A battery of = 2.70 V and internal resistance R = 0.700 is driving a motor. The

Question

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

Power = F.V = 2*0.5 = 1 W
The expression for the power is,

P =I^2r
r = P/I^2 = 1/I^2

v = I(R+r) = 2.7
I(0.7+1/I^2) = 2.5

0.7I^2 - 2.7 i + 1 = 0

The currents are, 3.44 A and 0.415 A

V for 3.44 = Ir = 0.2907 V

V for 0.415 = Ir = 2.4096 V

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