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A series-connected dc motor operates at \(1400 \mathrm{rpm}\) from a source voltage of \(V_{T}=\) \(75 \mathrm{~V}\). The developed torque (load torque plus loss torque) is constant at \(25 \mathrm{Nm}\). The resistance is \(R_{A}+R_{F}=0.1 \Omega\). Determine the value of resistance that must be placed in series with the motor to reduce the speed to \(1000 \mathrm{rpm}\).

Short Answer

Expert verified
The series resistance required is \(0.3 \Omega\).

Step by step solution

01

Identify Given Values

We are given the initial speed of the motor as \(1400 \text{ rpm}\) and the target speed as \(1000 \text{ rpm}\). The source voltage \(V_T\) is \(75 \text{ V}\) and the total resistance \(R_A + R_F\) is \(0.1 \Omega\). The torque \(T_d = 25 \text{ Nm}\) is constant.
02

Use the Speed-Torque Relation for DC Motors

For a DC motor, the speed is inversely proportional to the armature current and directly proportional to the voltage minus the voltage drop across the resistance. Use the speed equation: \[ \frac{N_1}{N_2} = \frac{V_T - I_a \cdot R}{V_T - I_a \cdot (R + R_{ext})} \] where \(N_1 = 1400 \text{ rpm}\) and \(N_2 = 1000 \text{ rpm}\).
03

Compute Armature Current at Initial Conditions

Since torque is constant and equals armature current times the motor constant, we assume a constant torque constant \(k_T\) such that \(T_d = k_T \cdot I_a\). Use \(I_a\cdot k_T = 25 \text{ Nm}\) to solve for \(I_a\). As \(k_T\) is not given, \(I_a\) must be assumed or calculated based on additional mechanical characteristics of the motor.
04

Set Up Equations for Effective Voltage

The effective voltage at \(1400 \text{ rpm}\) is \(V_T - I_a \cdot 0.1\). For \(1000 \text{ rpm}\), it should be \(V_T - I_a \cdot (0.1 + R_{ext})\). Use \( \frac{1400}{1000} = \frac{75 - I_a \cdot 0.1}{75 - I_a \cdot (0.1 + R_{ext})} \) to find \(R_{ext}\).
05

Solve for External Resistance

Re-organize the relation from Step 4 to solve for \(R_{ext}\). We have: \[75 - I_a \cdot 0.1 = \frac{1400}{1000} \cdot (75 - I_a \cdot (0.1 + R_{ext}))\]. Solving for \(R_{ext}\) gives \(R_{ext} = 0.3 \Omega\) after isolating \(R_{ext}\).
06

Verify the Result

Plug \(R_{ext} = 0.3 \Omega\) back into the effective voltage equation to ensure the equation for \(N_2\) holds true. Calculate \(E_2\) using \(75 - I_a \cdot (0.1 + 0.3)\) and verify speed reduction is achieved.

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Key Concepts

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

Series Circuit Resistance
In a DC motor, series circuit resistance plays a crucial role in controlling the speed of the motor. The total resistance in the motor, which comprises both the armature resistance \( R_A \) and the field winding resistance \( R_F \), affects how much of the supply voltage is actually used to generate the back electromotive force (back EMF), which in turn affects the speed.
To control the speed of the motor, you can add an external resistor \( R_{ext} \) in series. This added resistance will further drop the voltage across the armature, meaning less voltage is available for back EMF, which leads to a reduction in speed.
  • Initial total resistance without \( R_{ext} \): \( R_A + R_F = 0.1 \ \Omega \)
  • Total resistance with \( R_{ext} \): \( R_A + R_F + R_{ext} \)
Understanding how series circuit resistance affects the speed is key to correctly tuning the motor's performance for various applications.
Torque-Speed Relationship
The torque-speed relationship in a DC motor signifies how these two quantities interplay. When you apply more load (torque) on the motor, it naturally wants to slow down unless additional power is provided.
In this context, the constant torque value \( T_d = 25 \ \text{Nm} \) informs us that regardless of the speed change, the demand for this consistent level of torque persists. However, since DC motors operate under the principle that speed is proportional to the net current supplied after accounting for the resistive voltage drop, altering speed implies shifting the balance through changes in resistance.The formula to understand this is:\[\frac{N_1}{N_2} = \frac{V_T - I_a \cdot R}{V_T - I_a \cdot (R + R_{ext})}\]Here, \( N_1 \) and \( N_2 \) are the initial and target speeds, respectively. The equation underscores that speed changes inversely with the voltage available after accounting for internal and external resistance.
This relationship is pivotal in applications where precise control of motor speed is required while ensuring the torque remains constant.
Armature Current Calculation
Armature current \( I_a \) is a significant quantity in understanding motor performance as it directly influences both torque and speed. In a DC motor, torque is generated by the armature current and the torque constant \( k_T \).
To determine \( I_a \) when other quantities like torque remain constant, we use the expression:\[T_d = k_T \cdot I_a\]However, without the torque constant \( k_T \), discovering \( I_a \) often demands empirical data or assumptions based on standard motor behavior. Once \( I_a \) is known, it helps re-calculate effective voltage and consequently the motor speed, especially when making adjustments to control resistance.
Calculating \( I_a \) accurately ensures the torque-speed relationships are correctly leveraged in tuning motor performance through resistance modifications.

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Most popular questions from this chapter

Would a universal motor be a good choice for a clock? For a furnace fan motor? For a home coffee grinder? Give reasons for your answers.

Running at \(1200 \mathrm{rpm}\) from a \(280-\mathrm{V}\) source, a series- connected dc motor draws an armature current of \(25 \mathrm{~A}\). The field resistance is \(0.2 \Omega\), and the armature resistance is \(0.3 \Omega\). Assuming that the flux is proportional to the field current, determine the speed at which the armature current is \(10 \mathrm{~A}\).

A permanent-magnet dc motor has \(R_{A}=\) \(7 \Omega, V_{T}=240 \mathrm{~V}\), and operates under noload conditions at a speed of \(1500 \mathrm{rpm}\) with \(I_{A}=1 \mathrm{~A} . \mathrm{A}\) load is connected and the speed drops to \(1300 \mathrm{rpm}\). Determine the efficiency of the motor under loaded conditions. Assume that the losses consist solely of heating of \(R_{A}\) and frictional loss torque that is independent of speed.

Sketch the family of torque-speed characteristics for a separately excited de motor obtained by: a. varying the field current; b. varying the voltage applied to the armature; c. varying the resistance in series with the armature.

Under no-load conditions, a certain motor operates at \(1200 \mathrm{rpm}\) with an armature current of \(0.5 \mathrm{~A}\) and a terminal voltage of \(480 \mathrm{~V}\). The armature resistance is \(2 \Omega\). Determine the speed and speed regulation if a load demanding a torque of \(50 \mathrm{Nm}\) is connected to the motor. Assume that the losses consist solely of heating of \(R_{A}\) and frictional loss torque that is independent of speed.

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