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A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 \(\Omega\). When the motor is running at full load on a 120-V line, the emf in the rotor is 105 V. (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

Short Answer

Expert verified
(a) 4.69 A; (b) 562.8 W; (c) 492.45 W.

Step by step solution

01

Identify Known Values

We are given that the internal resistance \( R \) of the motor is 3.2 \( \Omega \), the line voltage \( V \) is 120 V, and the electromotive force (emf) \( E \) is 105 V.
02

Apply Ohm's Law to Find Current

Use the equation for a circuit with resistance, \( I = \frac{V - E}{R} \). Substitute the known values: \( I = \frac{120 \text{ V} - 105 \text{ V}}{3.2 \Omega} = \frac{15}{3.2} \approx 4.69 \text{ A}. \) Thus, the current drawn by the motor is approximately 4.69 A.
03

Calculate Power Delivered to the Motor

The power delivered to the motor \( P \) can be found using the formula \( P = VI \). Substitute the known values: \( P = 120 \, \text{V} \times 4.69 \text{ A} \approx 562.8 \text{ W}. \) This is the electrical power going into the motor.
04

Determine Mechanical Power Developed by the Motor

The mechanical power developed by the motor can be calculated as \( P_{mech} = EI \). Substitute the known values: \( P_{mech} = 105 \, \text{V} \times 4.69 \, \text{A} \approx 492.45 \, \text{W}. \) This is the output mechanical power of the motor.

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in electrical engineering that relates the voltage, current, and resistance in an electrical circuit. In simple terms, Ohm's Law states that the current (\( I \)) flowing through a conductor between two points is directly proportional to the voltage (\( V \)) across the two points and inversely proportional to the resistance (\( R \)) of the conductor. This can be mathematically expressed as:
  • \( I = \frac{V}{R} \)
In the context of DC motors, Ohm's Law helps us determine the current drawn by the motor from the supply line, taking into account any internal resistance and electromotive force present. By rearranging the formula to accommodate these additional factors, we can confirm the motor's current involvement in running efficiently.
Always remember to ensure all units are consistent when applying Ohm's Law and in practical circuit analysis scenarios.
Electromotive force
Electromotive force (emf) is a crucial concept in understanding how DC motors operate. It represents the energy provided by a power source, like a battery or generator, to a charge moving in a circuit. Despite its name, emf is not actually a force but rather a potential difference measured in volts (V). It is the driving factor that moves electrons around a circuit, allowing electric current to flow.
For DC motors, the emf can be thought of as the back electromotive force, which acts against the supply voltage when the motor is in operation. This is because the motor acts like a generator when it is rotating, producing its own voltage as a response to the motion, commonly known as back emf. Understanding emf is key to calculating how much voltage is effectively applied to overcome the internal resistance and perform work.
Power Calculation
Power calculation in electrical circuits is about determining the rate at which energy is transferred or converted. In terms of DC motors, we often calculate two types of power: electrical power supplied to the motor and mechanical power output by the motor.
  • Electrical power (\( P \)) is calculated using the formula \( P = VI \), where \( V \) is voltage and \( I \) is current.
  • For mechanical power, the formula \( P_{mech} = EI \) is used, where \( E \) is the electromotive force.
Accurately calculating these powers helps in assessing the motor efficiency and its operation under a given load. Understanding how to navigate between these types of power is fundamental to ensuring the motor's performance aligns with expectations.
Internal Resistance
Internal resistance is a natural resistance within electrical components that hinders current flow. In a DC motor, internal resistance refers to the cumulative resistance of the wires and components inside the motor.
It plays a vital role when analyzing circuits and can significantly affect motor performance. Increased internal resistance reduces the net voltage available for mechanical work, hence decreasing efficiency.
In the given exercise, identifying the internal resistance is crucial for calculating the current flow and the power delivered to the motor, as it enables us to precisely determine how much of the input energy is actually being converted to perform useful work.
Mechanical Power
Mechanical power in the context of DC motors is the useful power output that transforms electrical energy into mechanical energy. This power reflects how effectively the motor converts electrical inputs to perform work, such as rotating shafts or driving a load.
To find the mechanical power, we use the formula\[ P_{mech} = EI \]where \( E \) is the electromotive force of the motor and \( I \) is the current. High mechanical power indicates efficient conversion, while low values may point to energy losses.
Hence, understanding mechanical power helps in evaluating motor performance, ensuring that the motor meets design specifications and operates effectively in real-world applications.

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Most popular questions from this chapter

A conducting bar with mass m and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{B}\) fills the region between the rails (\(\textbf{Fig. P27.65}\)). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\), find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s). Let \(B =\) 0.80 T, \(I =\) 2.0 \(\times\) 10\(^3\) A, \(m =\) 25 kg, and \(L =\) 50 cm. For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

An electron in the beam of a cathode-ray tube is accelerated by a potential difference of 2.00 kV. Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.180 m. What is the magnitude of the field?

An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 cm between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force does this field exert on the wire?

A particle with initial velocity \(\vec{v}$$_0 =\) (5.85 \(\times\) 10\(^3\)m/s)\(\hat{\jmath}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\overrightarrow{B} =\) - (1.35 T)\(\hat{k}\). Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) -0.320 nC. You can ignore the weight of the particle.

A particle with charge -5.60 nC is moving in a uniform magnetic field \(\overrightarrow{B} =\) -(1.25 T)\(\hat{k}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) -(3.40 \(\times\) 10\(^{-7}\)N)\(\hat{\imath}\) + (7.40 \(\times\) 10\(^{-7}\)N)\(\hat{\jmath}\). (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{v}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\vec{v}\) and \(\overrightarrow{F}\)?

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