Question

A de motor is connected to a constant emf of \(12.0 \mathrm{V}\) The resistance of its windings is \(2.0 \Omega .\) At normal operating speed, the motor delivers \(6.0 \mathrm{W}\) of mechanical power. (a) What is the initial current drawn by the motor when it is first started up? (b) What current does it draw at normal operating speed? (c) What is the back emf induced in the windings at normal speed?

Short answer

Answer: 11 V

Step-by-step solution

(Step 1: Calculate initial current)

(When the motor is first started, it does not have any back emf, so the total emf across the motor is equal to the emf applied (12 V). We can use Ohm's law, which relates voltage (V), current (I), and resistance (R), to find the initial current: I = V / R Substitute the given values: I_initial = (12 V) / (2 Ω) = 6 A The initial current drawn by the motor is 6 A.)

(Step 2: Calculate current at normal operating speed)

(We are given that at normal operating speed, the motor delivers 6 W of mechanical power. We can use the power formula mentioned earlier: Power = Voltage * Current or P = V * I We can rearrange the formula to solve for the current at normal operating speed: I_normal = Power / Voltage Substitute the known values: I_normal = (6 W) / (12 V) = 0.5 A The current drawn by the motor at normal operating speed is 0.5 A.)

(Step 3: Calculate back emf at normal operating speed)

(At normal operating speed, the back emf (E_back) opposes the applied emf (V). To find the back emf, we will use Kirchhoff's law, which states that the sum of the voltages around a closed loop in a circuit is equal to zero: V - E_back - I_normal * R = 0 Rearrange the equation to solve for E_back: E_back = V - I_normal * R Substitute the known values: E_back = 12 V - (0.5 A) * (2 Ω) = 12 V - 1 V = 11 V The back emf induced in the windings at normal operating speed is 11 V.)