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A motor has a resistance of \(2.50 \Omega\) and is connected to a \(110-\mathrm{V}\) line. (a) Is the operating current of the motor (1) higher than \(44 \mathrm{~A},\) (2) \(44 \mathrm{~A},\) or (3) lower than 44 A? Why? (b) If the back emf of the motor at operating speed is \(100 \mathrm{~V}\), what is its operating current?

Short Answer

Expert verified
(a) 44 A; (b) 4 A.

Step by step solution

01

Understanding the Components

The motor has a resistance of \(2.50 \Omega\) and is connected to a \(110-\mathrm{V}\) line. There is also a back emf of \(100 \mathrm{~V}\). We need to determine if the operating current is higher, equal, or lower than \(44 \mathrm{~A}\), and find the operating current with the back emf.
02

Calculate Theoretical Current Without Back EMF

Using Ohm's Law \(V = IR\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance. Without considering back emf, the voltage across the motor is \(110 \mathrm{~V}\).Calculate the current: \[ I = \frac{V}{R} = \frac{110 \mathrm{~V}}{2.50 \Omega} \]\[ I = 44 \mathrm{~A} \] The theoretical current without back emf is \(44 \mathrm{~A}\).
03

Compare Theoretical Current to Given Values

We found the theoretical current to be exactly \(44 \mathrm{~A}\) in conditions without back emf. Thus, initially, the current is exactly \(44 \mathrm{~A}\), so option (2) is correct: it is \(44 \mathrm{~A}\).
04

Consider Back EMF in Calculation

Now, taking the back emf into account, the effective voltage across the motor is reduced by the back emf. Effective voltage is:\[ V_{\text{effective}} = 110 \mathrm{~V} - 100 \mathrm{~V} = 10 \mathrm{~V} \]The operating current is now:\[ I = \frac{V_{\text{effective}}}{R} = \frac{10 \mathrm{~V}}{2.50 \Omega} \] \[ I = 4 \mathrm{~A} \]
05

Conclusion

The actual operating current, considering the back emf, is \(4 \mathrm{~A}\). This is significantly lower than \(44 \mathrm{~A}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Current
Electric current is the flow of electric charge through a conductor. It is typically measured in amperes (A), which tells us how much charge passes through a point in the circuit per second.
In electrical circuits, current is often described in terms of direct current (DC) and alternating current (AC). DC is a constant flow of electric charge in one direction, while AC periodically reverses direction.
  • Current direction is usually defined from positive to negative, opposite to the direction of electron flow.
  • The amount of current in a circuit depends on the voltage and resistance present.
The motor problem illustrates how these factors interact. Without back emf, using Ohm's Law, we found the current to be equal to 44 A. However, with the back emf, the current significantly reduces to 4 A.
Resistance
Resistance is a measure of the opposition to the flow of electric current. It is measured in ohms (\(\Omega\)) and depends on the material, length, cross-sectional area, and temperature of the conductor.
Each material offers different levels of resistance, influencing current and power within the circuit. In the exercise, the motor has a resistance of \(2.50 \Omega\), which is an essential factor in our calculations.
  • Higher resistance means less current flows at a given voltage.
  • Resistance can cause electrical energy to convert into heat, sometimes undesirably, leading to energy losses.
The calculation of current using the resistance value shows how resistance impacts the overall current flow in both scenarios: with and without the back emf.
Voltage
Voltage refers to the electric potential difference between two points in a circuit. It is the force that pushes electric charges to move in a current and is measured in volts (V).
The voltage in a circuit is provided by a power source, like a battery or, in our example, a \(110 \mathrm{V}\) line. This initial voltage determines how much current could potentially flow through the circuit if connected directly across the motor.
  • Ohm's Law connects voltage (V), current (I), and resistance (R) in the formula \(V = IR\).
  • The difference in potential between two points causes current to flow in the circuit.
The exercise shows voltage as the driving factor for current, reduced by the back emf's effect, which is subtracted from the source voltage. Thus, back emf limits the current, shown by the reduced effective voltage of 10 V.
Back Electromotive Force (EMF)
Back electromotive force (emf) is a voltage produced by the motor itself, opposing the applied voltage from the power source. It is a natural part of many electrical devices that have moving parts, like motors.
The back emf occurs due to the motor's operation; as the motor spins, it generates a voltage in opposition to the supply voltage. This effect reduces the effective voltage across the motor.
  • Back emf depends on the motor's speed; as the speed rises, so does the back emf.
  • It stabilizes motor operation, preventing it from over-speeding.
In the exercise, the back emf is \(100 \mathrm{V}\), significantly reducing the effective voltage from \(110 \mathrm{V}\) to \(10 \mathrm{V}\), which results in a much lower operating current of \(4 \mathrm{A}\). This control is essential for keeping the motor's performance stable and efficient.

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Most popular questions from this chapter

A \(240-\mathrm{V}\) dc motor has an armature whose resistance is \(1.50 \Omega\). When running at its operating speed, it draws a current of 16.0 A. (a) What is the back emf of the motor when it is operating normally? (b) What is the starting current? (Assume that there is no additional resistance in the circuit.) (c) What series resistance would be required to limit the starting current to \(25 \mathrm{~A} ?\)

The primary coil of an ideal transformer has 720 turns, and the secondary coil has 180 turns. If the primary coil carries \(15 \mathrm{~A}\) at a voltage of \(120 \mathrm{~V},\) what are \((\) a) the voltage and (b) the output current of the secondary coil?

The secondary coil of an ideal transformer has 450 turns, and the primary coil has 75 turns. (a) Is this transformer a (1) step-up or (2) step-down transformer? Explain your choice. (b) What is the ratio of the current in the primary coil to the current in the secondary coil? (c) What is the ratio of the voltage across the primary coil to the voltage in the secondary coil?

When the magnetic flux through a single loop of wire increases by \(30 \mathrm{~T} \cdot \mathrm{m}^{2},\) an average current of \(40 \mathrm{~A}\) is induced in the wire. Assuming that the wire has a resistance of \(2.5 \Omega,\) (a) over what period of time did the flux increase? (b) If the current had been only \(20 \mathrm{~A},\) how long would the flux increase have taken?

A circular loop with an area of \(0.015 \mathrm{~m}^{2}\) is in a uniform magnetic field of \(0.30 \mathrm{~T}\). What is the flux through the loop's plane if it is (a) parallel to the field, (b) at an angle of \(37^{\circ}\) to the field, and (c) perpendicular to the field?

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