Question

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.

(a) Enter the lower current.

(b) Enter the higher current.

(c) Find the potential difference V across the terminals of the motor for the lower current.

(d) Find the potential difference V across the terminals of the motor for the higher current.

Answer 1

a)

power = force * velocity = 2 * 0.50 = 1.00 Watt

.

ohms law:       ? = I (r +R)         where   r is resistance of motor (unknown)

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and         Power = I²r         or        r = P /I²

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So...     ? = I(P/I² + R)         or

           

2.1= 1.00 / I    +   0.6*I

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multiply everything by I

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        2.1 I  = 1   +   0.6 I²

solution of this is

.

a)

I =   0.568 A . . .. . lower or

*b) 2.930 A . . . . higher   

in each case,   pot diff across motor = ? - current * R   so

.

(3)   pot diff for lower current =  2.1 - 0.568 * 0.60 =   1.7592 Volts

.

(4) pot diff for higher current = 2.1 -2.93 * 0.60 =    0.342 Volts