(8c27p110) A battery of = 2.70 V and internal resistance R = 0.500 is driving a

Question

(8c27p110) A battery of = 2.70 V and internal resistance R = 0.500 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. Submit Answer Tries 0/99

Enter the higher current. Submit Answer Tries 0/99

Find the potential difference V across the terminals of the motor for the lower current. Submit Answer Tries 0/99

Find the potential difference V across the terminals of the motor for the higher current. Submit Answer Tries 0/99

Discuss the fact that there are two solutions to this problem.

Could you please take extra care to show the alegerbra thank you

Explanation / Answer

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
2.7²R / (R +0.5)² = 1
2.7²R = (R +0.5)²
7.29R = R² + R + 0.25
R² - 6.29R + 0.25 = 0

quadratic equation:
to solve ax² + by + c = 0
x = [-b ±(b²-4ac)] / 2a

R = [6.29 ±(39.6-1)] / 2
R = [6.29 ±6.21] / 2
R = 6.25, 0.04 ohms

case 1, R = 6.25 ohms
I = 2.7/(R +0.5) = 0.4 amp

case 2, R = 0.04 ohms
I = 2.7/(R +0.5) = 5 amps

V (lower) = 2.7 - 0.5*0.4 = 2.5 V

V (higher) = 2.7 - 0.5*5 = 0.2 V

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