Question

A de motor has coils with a resistance of \(16 \Omega\) and is connected to an emf of \(120.0 \mathrm{V}\). When the motor operates at full speed, the back emf is \(72 \mathrm{V}\). (a) What is the current in the motor when it first starts up? (b) What is the current when the motor is at full speed? (c) If the current is \(4.0 \mathrm{A}\) with the motor operating at less than full speed, what is the back emf at that time?

Short answer

Answer: The initial current is 7.5 A, the full-speed current is 3.0 A, and the back emf at a current of 4.0 A is 56 V.

Step-by-step solution

(a) Find initial current in the motor

To find the current in the motor when it first starts up, we need to remember that the back emf is zero at this point. That's because the motor isn't spinning and isn't generating any opposing voltage. Using Ohm's law (\(V = IR\)) and the applied emf (\(120.0 \mathrm{V}\)): \(I = \frac{V}{R} = \frac{120.0}{16} = 7.5 \mathrm{A}\) So, the initial current in the motor when it first starts up is \(7.5 \mathrm{A}\).

(b) Find current at full speed

When the motor is running at full speed, the back emf is given as \(72 \mathrm{V}\). The total voltage across the coil will be the difference between the applied emf and the back emf: \(V_{coil} = \varepsilon - \varepsilon_b = 120.0 - 72 = 48 \mathrm{V}\) Now we can use Ohm's law to find the current at full speed: \(I_{full} = \frac{V_{coil}}{R} = \frac{48}{16} = 3.0 \mathrm{A}\) The current in the motor at full speed is \(3.0 \mathrm{A}\).

(c) Find back emf at \(4.0 \mathrm{A}\) current

With a given current of \(4.0 \mathrm{A}\) at less than full speed, we can use Ohm's law to find the voltage across the coil: \(V_{coil} = IR = 4.0 \times 16 = 64 \mathrm{V}\) Now let's find the back emf using the equation mentioned earlier: \(\varepsilon_b = \varepsilon - V_{coil} = 120.0 - 64 = 56 \mathrm{V}\) So, the back emf when the motor is operating at less than full speed with a current of \(4.0 \mathrm{A}\) is \(56 \mathrm{V}\).