Please solve using my numbers.

Explanation / Answer

R total = R +0.6
I total = 2.1/(R +0.6)
Voltage across R =
VL = 2.1R/(R +0.6)

power = 2.1/(R +0.6) x 2.1R/(R +0.6) = 1
2.1²R / (R +0.6)² = 1
2.1²R = (R +0.6)²
4.41R = R² + 1.2R + 0.36
R² - 3.21R + 0.36 = 0

quadratic equation:

R = 3.09, 0.116 ohms

case 1, R = 3.09 ohms
I = 2.1/(R +0.6) = 0.569amp

case 2, R = 0.116 ohms
I = 2.1/(R +0.6) = 2.932 amps

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