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Chapter 20: Problem 30

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

Expert verified
Answer: 11 V

Step by step solution

01

(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.)
02

(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.)
03

(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.)

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Most popular questions from this chapter

A transformer for an answering machine takes an ac voltage of amplitude $170 \mathrm{V}\( as its input and supplies a \)7.8-\mathrm{V}$ amplitude to the answering machine. The primary has 300 turns. (a) How many turns does the secondary have? (b) When idle, the answering machine uses a maximum power of \(5.0 \mathrm{W}\). What is the amplitude of the current drawn from the 170 -V line?
Calculate the equivalent inductance \(L_{\mathrm{eq}}\) of two ideal inductors, \(L_{1}\) and \(L_{2},\) connected in parallel in a circuit. Assume that their mutual inductance is negligible. [Hint: Imagine replacing the two inductors with a single equivalent inductor \(L_{\mathrm{eq}} .\) How is the emf in the parallel equivalent related to the emfs in the two inductors? What about the currents? \(]\)
A solenoid is \(8.5 \mathrm{cm}\) long, \(1.6 \mathrm{cm}\) in diameter, and has 350 turns. When the current through the solenoid is \(65 \mathrm{mA},\) what is the magnetic flux through one turn of the solenoid?
A transformer with 1800 turns on the primary and 300 turns on the secondary is used in an electric slot car racing set to reduce the input voltage amplitude of \(170 \mathrm{V}\) from the wall output. The current in the secondary coil is of amplitude \(3.2 \mathrm{A}\). What is the voltage amplitude across the secondary coil and the current amplitude in the primary coil?
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