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Chapter 22: Problem 43

A motor is designed to operate on \(117 \mathrm{~V}\) and draws a current of \(12.2 \mathrm{~A}\) when it first starts up. At its normal operating speed, the motor draws a current of \(2.30\) A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.

Short Answer

Expert verified
(a) 9.59 惟, (b) 94.94 V, (c) 8.90 A.

Step by step solution

01

Calculate Armature Coil Resistance

To find the resistance \( R \) of the armature coil, use Ohm's Law, which states \( V = IR \). At startup, the back EMF is zero, so the entire voltage appears across the armature.Given:- Voltage \( V = 117 \) V- Current at startup \( I = 12.2 \) AUsing Ohm's Law:\[ R = \frac{V}{I} = \frac{117}{12.2} = 9.59 hinspace \Omega. \]
02

Calculate Back EMF at Normal Speed

At normal speed, the back EMF \( E \) reduces the net voltage across the armature.Given:- Voltage \( V = 117 \) V- Current at normal speed \( I_n = 2.30 \) A- Resistance \( R = 9.59 \) \( \Omega \) (from Step 1)The effective voltage across the armature, \( V_{ ext{effective}}, \) is:\[ V_{ ext{effective}} = I_n \cdot R = 2.30 \times 9.59 = 22.06 ext{ V}. \]The back EMF \( E \) is then:\[ E = V - V_{ ext{effective}} = 117 - 22.06 = 94.94 ext{ V}. \]
03

Calculate Current at One-Third Normal Speed

At one-third normal speed, assume the back EMF is proportionally reduced. Hence, if normal speed has a back EMF of \( 94.94 \) V, at one-third speed the back EMF is:\[ E' = \frac{1}{3} \times 94.94 = 31.65 ext{ V}. \]The effective voltage \( V'_{ ext{effective}} \) is:\[ V'_{ ext{effective}} = V - E' = 117 - 31.65 = 85.35 ext{ V}. \]Using Ohm鈥檚 Law to find the current:\[ I' = \frac{V'_{ ext{effective}}}{R} = \frac{85.35}{9.59} = 8.90 ext{ A}. \]

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Key Concepts

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

Back EMF
When a motor is running, it generates something called back electromotive force, or back EMF. This is an important concept as it affects how motors work.
As the motor spins, the armature cuts through the magnetic lines of force, generating a voltage that opposes the voltage driving the motor.
This opposing voltage is the back EMF, and it rises with the speed of the motor. - At startup, the motor's back EMF is zero because the motor isn't spinning yet. Thus, the full supply voltage is across the armature coil. - Once the motor reaches its normal speed, back EMF increases, countering the original voltage and reducing the current drawn. Knowing how back EMF works helps you understand why a motor draws different currents at different speeds.
Armature Resistance
Armature resistance is the internal resistance within the motor's armature coils. It plays a crucial role in determining the current flow through a motor when operating.- Ohm's Law can be used to calculate this resistance. One can use the formula \( R = \frac{V}{I} \) where \( V \) is the voltage and \( I \) is the current.- At start-up, with the given voltage and full current, you determine the armature resistance since there's no back EMF.Finding the armature resistance is essential for calculating other aspects, such as back EMF and the current at different motor speeds. It tells us how much opposition the internal components provide to current flow.
Current Calculation
Current calculation in motors involves understanding both initial conditions and those during normal operation. The current changes as the speed changes because of impact from back EMF.- At full speed, the back EMF dramatically reduces the net voltage across the motor, thus lowering the current.- By using the effective voltage across the armature and known resistance, current can be recalculated for any speed.For example, if the motor runs at one-third of its normal speed, the back EMF is significantly decreased. Knowing the new effective voltage, you can use \( I = \frac{V_{\text{effective}}}{R} \) to find the new current.
This calculation is vital for predicting motor behavior under different operating conditions.

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Most popular questions from this chapter

A generator has a square coil consisting of 248 turns. The coil rotates at \(79.1 \mathrm{rad} / \mathrm{s}\) in a \(0.170\) -T magnetic field. The peak out put of the generator is \(75.0 \mathrm{~V}\). What is the length of one side of the coil?

A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius \(r=0.20 \mathrm{~m} .\) The normal to the plane of the loop is parallel to a constant magnetic field \(\left(\phi=0^{\circ}\right)\) of magnitude \(0.75 \mathrm{~T}\). What is the change \(\Delta \Phi\) in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?

Multiple-Concept Example 13 reviews the concepts used in this problem. A long solenoid (cross-sectional area \(=1.0 \times 10^{-6} \mathrm{~m}^{2}\), number of turns per unit length \(=2400\) turns \(/ \mathrm{m}\) ) is bent into a circular shape so it looks like a doughnut. This wire-wound doughnut is called a toroid. Assume that the diameter of the solenoid is small compared to the radius of the toroid, which is \(0.050 \mathrm{~m}\). Find the emf induced in the toroid when the current decreases to \(1.1\) A from \(2.5 \mathrm{~A}\) in a time of \(0.15 \mathrm{~s}\).

In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a wall receptacle. (a) How is the power delivered by the receptacle to the primary related to the power delivered by the secondary to the picture tube? Give your answer in the form of an equation, and explain what assumptions are implied when this equation is used. (b) How is the turns ratio of the transformer related to the currents in the primary and the secondary? (c) How is the turns ratio of the transformer related to the voltage across the primary and the voltage across the secondary? (d) Express the turns ratio \(N_{\mathrm{s}} /\) \(N_{\mathrm{p}}\) of the transformer in terms of the power \(P\) used by the picture tube, the voltage \(V_{\mathrm{p}}\) across the primary, and the current \(I_{\mathrm{s}}\) in the secondary. The primary of the transformer is connected to a \(120-\mathrm{V}\) receptacle. The picture tube of a television set uses \(91 \mathrm{~W}\), and there is \(5.5 \mathrm{~mA}\) of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio \(N_{\mathrm{s}} / N_{\mathrm{p}}\) of the transformer.

The batteries in a portable CD player are recharged by a unit that plugs into a wall socket. Inside the unit is a step-down transformer with a turns ratio of \(1: 13 .\) The wall socket provides \(120 \mathrm{~V}\). What voltage does the secondary coil of the transformer provide?
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