DC Motor — Back EMF, Torque, Speed and Power

i Learning Objectives

By the end of this chapter you should be able to:

Explain how back EMF arises and why it opposes the applied voltage.
Trace the self-regulating sequence by which a motor matches its current to its load.
Derive the condition for maximum power and explain why it is never used.
Derive the torque equation \(T_a = K_a\Phi I_a\) and distinguish armature from shaft torque.
State how torque varies with current in shunt and series motors.
Derive the speed equation and use the ratio method for speed changes.
Explain the flux-weakening sequence and why reducing flux raises the speed.
Compute speed regulation.

Section 34-1

Back EMF in a DC Motor

↩

How It Arises

The motor cannot help generating

\[I \text{ carrying conductor} \Rightarrow B \Rightarrow T \text{ induced} \Rightarrow \text{rotates conductor} \Rightarrow \text{cuts } \Phi \Rightarrow \text{EMF induced}\]

The direction of the induced EMF is opposite to \(V\), so it is called the counter or back EMF.
\(E_b\) is in series with \(V\) but opposite in direction — it opposes the very current that causes it.

The last point is Lenz's law from Chapter 11, applied to a rotating machine. A motor is a generator that happens to be running backwards, and the EMF it generates is what limits the current it takes.

\[E_b = V - I_aR_a \qquad\Longleftrightarrow\qquad I_a = \frac{V - E_b}{R_a}\]

Video · DC Motor Basic Operation

Section 34-2

Advantage of Back EMF

Electrical to mechanical energy conversion, \(E_bI_a\), is possible only because of \(E_b\).
\(E_b\) makes the DC motor self-regulating, through \(I_a = (V - E_b)/R_a\).

The first point deserves more than a passing mention. If \(E_b\) were zero the whole input \(VI_a\) would go into \(I_a^{2}R_a\) and nothing would be converted — which is exactly the condition of a stalled motor, drawing enormous current and delivering no power at all. The converted power is \(E_bI_a\) and nothing else, so a motor that is not turning is not a motor but a resistor.

! The Starting Problem

At the instant of switching on, the armature is stationary, so \(E_b = 0\) and

\[I_a = \frac{V}{R_a}\]

which for a typical machine is twenty to thirty times the rated current. Chapter 21 raised this; Chapter 36 solves it with a starter. Example 34.1 quantifies it.

Section 34-3

The Self-Regulating Action

A DC motor draws exactly the current its load requires, without any control action. The mechanism is a feedback loop through \(E_b\).

Load increases

\[\text{load}\uparrow \Rightarrow \left(T_m \lt T_L\right) \Rightarrow N\downarrow \Rightarrow E_b\downarrow \Rightarrow I_a\uparrow \Rightarrow T\uparrow\]

The motor slows a little, which lowers \(E_b\), which admits more current, which raises the torque until it again matches the load.

Load decreases

\[\text{load}\downarrow \Rightarrow \left(T_m \gt T_L\right) \Rightarrow N\uparrow \Rightarrow E_b\uparrow \Rightarrow I_a\downarrow \Rightarrow T\downarrow\]

The motor speeds up a little, which raises \(E_b\), which admits less current, which lowers the torque until it again matches.

⚖

On No Load

Small torque, small current, high back EMF

On no load the motor needs only a small torque — enough to overcome friction and windage — so it withdraws a small current, and therefore

\[E_b = V - I_aR_a \approx V\]

The back EMF is at its highest when the current is at its lowest. The two move in opposite directions, which is precisely what makes the loop self-correcting.

This is worth stating carefully because it is easy to get backwards. Light load means high \(E_b\) and low \(I_a\); heavy load means low \(E_b\) and high \(I_a\). The extreme case is a stalled motor: \(E_b = 0\) and \(I_a\) at its maximum possible value.

1 Worked Example 34.1 — Back EMF and Starting Current

Problem. A 250 V shunt motor takes 40 A from the line. Its armature resistance is 0.20 \(\Omega\) and shunt field resistance 125 \(\Omega\). Find the back EMF while running, and the current that would flow at the instant of switching on if it were connected directly to the supply.

Currents while running.

\[I_{sh} = \frac{250}{125} = 2~\mathrm{A}, \qquad I_a = 40 - 2 = 38~\mathrm{A}\]

Back EMF.

\[E_b = V - I_aR_a = 250 - (38)(0.20) = 250 - 7.6 = 242.4~\mathrm{V}\]

At the instant of starting. The armature is stationary, so \(E_b = 0\):

\[I_a = \frac{V}{R_a} = \frac{250}{0.20} = 1250~\mathrm{A}\]

Ratio.

\[\frac{1250}{38} = 32.9\]

Comment. Nearly 33 times the running current. The back EMF absorbs 97 % of the applied voltage in normal running, leaving only 7.6 V to drive the current through the armature — which is why the armature resistance can be so small without the current being enormous.

Remove that 242.4 V and only the 0.20 \(\Omega\) is left. The winding would be destroyed in seconds, and the supply would very likely trip. Chapter 36 inserts a starting resistance to limit the current until \(E_b\) has built up.

Section 34-4

Condition for Maximum Power

The mechanical power developed is \(P_m = E_bI_a\). Expressing it in terms of the current alone:

\[\begin{aligned} P_m &= VI_a - I_a^{2}R_a \\[4pt] \Rightarrow \frac{\mathrm{d}P_m}{\mathrm{d}I_a} &= V - 2I_aR_a = 0 \quad\Rightarrow\quad I_aR_a = \frac{V}{2} \\[4pt] \therefore\ V &= E_b + I_aR_a = E_b + \frac{V}{2} \quad\Rightarrow\quad \boxed{E_b = \frac{V}{2}} \end{aligned}\]

Developed power against armature current for a 250 V machine with \(R_a = 0.20~\Omega\).

! A Condition Never Used in Practice

At \(E_b = V/2\) exactly half the input is converted and half is lost in the armature, so the armature-circuit efficiency is 50 %.

Worse, the current is \(V/2R_a\), which for a real machine is many times the rated value. The winding would burn out long before the condition could be reached, and the commutation would have failed already, since Chapter 31's reactance voltage scales with \(I_a\).

The result is nonetheless worth knowing, because it establishes that a motor's power output has a maximum — it does not increase indefinitely with current — and it identifies where the theoretical ceiling lies.

2 Worked Example 34.2 — The Maximum-Power Point

Problem. For the motor of Example 34.1, find the armature current, back EMF and developed power at the condition for maximum power. Compare with the rated condition and comment.

At maximum power.

\[I_a = \frac{V}{2R_a} = \frac{250}{(2)(0.20)} = 625~\mathrm{A}, \qquad E_b = \frac{V}{2} = 125~\mathrm{V}\]

\[P_m = E_bI_a = (125)(625) = 78\,125~\mathrm{W}\]

Armature copper loss there.

\[I_a^{2}R_a = (625)^{2}(0.20) = 78\,125~\mathrm{W}\]

Exactly equal to the power developed, as \(E_b = V/2\) requires.

At the rated condition.

\[P_m = E_bI_a = (242.4)(38) = 9211~\mathrm{W}\]

Comparison.

\[\frac{78\,125}{9211} = 8.48\ \text{times the power, at}\ \frac{625}{38} = 16.4\ \text{times the current}\]

Comment. The power has risen by a factor of 8.5 but the current by 16.4 — and the losses by the square of that, a factor of 270. Doubling the current beyond the rated point buys progressively less power for rapidly more heat, until at 625 A the machine gains nothing at all from further current.

This is the same diminishing return that gave Chapter 33 its efficiency curve, seen from the other side. A machine is rated by what it can cool, not by what it can theoretically produce.

Section 34-5

Torque Equation

Torque production in a DC motor — Torque in a DC motor.

Equating the converted electrical power to the mechanical power:

\[\begin{aligned} P_m &\Rightarrow E_bI_a = T_a\omega \\[4pt] &\Rightarrow \left(\frac{\Phi PN}{60}\times\frac{Z}{A}\right)I_a = T_a\left(\frac{2\pi N}{60}\right) \\[4pt] &\Rightarrow T_a = \frac{1}{2\pi}\times\frac{\Phi PZ}{A}\times I_a \\[4pt] &\Rightarrow \boxed{T_a = K_a\Phi I_a} \end{aligned}\]

\[T_{sh} = T_a - \text{mechanical losses}\]

🔗

The Constant Is Chapter 27's

Same number, both directions

\[K_a = \frac{ZP}{2\pi A}\]

This is exactly the \(k_a\) of Chapter 27, where it appeared in \(E_g = k_a\Phi\omega\). The derivation above shows it could not have been otherwise — the two equations come from the same power balance, read in opposite directions.

\[E_b = K_a\Phi\omega \qquad\qquad T_a = K_a\Phi I_a\]

The distinction between armature torque \(T_a\) and shaft torque \(T_{sh}\) matters in problems. \(T_a\) is what the conductors produce; \(T_{sh}\) is what emerges at the coupling after friction, windage and iron losses have taken their share — the last stage of Chapter 33's power-flow diagram.

3 Worked Example 34.3 — Armature and Shaft Torque

Problem. The motor of Example 34.1 is 4-pole lap-wound with 500 armature conductors and a flux of 25 mWb per pole. Find the machine constant, the speed, the armature torque and the shaft torque, taking the mechanical losses as 500 W.

Machine constant. Lap-wound, so \(A = P = 4\):

\[K_a = \frac{ZP}{2\pi A} = \frac{(500)(4)}{2\pi(4)} = \frac{500}{2\pi} = 79.58\]

Speed. From \(E_b = \Phi PNZ/60A\):

\[N = \frac{60AE_b}{\Phi ZP} = \frac{(60)(4)(242.4)}{(0.025)(500)(4)} = \frac{58\,176}{50} = 1163.5~\mathrm{rev/min}\]

Armature torque.

\[T_a = K_a\Phi I_a = (79.58)(0.025)(38) = 75.60~\mathrm{N\,m}\]

Check via power: \(\omega = 2\pi(1163.5)/60 = 121.84\) rad/s, and

\[T_a = \frac{E_bI_a}{\omega} = \frac{(242.4)(38)}{121.84} = \frac{9211}{121.84} = 75.60~\mathrm{N\,m} \ \checkmark\]

Shaft torque. The mechanical losses correspond to a torque of

\[T_{\text{loss}} = \frac{500}{121.84} = 4.10~\mathrm{N\,m}\]

\[T_{sh} = 75.60 - 4.10 = 71.50~\mathrm{N\,m}\]

Comment. The shaft torque is 5.4 % below the armature torque, which is the mechanical loss expressed as a fraction. Note that the loss torque is nearly constant while the armature torque varies with load, so the discrepancy is proportionally far worse at light load — at a tenth of this current, \(T_a\) would be 7.56 N·m and the loss torque still 4.10 N·m, leaving only 3.46 N·m at the shaft.

Section 34-6

Torque in Shunt and Series Motors

Series motor

\[\Phi \propto I_a \quad\Longrightarrow\quad T_a \propto I_a^{2}\]

Because the field carries the armature current. Below saturation only — beyond it the flux is fixed and \(T_a \propto I_a\) again.

Shunt motor

\[\Phi = \text{constant} \quad\Longrightarrow\quad T_a \propto I_a\]

Because the field is across the supply and takes a current fixed by \(V/R_{sh}\).

This single difference generates the whole of Chapter 35. A series motor's square-law torque gives it enormous pull at starting, which is why traction and hoists use it — but the same relation, run backwards, means the flux vanishes when the load does, and \(N \propto 1/\Phi\) then sends the speed towards infinity. The characteristic that makes a series motor excellent under load makes it dangerous without one.

Section 34-7

Speed of a DC Motor

\[\begin{aligned} E_b &= V - I_aR_a \\[4pt] \Rightarrow \frac{\Phi PN}{60}\times\left(\frac{Z}{A}\right) &= V - I_aR_a \\[4pt] \Rightarrow N &= \frac{V - I_aR_a}{\Phi}\times\left(\frac{60A}{ZP}\right) \\[4pt] \Rightarrow N &= \frac{E_b}{\Phi}\times\left(\frac{60A}{ZP}\right) \\[4pt] \Rightarrow &\boxed{N = K\frac{E_b}{\Phi}} \end{aligned}\]

📏

The Ratio Method

Avoid the constants entirely

Because \(K\) is fixed for a given machine, most problems are best done as a ratio:

\[\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}}\times\frac{\Phi_1}{\Phi_2}\]

Neither \(Z\), \(P\) nor \(A\) need be known — only the two back EMFs and the flux ratio. This is by far the commonest calculation in DC motor work.

Both quantities in the ratio deserve care. \(E_b\) must be recomputed from \(V - I_aR_a\) at each condition, and the flux ratio is found from the field current — via the magnetisation curve of Chapter 28 if the machine is saturated, or proportionally if it is not.

4 Worked Example 34.4 — Speed Changes by the Ratio Method

Problem. A 220 V shunt motor runs at 1000 rev/min with an armature current of 40 A and an armature resistance of 0.25 \(\Omega\). Find the new speed if (a) the load increases so that the armature current becomes 60 A, and (b) the flux is reduced by 20 % with the armature current unchanged at 40 A.

Initial back EMF.

\[E_{b1} = 220 - (40)(0.25) = 220 - 10 = 210~\mathrm{V}\]

(a) Higher load, same flux.

\[E_{b2} = 220 - (60)(0.25) = 220 - 15 = 205~\mathrm{V}\]

\[N_2 = N_1\frac{E_{b2}}{E_{b1}} = 1000\left(\frac{205}{210}\right) = 976.2~\mathrm{rev/min}\]

A drop of only 2.4 % for a 50 % increase in armature current — the shunt motor's near-constant speed.

(b) Flux reduced 20 %, same current. The back EMF is unchanged at 210 V, but the flux ratio is now \(\Phi_2/\Phi_1 = 0.80\):

\[N_2 = N_1\frac{E_{b2}}{E_{b1}}\times\frac{\Phi_1}{\Phi_2} = 1000\left(\frac{210}{210}\right)\left(\frac{1}{0.80}\right) = 1250~\mathrm{rev/min}\]

A rise of 25 % — far more than the load change achieved.

Comment. Compare the two. Changing the load by half moved the speed 2.4 %; weakening the field by a fifth moved it 25 %. The flux is by far the more powerful lever on speed, which is why field control is the principal method above rated speed in Chapter 37.

Note also that part (b) is not a steady state. At the instant the flux is reduced the current is still 40 A, but the sequence of Section 34-8 then raises it sharply before the speed settles.

Section 34-8

Relation Between Torque and Speed

\[\boxed{N = K\frac{E_b}{\Phi}} \qquad\qquad \boxed{T_a \propto \Phi I_a}\]

\(T_a\) is a function of \(\Phi\) and \(I_a\) but is independent of \(N\).
\(N\) depends on \(T\), and not vice versa.

? The Apparent Paradox

The speed equation says \(\Phi\uparrow \Rightarrow N\downarrow\), and the torque equation says \(\Phi\uparrow \Rightarrow T_a\uparrow\). Both together seem to say that more torque produces less speed.

That is not possible, because torque always tends to produce rotation. If \(T\uparrow\) then \(N\) must rise, not fall. The resolution is that the two equations cannot be applied simultaneously as though the other variables held still.

🔁

The Flux-Weakening Sequence

What actually happens, step by step

Suppose \(\Phi\) is reduced by lowering \(I_f\):

Since \(E_b \propto N\Phi\), the back EMF falls — the speed \(N\) is momentarily constant because of the inertia of the heavy armature.
\(E_b\downarrow \Rightarrow I_a = (V - E_b)/R_a\) rises. Moreover, because \(R_a\) is very small, a small fall in \(E_b\) produces a large rise in \(I_a\).
In \(T_a \propto \Phi I_a\), the fall in \(\Phi\) is more than counterbalanced by the large rise in \(I_a\), so \(T_a\) rises.
\(T_a\uparrow \Rightarrow N\uparrow\), and the motor accelerates until the rising \(E_b\) brings the current back down.

The paradox dissolves because \(I_a\) is not a free variable. It is determined by \(E_b\), which is determined by \(N\) and \(\Phi\) together.

Step 2 is the crux, and it is worth putting numbers to it — Example 34.5 does so. The smallness of \(R_a\), which seemed a mere detail in Chapter 33's loss accounting, is what makes the whole sequence work.

5 Worked Example 34.5 — The Sequence in Numbers

Problem. The motor of Example 34.4 runs at 1000 rev/min with \(I_a = 40\) A. The field is suddenly weakened by 20 %. Find the armature current and torque at the instant before the speed has had time to change, and compare with the initial values.

Before weakening.

\[E_{b1} = 210~\mathrm{V}, \qquad I_a = 40~\mathrm{A}, \qquad T_a \propto \Phi_1(40) = 40\Phi_1\]

The instant after weakening. The speed is still 1000 rev/min but the flux is 0.80 of its previous value, so

\[E_{b2} = (210)(0.80) = 168~\mathrm{V}\]

New armature current.

\[I_a = \frac{V - E_b}{R_a} = \frac{220 - 168}{0.25} = \frac{52}{0.25} = 208~\mathrm{A}\]

New torque.

\[T_a \propto \left(0.80\Phi_1\right)(208) = 166.4\,\Phi_1\]

\[\frac{T_{a2}}{T_{a1}} = \frac{166.4}{40} = 4.16\]

Comment. A 20 % reduction in flux produced a 5.2-fold rise in armature current and a 4.2-fold rise in torque. The flux fell by a fifth and the current rose by more than five times, which is exactly the "counterbalanced by \(I_a\uparrow\uparrow\)" of step 3 — and it shows why the double arrow is not rhetorical.

The reason is the smallness of \(R_a\). The back EMF fell by 42 V, but the driving voltage \(V - E_b\) rose from 10 V to 52 V — a factor of 5.2 — because it was small to begin with.

The motor now accelerates, raising \(E_b\) again, and settles at the 1250 rev/min found in Example 34.4(b). The transient current is the reason field rheostats are moved slowly, and why a sudden open circuit in a shunt field is dangerous: with \(\Phi\) approaching zero, the speed rises without limit.

Section 34-9

Speed Regulation

The change in speed when the load on the motor is reduced from its rated value to zero, expressed as a percentage of the rated-load speed.

\[\%\ \text{speed regulation} = \frac{N_{\text{no load}} - N_{\text{full load}}}{N_{\text{full load}}} \times 100\]

Table 34.1 — Typical speed regulation.
Motor	Regulation	Because
Shunt	2–8 %	flux constant; only \(I_aR_a\) changes
Cumulative compound	10–25 %	series field weakens as load falls
Series	very large	flux falls with load, so \(N \propto 1/\Phi\) rises steeply

Note the parallel with Chapter 29. Voltage regulation measured how well a generator held its voltage; speed regulation measures how well a motor holds its speed, and both are defined as a change divided by the full-load value.

Section 34-10

Summary and Key Formulas

Back EMF arises because the rotating armature cuts flux. It opposes \(V\), so \(E_b = V - I_aR_a\) and \(I_a = (V - E_b)/R_a\).
Conversion of power, \(E_bI_a\), is possible only because of \(E_b\). A stalled motor converts nothing.
\(E_b\) makes the motor self-regulating: more load \(\Rightarrow\) \(N\downarrow\) \(\Rightarrow\) \(E_b\downarrow\) \(\Rightarrow\) \(I_a\uparrow\) \(\Rightarrow\) \(T\uparrow\). On no load \(E_b\) is at its highest and \(I_a\) at its lowest.
Maximum power occurs at \(E_b = V/2\), that is \(I_a = V/2R_a\) — never used, since the armature-circuit efficiency is then 50 %.
Torque: \(T_a = K_a\Phi I_a\) with \(K_a = ZP/2\pi A\) — the same constant as in \(E_b = K_a\Phi\omega\). Shaft torque is \(T_a\) less the mechanical losses.
Series motor: \(T_a \propto I_a^{2}\). Shunt motor: \(T_a \propto I_a\).
Speed: \(N = KE_b/\Phi\), most usefully applied as the ratio \(N_2/N_1 = (E_{b2}/E_{b1})(\Phi_1/\Phi_2)\).
\(T_a\) is independent of \(N\); \(N\) depends on \(T\) and not the reverse.
Weakening the flux raises the speed, because the resulting large rise in \(I_a\) more than offsets the fall in \(\Phi\), so the torque rises and the motor accelerates.
Speed regulation \(= (N_{NL} - N_{FL})/N_{FL} \times 100\), typically 2–8 % for a shunt motor.

Table 34.2 — Formulas of this chapter.
Quantity	Formula	Notes
Back EMF	\(E_b = V - I_aR_a - 2v_b\)	brush drop if given
Armature current	\(I_a = (V - E_b)/R_a\)	the self-regulating relation
Starting current	\(V/R_a\)	20–30× rated
Converted power	\(P_m = E_bI_a = VI_a - I_a^{2}R_a\)	—
Maximum power	\(E_b = V/2\), \(I_a = V/2R_a\)	theoretical only
Machine constant	\(K_a = ZP/2\pi A\)	same as Chapter 27
Armature torque	\(T_a = K_a\Phi I_a = E_bI_a/\omega\)	—
Shaft torque	\(T_{sh} = T_a - T_{\text{loss}}\)	\(T_{\text{loss}} = P_{\text{mech}}/\omega\)
Speed	\(N = KE_b/\Phi\)	\(K = 60A/ZP\)
Speed ratio	\(\dfrac{N_2}{N_1} = \dfrac{E_{b2}}{E_{b1}}\dfrac{\Phi_1}{\Phi_2}\)	the working formula
Speed regulation	\(\dfrac{N_{NL} - N_{FL}}{N_{FL}}\times 100\)	2–8 % shunt

Section 34-11

Common Mistakes

Thinking \(E_b\) falls on no load. It is at its highest then, because \(I_a\) is smallest. The two always move oppositely.
Using \(I_L\) instead of \(I_a\). For a shunt motor \(I_a = I_L - I_{sh}\).
Writing \(E_b = V + I_aR_a\). That is the generator form. A motor's back EMF is less than the applied voltage.
Treating the maximum-power condition as a design target. It implies 50 % armature efficiency and a destructive current.
Confusing armature and shaft torque. The mechanical losses lie between them.
Applying \(T_a \propto I_a^{2}\) to a series motor past saturation. Beyond the knee the flux is fixed and the law reverts to \(T_a \propto I_a\).
Forgetting to recompute \(E_b\) in a ratio problem. A change in \(I_a\) changes \(E_b\) as well.
Inverting the flux ratio. Since \(N \propto 1/\Phi\), the ratio is \(\Phi_1/\Phi_2\), not \(\Phi_2/\Phi_1\).
Concluding that raising the flux raises both torque and speed. It raises torque and lowers speed; the sequence of Section 34-8 explains why there is no contradiction.
Opening a shunt field while running. With \(\Phi \to 0\) and \(N \propto 1/\Phi\), the motor races.

Section 34-12

Chapter Review

Practice Problems

Find \(I_a = I_L - I_{sh}\) first, then \(E_b\), then whatever is asked. For speed changes use the ratio form.

P34.1 A 220 V shunt motor takes 30 A with \(R_a = 0.30~\Omega\) and \(R_{sh} = 110~\Omega\). Find \(E_b\) and the converted power.

Show answer

\[I_{sh} = 2~\mathrm{A}, \qquad I_a = 28~\mathrm{A}\]

\[E_b = 220 - (28)(0.30) = 220 - 8.4 = 211.6~\mathrm{V}\]

\[P_m = (211.6)(28) = 5925~\mathrm{W}\]
P34.2 For P34.1, find the starting current if connected directly across the supply, as a multiple of the running armature current.

Show answer

\[I_a = \frac{220}{0.30} = 733.3~\mathrm{A}, \qquad \frac{733.3}{28} = 26.2\ \text{times}\]
P34.3 The motor of P34.1 runs at 900 rev/min. Find the armature torque and, if mechanical losses are 300 W, the shaft torque.

Show answer

\[\omega = \frac{2\pi(900)}{60} = 94.25~\mathrm{rad/s}\]

\[T_a = \frac{E_bI_a}{\omega} = \frac{5925}{94.25} = 62.87~\mathrm{N\,m}\]

\[T_{\text{loss}} = \frac{300}{94.25} = 3.18~\mathrm{N\,m}, \qquad T_{sh} = 59.69~\mathrm{N\,m}\]
P34.4 A 240 V motor with \(R_a = 0.40~\Omega\) runs at 800 rev/min with \(I_a = 25\) A. Find the speed if the load raises the current to 50 A, flux constant.

Show answer

\[E_{b1} = 240 - 10 = 230~\mathrm{V}, \qquad E_{b2} = 240 - 20 = 220~\mathrm{V}\]

\[N_2 = 800\left(\frac{220}{230}\right) = 765.2~\mathrm{rev/min}\]
P34.5 For P34.4, find the speed if instead the flux is reduced to 75 % with \(I_a\) settling back to 25 A.

Show answer

\[N_2 = 800\left(\frac{230}{230}\right)\left(\frac{1}{0.75}\right) = 1066.7~\mathrm{rev/min}\]
P34.6 Find the maximum power condition for a 400 V motor with \(R_a = 0.15~\Omega\), and comment on its practicality.

Show answer

\[I_a = \frac{400}{(2)(0.15)} = 1333~\mathrm{A}, \qquad E_b = 200~\mathrm{V}\]

\[P_m = (200)(1333) = 266.7~\mathrm{kW}\]
Entirely impractical: 1333 A would be many times the rating of any machine with this armature resistance, and half the input — 267 kW — would be dissipated as heat in the armature.
P34.7 A shunt motor runs at 1020 rev/min on no load and 980 rev/min at full load. Find the speed regulation.

Show answer

\[\frac{1020 - 980}{980} \times 100 = \frac{40}{980} \times 100 = 4.08\,\%\]
Well inside the 2–8 % typical of shunt motors.
P34.8 A 240 V motor with \(R_a = 0.30~\Omega\) takes 3 A on no load and 45 A at full load, flux constant. Find the speed regulation.

Show answer

\[E_{b,NL} = 240 - 0.9 = 239.1~\mathrm{V}, \qquad E_{b,FL} = 240 - 13.5 = 226.5~\mathrm{V}\]
Since \(N \propto E_b\) at constant flux,
\[\text{regulation} = \frac{239.1 - 226.5}{226.5}\times 100 = 5.56\,\%\]
The speeds themselves are not needed — only their ratio.
P34.9 Explain why weakening the field of a running motor makes it speed up, given that \(T_a \propto \Phi I_a\).
Show answer
Because \(I_a\) is not independent. The sequence is:
1. \(\Phi\downarrow\) and \(N\) momentarily constant (armature inertia), so \(E_b \propto N\Phi\) falls.
2. \(I_a = (V - E_b)/R_a\) rises, and because \(R_a\) is very small the rise is large.
3. In \(T_a \propto \Phi I_a\) the large rise in \(I_a\) outweighs the modest fall in \(\Phi\), so \(T_a\) rises.
4. The excess torque accelerates the motor until the rising \(E_b\) restores the balance.
Example 34.5 puts numbers to it: a 20 % flux reduction gave a 5.2-fold current rise and a 4.2-fold torque rise.
P34.10 Why is an open circuit in the shunt field of a running motor dangerous?

Show answer
Because \(N = KE_b/\Phi\), so as \(\Phi \to 0\) the speed rises without limit. Only the residual flux remains, and the motor accelerates until something fails mechanically — usually the armature banding, at which point the winding is thrown out of its slots.
The danger is greatest on light load, since there is little load torque to restrain the acceleration. A loaded motor may simply stall instead, drawing heavy current.

Shunt and compound motors are therefore protected by a no-volt release with a field failure trip, which Chapter 36 describes as part of the starter.

Multiple-Choice Questions

MCQ 1. The back EMF of a DC motor:
(a) aids the applied voltage (b) opposes the applied voltage (c) equals it (d) is zero while running

Show answer
(b) opposes the applied voltage — hence "counter" or "back" EMF.
MCQ 2. On no load, the back EMF is:
(a) lowest (b) highest (c) zero (d) unchanged

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(b) highest, because \(I_a\) and therefore \(I_aR_a\) are smallest.
MCQ 3. The power converted from electrical to mechanical form is:
(a) \(VI_a\) (b) \(E_bI_a\) (c) \(I_a^{2}R_a\) (d) \(VI_L\)

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(b) \(E_bI_a\) — conversion is possible only because of the back EMF.
MCQ 4. Maximum mechanical power occurs when:
(a) \(E_b = V\) (b) \(E_b = V/2\) (c) \(E_b = 0\) (d) \(E_b = 2V\)

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(b) \(E_b = V/2\), giving \(I_a = V/2R_a\) — theoretical only.
MCQ 5. The torque of a DC motor is:
(a) \(K_a\Phi\omega\) (b) \(K_a\Phi I_a\) (c) \(K_aI_a\omega\) (d) \(K_a\Phi/I_a\)

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(b) \(K_a\Phi I_a\), with the same \(K_a\) as in the back-EMF equation.
MCQ 6. For a series motor below saturation, torque varies as:
(a) \(I_a\) (b) \(I_a^{2}\) (c) \(\sqrt{I_a}\) (d) \(1/I_a\)

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(b) \(I_a^{2}\), since \(\Phi \propto I_a\) as well.
MCQ 7. The speed of a DC motor is proportional to:
(a) \(E_b\Phi\) (b) \(E_b/\Phi\) (c) \(\Phi/E_b\) (d) \(E_b\) alone

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(b) \(E_b/\Phi\).
MCQ 8. Weakening the field of a running motor causes the speed to:
(a) fall (b) rise (c) stay the same (d) oscillate

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(b) rise, by the sequence of Section 34-8.
MCQ 9. The armature torque of a DC motor is:
(a) proportional to \(N\) (b) inversely proportional to \(N\) (c) independent of \(N\) (d) proportional to \(N^{2}\)

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(c) independent of \(N\). It is a function of \(\Phi\) and \(I_a\) only; the speed depends on the torque, not the reverse.
MCQ 10. Typical speed regulation of a shunt motor is:
(a) 2–8 % (b) 20–30 % (c) 50 % (d) zero

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(a) 2–8 %, which is why it is called a constant-speed motor.

Conceptual Questions

Explain how back EMF arises in a DC motor and why it opposes the supply.
State the advantages of back EMF and explain why conversion is impossible without it.
Trace the self-regulating sequence for an increase and a decrease of load.
Derive the condition for maximum power and explain why it has no practical use.
Derive the torque equation and show that its constant is the same as that of the back-EMF equation.
Distinguish armature torque from shaft torque.
Derive the speed equation and explain the advantage of the ratio form.
Resolve the apparent paradox that increasing flux raises torque but lowers speed.

Looking Ahead

Section 34-6 gave the two torque laws — \(T_a \propto I_a\) for a shunt motor and \(T_a \propto I_a^{2}\) for a series motor — and Section 34-7 the speed equation. Chapter 35 combines them into the characteristics of each motor type: \(T_a\) against \(I_a\), \(N\) against \(I_a\), and \(N\) against \(T_a\). They turn out to be the mirror images of the generator characteristics of Chapter 29, and they explain every application in Chapter 26's table.

Chapter 36 then answers the question Example 34.1 raised. A starting current of 1250 A cannot be permitted, so a starter inserts resistance in the armature circuit and cuts it out in steps as \(E_b\) builds up — together with the no-volt and overload releases that also protect against the field failure of P34.10.