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Schaum’s Outline of College Physics

Eugene Hecht

Chapter 33

Electric Generators and Motors - all with Video Answers

Educators

Chapter Questions

01:32

Problem 1

An ac generator produces an output voltage of $\varepsilon=170 \sin 377 t$ volts, where $t$ is in seconds. What is the frequency of the ac voltage?
A sine curve plotted as a function of time is no different from a cosine curve, except for the location of $t=0 .$ Since $\varepsilon=2 \pi \mathrm{NBA} f$ $\cos 2 \pi f t$, we have $377 \mathrm{t}=2 \pi f t$, from which we find that the frequency $f=60 \mathrm{~Hz}$.

Paul Gabriel
Paul Gabriel
Numerade Educator
02:55

Problem 2

How fast must a 1000 -turn coil (each with a $20 \mathrm{~cm}^{2}$ area) turn in the Earth's magnetic field of $0.70 \mathrm{G}$ to generate a voltage that has a maximum value (i.e., an amplitude) of $0.50 \mathrm{~V}$ ?

We assume the coil's axis to be oriented in the field so as to give maximum flux change when rotated. Then $B=7.0 \times 10^{-5} \mathrm{~T}$ in the expression
$$
\mathscr{E}=2 \pi N A B f \cos 2 \pi f t
$$
Because $\cos 2 \pi f t$ has a maximum value of unity, the amplitude of the voltage is 2\piNBAf. Therefore,
$$
f=\frac{0.50 \mathrm{~V}}{2 \pi N A B}=\frac{0.50 \mathrm{~V}}{(2 \pi)(1000)\left(20 \times 10^{-4} \mathrm{~m}^{2}\right)\left(7.0 \times 10^{-5} \mathrm{~T}\right)}=0.57 \mathrm{kHz}
$$

Paul Gabriel
Paul Gabriel
Numerade Educator
01:29

Problem 3

When turning at $1500 \mathrm{rev} / \mathrm{min}$, a certain generator produces $100.0$ V. What must be its frequency in rev/min if it is to produce $120.0$ V?
Because the amplitude of the emf is proportional to the frequency, we have, for two frequencies $f_{1}$ and $f_{2}$,
$$
\frac{\delta_{1}}{\delta_{2}}=\frac{f_{1}}{f_{2}} \quad \text { or } \quad f_{2}=f_{1} \frac{\delta_{2}}{\varepsilon_{1}}=(1500 \mathrm{rev} / \min )\left(\frac{120.0 \mathrm{~V}}{100.0 \mathrm{~V}}\right)=1800 \mathrm{rev} / \mathrm{min}
$$

Paul Gabriel
Paul Gabriel
Numerade Educator
01:07

Problem 4

A certain generator has armature resistance $0.080 \Omega$ and develops an induced emf of $120 \mathrm{~V}$ when driven at its rated speed. What is its terminal voltage when $50.0 \mathrm{~A}$ is being drawn from it? The generator acts like a battery with emf $=120 \mathrm{~V}$ and internal resistance $r=0.080 \Omega$. As with a battery, Terminal p.d. $=(\mathrm{emf})-I r=120 \mathrm{~V}-(50.0 \mathrm{~A})(0.080 \Omega)=116 \mathrm{~V}$

Paul Gabriel
Paul Gabriel
Numerade Educator
01:50

Problem 5

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 P = VI,
$$
\begin{array}{l}
\text { Current to the external circuit }=I_{x}=\frac{\mathrm{P}}{V}=\frac{40000 \mathrm{~W}}{250 \mathrm{~V}}=160 \mathrm{~A} \\
\text { Field current }=I_{f}=\frac{V_{f}}{r_{f}}=\frac{250 \mathrm{~V}}{100 \Omega}=2.5 \mathrm{~A} \\
\text { Armature current }=I_{a}=I_{x}+I_{f}=162.5 \mathrm{~A} \\
\text { Total induced emf }=|\mathscr{E}|=\left(250 \mathrm{~V}+I_{a} r_{a} \mathrm{drop}\right. \text { in armature } \\
=250 \mathrm{~V}+(162.5 \mathrm{~A})(0.06 \Omega)=260 \mathrm{~V} \\
\text { Armature power }=I_{a}|\mathscr{E}|=(162.5 \mathrm{~A})(260 \mathrm{~V})=42 \mathrm{~kW}
\end{array}
$$
Armature current
$$
\begin{aligned}
\text { Power loss in the armature } &=I_{a}^{2} r_{a}=(162.5 \mathrm{~A})^{2}(0.06 \Omega)=1.6 \mathrm{~kW} \\
\text { Power loss in the field } &=I_{f}^{2} r_{f}=(2.5 \mathrm{~A})^{2}(100 \Omega)=0.6 \mathrm{~kW} \\
\text { Power developed } &=(\text { Power delivered })+\text { (Power loss in armature) }+(\text { Power loss in field) }\\
&=40 \mathrm{~kW}+1.6 \mathrm{~kW}+0.6 \mathrm{~kW}=42 \mathrm{~kW}
\end{aligned}
$$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
01:18

Problem 6

The resistance of the armature in the motor shown in Fig. $33-2$ is $2.30 \Omega$. It draws a current of $1.60$ 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,
or
$$
\begin{array}{l}
\text { Line voltage }=\text { back emf }+I r \\
\qquad \text { Back emf }=120 \mathrm{~V}-(1.60 \mathrm{~A})(2.30 \Omega)=116 \mathrm{~V}
\end{array}
$$

Paul Gabriel
Paul Gabriel
Numerade Educator
02:37

Problem 7

A 0.250-hp motor (like that in Fig. $33-2$ ) has a resistance of $0.500$ $\Omega$. (a) How much current does it draw on $110 \mathrm{~V}$ when its output is $0.250 \mathrm{hp}$ ? (b) What is its back emf?
(a) Assume the motor to be 100 percent efficient so that the input power $V I$ equals its output power $(0.250 \mathrm{hp})$. Then
$$
(110 \mathrm{~V})(I)=(0.250 \mathrm{hp})(746 \mathrm{~W} / \mathrm{hp}) \text { or } I=1.695 \text { A }
$$
(b) Back emf $=($ line voltage $)-I r=110 \mathrm{~V}-(1.695 \mathrm{~A})(0.500 \Omega)=$
$109 \mathrm{~V}$

Paul Gabriel
Paul Gabriel
Numerade Educator
06:19

Problem 8

In a shunt motor, the permanent magnet is replaced by an electromagnet activated by a field coil that shunts the armature. The shunt motor shown in Fig. 33-4 has an armature resistance of $0.050$ and is connected to a $120 \mathrm{~V}$ line. $(a)$ What is the armature current at the starting instant, i.e., before the armature develops any back emf? (b) What starting rheostat resistance $R$, in series with the armature, will limit the starting current to $60 \mathrm{~A}$ ? $(c)$ With no starting resistance, what back emf is generated when the armature current is $20 \mathrm{~A}$ ? $(d)$ If this machine were running as a generator, what would be the total induced emf developed by the armature when the armature is delivering $20 \mathrm{~A}$ at $120 \mathrm{~V}$ to the shunt field and external circuit?
(a) Armature current $=\frac{\text { Impressed voltage }}{\text { Armature resistance }}=\frac{120 \mathrm{~V}}{0.050 \Omega}=2.4 \mathrm{kA}$
(b) Armature current $=\frac{\text { Impressed voltage }}{0.050 \Omega+R} \quad$ or $\quad 60 \mathrm{~A}=\frac{120 \mathrm{~V}}{0.050 \Omega+R}$
from which $R=2.0 \Omega$.
(c) Back emf $\begin{aligned}=&(\text { Impressed voltage })-(\text { Voltage drop in armature resistance }) \\ &=120 \mathrm{~V}-(20 \mathrm{~A})(0.050 \Omega)=119 \mathrm{~V}=0.12 \mathrm{kV} \end{aligned}$
(d) $\begin{aligned} \text { Induced emf } &=(\text { Terminal voltage })+(\text { Voltage drop in armature resistance) }\\ &=120 \mathrm{~V}+(20 \mathrm{~A})(0.050 \Omega)=121 \mathrm{~V}=0.12 \mathrm{kV} \end{aligned}$

Prabhat Tyagi
Prabhat Tyagi
Numerade Educator
03:25

Problem 9

The shunt motor shown in Fig. 33-5 has an armature resistance of $0.25 \Omega$ and a field resistance of $150 \Omega .$ It is connected across $120-$ $\mathrm{V}$ mains and is generating a back emf of $115 \mathrm{~V}$. Compute: $(a)$ the armature current $I_{a}$, the field current $I_{f}$, and the total current $I_{t}$ taken by the motor; $(b)$ the total power taken by the motor; $(c)$ the power lost in heat in the armature and field circuits; $(d)$ the electrical efficiency of this machine (when only heat losses in the armature and field are considered).
(a) $I_{a}=\frac{(\text { Impressed voltage })-(\text { Back emf })}{\text { Armature resistance }}=\frac{(120-115)}{0.25 \Omega}=20 \mathrm{~A}$
$I_{f}=\frac{\text { Impressed voltage }}{\text { Field resistance }}=\frac{120 \mathrm{~V}}{150 \Omega}=0.80 \mathrm{~A}$
$I_{t}=I_{a}+I_{f}=20.80 \mathrm{~A}=21 \mathrm{~A}$
(b) Power input $=(120 \mathrm{~V})(20.80 \mathrm{~A})=2.5 \mathrm{~kW}$
(c) $I_{a}^{2} r_{a}$ loss in armature $=(20 \mathrm{~A})^{2}(0.25 \Omega)=0.10 \mathrm{~kW}$
$$
I_{f}^{2} r_{f} \text { loss in field }=(0.80 \mathrm{~A})^{2}(150 \Omega)=96 \mathrm{~W}
$$
(d) Power output $=($ Power input $)-($ Power losses $)=2496-(100+$ 96 ) $=2.3 \mathrm{~kW}$
Alternatively,
Power output $=($ Armature current $)($ Back emf $)=(20 \mathrm{~A})(115 \mathrm{~V})=$
$2.3 \mathrm{~kW}$
Then $\quad$ Efficiency $=\frac{\text { Power output }}{\text { Power input }}=\frac{2300 \mathrm{~W}}{2496 \mathrm{~W}}=0.921=92 \%$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
02:29

Problem 10

A motor has a back emf of $110 \mathrm{~V}$ and an armature current of 90 A when running at 1500 rpm. Determine the power and the torque developed within the armature.
Power $=($ Armature current $)($ Back emf $)=(90 \mathrm{~A})(110 \mathrm{~V})=9.9 \mathrm{~kW}$
From Chapter 10, power $=\tau \omega$ where $\omega=2 \pi f=2 \pi(1500 \times 1 / 60)$ $\mathrm{rad} / \mathrm{s}$
$$
\text { Torque }=\frac{\text { Power }}{\text { Angular speed }}=\frac{9900 \mathrm{~W}}{(2 \pi \times 25) \mathrm{rad} / \mathrm{s}}=63 \mathrm{~N} \cdot \mathrm{m}
$$

Paul Gabriel
Paul Gabriel
Numerade Educator
02:49

Problem 11

A motor armature develops a torque of $100 \mathrm{~N} \cdot \mathrm{m}$ when it draws 40 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, $T_{f} / T_{i}=I_{f} B_{f} / I_{i} B_{i}$, hence,
$$
\tau_{f}=(100 \mathrm{~N} \cdot \mathrm{m})\left(\frac{70}{40}\right)(0.80)=0.14 \mathrm{kN} \cdot \mathrm{m}
$$

Paul Gabriel
Paul Gabriel
Numerade Educator
04:12

Problem 12

Determine the separate effects on the induced emf of a generator if $(a)$ the flux per pole is doubled, and $(b)$ the speed of the armature is doubled.

Vishal Gupta
Vishal Gupta
Numerade Educator
02:56

Problem 13

The emf induced in the armature of a shunt generator is $596 \mathrm{~V}$. The armature resistance is $0.100 \Omega .$ (a) Compute the terminal voltage when the armature current is 460 A. (b) The field resistance is $110 \Omega$. Determine the field current, and the current and power delivered to the external circuit.

Paul Gabriel
Paul Gabriel
Numerade Educator
02:14

Problem 14

A dynamo (generator) delivers $30.0 \mathrm{~A}$ at $120 \mathrm{~V}$ to an external circuit when operating at $1200 \mathrm{rpm} .$ What torque is required to drive the generator at this speed if the total power losses are 400 W?

Paul Gabriel
Paul Gabriel
Numerade Educator
02:58

Problem 15

A $75.0$ -kW, 230 - $V$ 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?

Paul Gabriel
Paul Gabriel
Numerade Educator
02:34

Problem 16

A $120-V$ generator is run by a windmill that has blades $2.0 \mathrm{~m}$ long. The wind, moving at $12 \mathrm{~m} / \mathrm{s}$, is slowed to $7.0 \mathrm{~m} / \mathrm{s}$ after passing the windmill. The density of air is $1.29 \mathrm{~kg} / \mathrm{m}^{3}$. If the system has no losses, what is the largest current the generator can produce? [Hint: How much energy does the wind lose per second?]

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
03:00

Problem 17

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 rpm.

Vishal Gupta
Vishal Gupta
Numerade Educator
01:59

Problem 18

The active length of each armature conductor of a motor is 30 $\mathrm{cm}$, and the conductors are in a field of $0.40 \mathrm{~Wb} / \mathrm{m}^{3}$. A current of 15 A flows in each conductor. Determine the force acting on each conductor.

Vishal Gupta
Vishal Gupta
Numerade Educator
03:15

Problem 19

A shunt motor with armature resistance $0.080 \Omega$ is connected to $120 \mathrm{~V}$ mains. With $50 \mathrm{~A}$ in the armature, what are the back emf and the mechanical power developed within the armature?

Vishal Gupta
Vishal Gupta
Numerade Educator
01:55

Problem 20

A shunt motor is connected to a $110-V$ line. When the armature generates a back emf of $104 \mathrm{~V}$, the armature current is $15 \mathrm{~A}$. Compute the armature resistance.

Vishal Gupta
Vishal Gupta
Numerade Educator
02:27

Problem 21

A shunt dynamo has an armature resistance of $0.120 \Omega .(a)$ If it is connected across 220 -V mains and is running as a motor, what is the induced (back) emf when the armature current is $50.0 \mathrm{~A}$ ?
(b) If this machine is running as a generator, what is the induced emf when the armature is delivering $50.0 \mathrm{~A}$ at $220 \mathrm{~V}$ to the shunt field and external circuit?

Vishal Gupta
Vishal Gupta
Numerade Educator
01:42

Problem 22

A shunt motor has a frequency of $900 \mathrm{rpm}$ when it is connected to 120 - $V$ mains and delivering 12 hp. The total losses are 1048 W. Compute the power input, the line current, and the motor torque.

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
02:09

Problem 23

A shunt motor has armature resistance $0.20 \Omega$ and field resistance $150 \Omega$, and draws 30 A when connected to a 120-V supply line. Determine the field current, the armature current, the back emf, the mechanical power developed within the armature, and the electrical efficiency of the machine.

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
01:47

Problem 24

A shunt motor develops $80 \mathrm{~N} \cdot \mathrm{m}$ of torque when the flux density in the air gap is $1.0 \mathrm{~Wb} / \mathrm{m}^{2}$ and the armature current is $15 \mathrm{~A}$. What is the torque when the flux density is $1.3 \mathrm{~Wb} / \mathrm{m}^{2}$ and the armature current is $18 \mathrm{~A}$ ?

Vishal Gupta
Vishal Gupta
Numerade Educator
03:15

Problem 25

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?

Vishal Gupta
Vishal Gupta
Numerade Educator