Question

A motor with an internal resistance of R(motor) = 50 Ohms requires a potential difference of V(motor)=20V to work properly. As you don't have a 20V power source, you place two 12V batteries in series with a custom-built resistor with resistance R1.
(a) How much current moves through the motor when it is working properly?
(b) What must be the resistance of your custom-built resistor so that the potential difference across the motor is 20V?
(c) If you built your resistor from aluminium wire with diameter of 0.03mm and a resisticity of 2.65*10^-8 Ohmes meter, what length of wire must you use to create the necessary resistance of R1?
(d) What is the ratio of the power used by the motor to the power dissipated by the resistor, P(motor)/P(R1)?

Answer 1

Internal resistance of the motor = R = 50

Potential difference across the motor = V_m = 20 V

Current moving through the motor = I_m

V_m = I_mR

20 = I_m(50)

I_m = 0.4 A

Resistance of custom built resistor = R₁

Two 12V batteries are used as power source in series with a custom built resistor.

Voltage = V = 2x12 = 24 V

For the motor to function properly a current(potential difference across it is 20V) of I_m should flow through the circuit.

V = I_m(R + R₁)

24 = (0.4)(50 + R₁)

R₁ = 10

The custom built resistor is made from aluminium.

Resistivity of aluminium = = 2.65 x 10^-8m

Diameter of aluminium wire used = D = 0.03 mm = 0.03 x 10^-3 m

Length of aluminium wire needed = L

Cross section area = A = D²/4

L = 0.267 m

Power of the motor = P_m

P_m = I_m²R

P_m = (0.4)²(50)

P_m = 8 W

Power dissipated by the resistor = P₁

P₁ = I_m²R₁

P₁ = (0.4)²(10)

P₁ = 1.6 W

P_m/P₁ = 8/1.6 = 5

a) Current through the motor when it is working properly = 0.4 A

b) Resistance of custom built resistor = 10

c) Length of wire used for the resistor = 0.267 m

d) P_motor/P_R1 = 5