Problem 51 - University Physics with Modern Physics

Question

In a shunt-wound dc motor with the field coils and rotor connected in parallel (Fig. E27.51), the resistance \(R _ { \text { f of the } }\) field coils is \(106 \Omega ,\) and the resistance \(R _ { r }\) of the rotor is 5.9\(\Omega .\) When a potential difference of 120\(\mathrm { V }\) is applied to the brushes and the motor is running at full speed delivering mechani- cal power, the current supplied to it is 4.82 A. (a) What is the current in the field coils? (b) What is the current in the rotor? (c) What is the induced emf developed by the motor? (d) How much mechanical power is developed by this motor?

Step-by-Step Solution

Answer

(a) 1.1321 A, (b) 3.6879 A, (c) 97.26 V, (d) 358.66 W

Step 1: Find Current in Field Coils

The current through the field coils (shunt field) can be calculated using Ohm's Law for the field circuit. Since the coils are connected in parallel, apply Ohm's Law: \[I_f = \frac{V}{R_f} = \frac{120\, \text{V}}{106\, \Omega} \approx 1.1321\, \text{A}\]The current in the field coils is approximately \(1.1321\, \text{A}\).

Step 2: Find Current in Rotor

The total current supplied to the motor is given as \(I = 4.82\, \text{A}\). The total current is the sum of the currents through the field coils and the rotor because they are in parallel. Therefore:\[I_r = I - I_f = 4.82\, \text{A} - 1.1321\, \text{A} \approx 3.6879\, \text{A}\]The current in the rotor is approximately \(3.6879\, \text{A}\).

Step 3: Calculate Induced EMF in Motor

The induced electromotive force (emf) can be determined from the voltage drop across the rotor which overcomes the resistance of the rotor. Starting from Ohm’s Law for the rotor’s circuit:\[V = E + I_r \cdot R_r\]Solve for \(E\):\[E = V - I_r \cdot R_r = 120\, \text{V} - 3.6879\, \text{A} \times 5.9\, \Omega \approx 97.26\, \text{V}\]The induced emf is approximately \(97.26\, \text{V}\).

Step 4: Calculate Mechanical Power Developed

The mechanical power developed by the motor is the power associated with the induced emf and the current through the rotor. Therefore, the mechanical power \(P\) is:\[P = E \cdot I_r = 97.26\, \text{V} \times 3.6879\, \text{A} \approx 358.66\, \text{W}\]The mechanical power developed is approximately \(358.66\, \text{W}\).

Key Concepts

Shunt-Wound Motor

A shunt-wound motor is a type of DC motor where the field windings, or coils, are connected in parallel with the armature or rotor circuit. This configuration offers specific benefits in terms of motor control and speed consistency. Understanding these components and their arrangement helps to comprehend how this motor achieves its operation.

In the shunt-wound setup:

• The field coils generate a magnetic field that is essential for motor operation.
• The parallel connection ensures that the field current (current through the field coils) remains relatively constant, even with varying load conditions. This helps maintain a consistent motor speed.
• Such motors are useful in applications requiring steady speed and quick response to speed changes without abrupt fluctuations.

By analyzing these characteristics, you can see why shunt-wound motors are often used in industrial and mechanical systems where control and reliability are paramount.

Ohm's Law

Ohm's Law is a fundamental principle used to calculate the relationships between voltage (V), current (I), and resistance (R) in electrical circuits. It is given by the equation: \[ V = I \times R \].

This law is critical when analyzing circuits, especially those involving shunt-wound motors.

• By using Ohm's Law, you can calculate the current flowing through the field coils (\(I_f\)) by rearranging the equation to \[ I_f = \frac{V}{R_f} \].
• For the rotor current, subtract the field current from the total supplied current, since both field coils and rotor are in parallel.
• Understanding this relationship helps to analyze how different components contribute to the overall operation of the motor.

Ohm's Law is indispensable in solving numerous electrical problems and forms the backbone of electrical circuit theory.

Electromotive Force (EMF)

Electromotive Force (EMF) represents the voltage generated by an electrical source, such as a motor, over overcoming its inherent resistive elements. In the context of DC motors, induced EMF is generated due to the movement of the rotor within a magnetic field.

In calculations involving EMF:

• The voltage across the rotor can be expressed as the sum of the EMF and the product of current through the rotor (\(I_r\)) and resistance of the rotor (\(R_r\)).
• Rearranging gives us the formula: \[ E = V - I_r \cdot R_r \].
• This allows calculating the EMF by considering how much of the supplied voltage is used to overcome the rotor resistance.

Understanding and calculating EMF is essential in designing and analyzing motor performance, as it affects how much electrical energy is converted into mechanical power.

Mechanical Power Calculation

Calculating the mechanical power developed by a motor is crucial to understanding its efficiency and performance. Mechanical power tells us how much of the electrical power is converted into useful mechanical energy to do work.To calculate mechanical power in a DC motor:

• Use the formula \[ P = E \cdot I_r \], where \(E\) is the induced EMF and \(I_r\) is the rotor current.
• This equation signifies the power output due to the interaction of electromagnetic fields within the motor.
• Efficient energy conversion happens when the majority of electrical input power turns into mechanical power, minimizing losses.

Mechanical power calculations are vital in motor selection for applications requiring a specific power rating, ensuring that the motor can handle the demanded workload effectively.