The motor in a toy car is powered by four batteries in series, which produce a total emf of 6.3 V. The motor draws 3.1 A and develops a 2.1 V back emf at normal speed. Each battery has a 0.18 Ω internal resistance. what is the resistance of the motor?

0.635 Ω

Explanation:

Using

E-E' = IR Equation 1

Where E = total Emf of the of the batteries, E' = back emf of the motor, I = current of the motor, R = combined resistance of the battery and the motor.

Make R the subject of the equation

R = E-E'/I Equation 2

Given: E = 6.3 V, E' = 2.1 V, I = 3.1 A.

Substitute into equation 2

R = (6.3-2.1)/3.1

R = 4.2/3.1

R = 1.355 Ω

Since the motor and the batteries are connected in series,

R = R'+r' Equation 3

Where R' = Resistance of the motor, r' = resistance of the batteries.

make R' the subject of the equation

R' = R-r' Equation 4

Given: R = 1.355 Ω, r' = 0.18×4 (four batteries) = 0.72 Ω

Substitute into equation 4

R' = 1.355-0.72

R' = 0.635 Ω

Hence the resistance of the motor = 0.635 Ω

car a and b are moving in opposite directions

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