(8c27p110) A battery of = 2.20 V and internal resistance R = 0.500 is driving a

Question

(8c27p110) A battery of = 2.20 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.
5.15×10-1 A

Enter the higher current.
3.89 A

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.

Explanation / Answer

Power output is P = 2*0.5 = 1W

Let V be the voltage across the motor terminals, and I be the current.

P = 1 = I*V

V = 2.2 - I*0.5 (Kirchhoff's Voltage rule)

giving 1= 2.2*I - 0.5*I^2

=> I^2 - 4.4*I + 2 = 0

=> I = [4.4 + sqrt(11.36)]/2 = 3.88 A and second value of I = [4.4 - sqrt(11.36)]/2 = 0.51 A

(1) Hence the lower current = 0.51 A

(2) and the higher current = 3.88 A

(3) V = 2.2 - 0.51*0.5 = 1.945 V

(4) V = 2.2 - 3.88*0.5 = 0.26 V

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