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Chapter 33: Problem 6

The resistance of the armature in the motor shown in Fig. \(33-2\) is \(2.30 \Omega\). It draws a current of \(1.60 \mathrm{~A}\) when operating on \(120 \mathrm{~V}\). What is its back emf under these circumstances? The motor acts like a back emf in series with an \(I R\) drop through its internal resistance. Therefore, $$\text { Line voltage }=\text { back emf }+I r$$ or $$ \text { Back } \mathrm{emf}=120 \mathrm{~V}-(1.60 \mathrm{~A})(2.30 \Omega)=116 \mathrm{~V} $$

Short Answer

Expert verified
The back emf is approximately 116.32 V.

Step by step solution

01

Understanding the Problem

We need to find the back electromotive force (emf) in a motor given its armature resistance, the current it draws, and the voltage it operates on. The motor's line voltage is equal to the sum of the back emf and the voltage drop across its internal resistance.
02

Identify Given Values

The resistance of the armature is provided as \( r = 2.30 \, \Omega \), the current drawn by the motor is \( I = 1.60 \, \mathrm{A} \), and the line voltage is \( V = 120 \, \mathrm{V} \).
03

Apply the Formula

The voltage across the resistance due to current is given by Ohm's Law: \( I r = (1.60 \, \mathrm{A})(2.30 \, \Omega) \). The formula for back emf is: \[ \text{Back emf} = V - I r \].
04

Calculate IR Drop

Compute the voltage drop across the internal resistance using \( I r = (1.60 \, \mathrm{A})(2.30 \, \Omega) = 3.68 \, \mathrm{V} \).
05

Compute the Back EMF

Substitute the values into the back emf formula: \[ \text{Back emf} = 120 \, \mathrm{V} - 3.68 \, \mathrm{V} = 116.32 \, \mathrm{V} \].
06

Interpret the Result

The back emf of the motor is approximately \( 116.32 \, \mathrm{V} \), which opposes the applied line voltage in the motor circuit.

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Key Concepts

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

Ohm's Law in Motor Circuits
Ohm's Law is fundamental for understanding how electrical circuits work. It relates voltage (V), current (I), and resistance (R) in an equation: \( V = I \times R \).
When applied to a motor, Ohm's Law helps explain how voltage drop occurs across the motor's internal resistance. This voltage drop is due to the current flowing through the motor's resistance.
For instance, using the given motor problem, the current flowing through the motor's resistance creates a voltage drop, calculated by \( I \times R \). Here, the given values are current \( I = 1.60 \, \mathrm{A} \) and resistance \( R = 2.30 \, \Omega \). The resulting voltage drop is \( 3.68 \, \mathrm{V} \).
Understanding this concept is critical as it greatly influences the performance of electrical devices, including motors.
Understanding Motor Resistance
Motor resistance is a key component in determining how much current a motor draws. It is essentially the opposition that the motor's armature presents to the electric current.
The resistance can cause an internal voltage drop when current passes through it, calculated by \( I \times R \).
The specified armature resistance of the motor in question is \( 2.30 \, \Omega \). This value is crucial for calculating the voltage drop (or IR drop). The resistance value determines the proportion of the total voltage that is consumed within the motor.
This internal consumption of voltage ultimately affects the back electromotive force (emf) which determines the net voltage available to produce useful work.
Line Voltage and Back EMF Calculation
The line voltage applied to the motor splits into two parts: the back emf and the voltage drop across the motor's internal resistance. Back emf is an induced voltage that opposes the original voltage applied to the motor.
To find the back emf, the formula used is: \[ \text{Back emf} = V - I \times R \] where \( V \) is the line voltage, \( I \times R \) is the voltage drop calculated with Ohm's Law.
In the example, the line voltage (applied voltage) is \( 120 \, \mathrm{V} \), and the drop across the internal resistance is \( 3.68 \, \mathrm{V} \). Subtracting this drop from the line voltage gives a back emf of \( 116.32 \, \mathrm{V} \).
Understanding back emf is vital for efficiency in motor operations as it signifies how much of the input voltage is effectively converted into mechanical energy.

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Most popular questions from this chapter

A \(75.0-\mathrm{kW}, 230-\mathrm{V}\) shunt generator has a generated emf of \(243.5 \mathrm{~V}\). If the field current is \(12.5 \mathrm{~A}\) at rated output, what is the armature resistance?

A generator has an armature with 500 turns, which cut a flux of \(8.00 \mathrm{mWb}\) during each rotation. Compute the back emf it develops when run as a motor at \(1500 \mathrm{rpm}\).

Some generators, called shunt generators, use electromagnets in place of permanent magnets, with the field coils for the electromagnets activated by the induced voltage. The magnet coil is in parallel with the armature coil (it shunts the armature). As shown in Fig. \(33-3\), a certain shunt generator has an armature resistance of \(0.060 \Omega\) and a shunt resistance of \(100 \Omega\). What power is developed in the armature when it delivers \(40 \mathrm{~kW}\) at \(250 \mathrm{~V}\) to an external circuit? From \(\mathrm{P}=V I\) $$ \begin{array}{l} \text { Current to the external circuit }=I_{x}=\frac{\mathrm{P}}{V}=\frac{40000 \mathrm{~W}}{250 \mathrm{~V}}=160 \mathrm{~A} \\ \text { Field current }=I_{f}=\frac{V_{f}}{r_{f}}=\frac{250 \mathrm{~V}}{100 \Omega}=2.5 \mathrm{~A} \\ \text { Armature current }=I_{a}=I_{x}+I_{f}=162.5 \mathrm{~A} \end{array} $$ Total induced emf \(=|\mathscr{E}|=\left(250 \mathrm{~V}+I_{a} r_{a}\right.\) drop in armature $$ \begin{aligned} &=250 \mathrm{~V}+(162.5 \mathrm{~A})(0.06 \Omega)=260 \mathrm{~V} \\ \text { Armature power } &=I_{a}|\mathscr{E}|=(162.5 \mathrm{~A})(260 \mathrm{~V})=42 \mathrm{~kW} \end{aligned} $$ $$ \begin{array}{l} \text { Alternative Method }\\\ \begin{array}{l} \text { Power loss in the armature }=I_{a}^{2} r_{a}=(162.5 \mathrm{~A})^{2}(0.06 \Omega)=1.6 \mathrm{~kW} \\ \qquad \begin{aligned} \text { Power loss in the field } &=I_{f}^{2} r_{f}=(2.5 \mathrm{~A})^{2}(100 \Omega)=0.6 \mathrm{~kW} \\ \text { Power developed } &=(\text { Power delivered })+(\text { Power loss in armature })+(\text { Power loss in field) }\\\ &=40 \mathrm{~kW}+1.6 \mathrm{~kW}+0.6 \mathrm{~kW}=42 \mathrm{~kW} \end{aligned} \end{array} \end{array} $$

A motor armature develops a torque of \(100 \mathrm{~N} \cdot \mathrm{m}\) when it draws \(40 \mathrm{~A}\) from the line. Determine the torque developed if the armature current is increased to \(70 \mathrm{~A}\) and the magnetic field strength is reduced to 80 percent of its initial value. The torque developed by the armature of a given motor is proportional to the armature current and to the field strength (see Chapter 30). In other words, the ratio of the torques equals the ratio of the two sets of values of \(|N I A B|\). Using subscripts \(i\) and \(f\) for initial and final values, \(\tau_{f} / \tau_{i}=I_{f} B_{f} / I_{i} B_{i}\), hence, $$ \tau_{f}=(100 \mathrm{~N} \cdot \mathrm{m})\left(\frac{70}{40}\right)(0.80)=0.14 \mathrm{kN} \cdot \mathrm{m} $$

A shunt motor has a field resistance of \(200 \Omega\) and an armature resistance of \(0.50 \Omega\) and is connected to 120-V mains. The motor draws a current of \(4.6\) A when running at full speed. What current will be drawn by the motor if the speed is reduced to 90 percent of full speed by application of a load?
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