Electrical vs Mechanical Generator Frequency Explained

A generator shaft completes 6 revolutions per second, but an oscilloscope shows 12 positive voltage pulses every second. Is the generator producing electricity at 12 Hz, or 6 Hz?

Both numbers describe something real, but they describe different things. With one north-south magnetic pole pair and a split-ring commutator, the coil generates 6 complete AC cycles per second. The commutator then turns the negative half of each cycle into another positive pulse, giving 12 pulses per second at the terminals.

Mechanical frequency is the number of complete shaft revolutions per second. Electrical frequency is the number of complete cycles of alternating emf induced in the winding per second.

For a generator with \(p\) magnetic pole pairs:

\[
f_{\mathrm e}=p f_{\mathrm m}
\]

Here, \(f_{\mathrm e}\) is electrical frequency in hertz (Hz), \(f_{\mathrm m}\) is mechanical frequency in hertz, and \(p\) is the number of magnetic pole pairs. One hertz means one complete cycle per second.

With one pole pair, the frequencies are equal. With two pole pairs, the electrical frequency is twice the mechanical frequency.

A split-ring commutator introduces a third quantity: the frequency of the voltage pulses at the external terminals. That is not necessarily the same as the electrical frequency inside the coil.

01Measuring mechanical and electrical frequency

Imagine investigating a simple generator with one rectangular coil rotating between two magnetic poles.

The north pole sits on the left, the south pole on the right, and the magnetic field points horizontally across the gap. A motor rotates the coil around a shaft perpendicular to that field.

We need two measurements.

First, put a visible mark on the shaft. A rotation sensor counts how often the mark completes a revolution. This gives the mechanical frequency.

Second, connect the coil to an oscilloscope. The oscilloscope records how the voltage across two fixed terminals changes with time.

There is a practical problem: the coil rotates, but the oscilloscope does not. Connecting ordinary wires directly would twist them around the shaft.

The first solution is to use slip rings.

Each end of the coil connects to its own complete metal ring, mounted on the rotating shaft. A stationary conducting brush presses against each ring. The brushes connect to the external circuit and oscilloscope.

This arrangement transfers the electrical signal without twisting the wires.

But it also lets us investigate a more interesting question: why should one revolution produce an alternating voltage at all?

02From coil geometry to the frequency of induced emf

Picture the rectangular coil with its broad face perpendicular to the magnetic field. The magnetic field passes directly through the area enclosed by the coil.

Would you expect the induced voltage to be greatest at that position?

It is tempting to say yes because the magnetic flux is greatest. But at that exact instant, the flux is momentarily not changing. The induced emf is therefore zero.

A quarter-turn later, the coil’s face is parallel to the magnetic field. The flux through the coil is zero, but it is changing at its greatest rate. The induced emf now has its greatest magnitude.

This is the connection between geometry and frequency.

Deriving the waveform

We describe the coil’s orientation using an angle \(\theta\), measured between the magnetic field and the normal to the coil.

The normal is an imaginary line perpendicular to the coil’s flat surface.

For a coil with \(N\) identical turns, each enclosing area \(A\), rotating in a uniform magnetic field of strength \(B\), the magnetic flux linkage is:

\[
\lambda=NBA\cos\theta
\]

Here:

• \(\lambda\) is magnetic flux linkage, measured in weber-turns (Wb turns).
• \(N\) is the number of turns, with no unit.
• \(B\) is magnetic flux density, measured in tesla (T).
• \(A\) is the area of one turn, measured in square metres (m²).
• \(\theta\) is the angle between the magnetic field and the coil’s normal.

Flux linkage accounts for the magnetic flux through every turn of the coil.

If the shaft rotates at constant angular speed \(\omega\), measured in radians per second (rad/s), we can choose the starting position so that:

\[
\theta=\omega t
\]

The symbol \(t\) represents time in seconds.

Faraday’s law tells us that the induced emf depends on the rate of change of flux linkage:

\[
\mathcal E=-\frac{d\lambda}{dt}
\]

The minus sign represents Lenz’s law: the induced emf has a direction that opposes the change producing it.

Substituting the rotating-coil expression gives:

\[
\begin{aligned}
\lambda&=NBA\cos(\omega t)\\
\mathcal E&=-\frac{d\lambda}{dt}\\
&=NBA\omega\sin(\omega t)
\end{aligned}
\]

The induced emf is sinusoidal. We have chosen the coil’s reference polarity and increasing rotation angle so that the first half-cycle is positive.

Reversing the reference polarity would reverse the signs, but not the frequency.

Follow one complete revolution

For the simple one-pole-pair generator, consider four positions.

The flux starts positive, becomes negative, and returns to its original value after one complete revolution.

Meanwhile, the induced emf completes one positive half-cycle and one negative half-cycle.

That is one complete AC cycle per mechanical revolution.

Therefore, with one magnetic pole pair:

\[
f_{\mathrm e}=f_{\mathrm m}
\]

The shaft speed determines how quickly this entire sequence repeats.

03The experiment: change the rings, not the rotation

Now return to the electrical contacts.

Keep the same motor, magnets, shaft, and rectangular coil. Change only the mechanism connecting the rotating coil to the stationary brushes.

This comparison separates changes in the generated emf from changes in the voltage delivered to the external circuit.

First arrangement: two slip rings

Each end of the coil connects permanently to a different continuous metal ring.

Call the coil ends X and Y, and the stationary brushes P and Q.

With slip rings, X always connects to P, and Y always connects to Q.

The induced emf reverses direction every half-turn. Because the external connections do not reverse, the oscilloscope displays positive and negative half-cycles.

The output is alternating voltage.

With one pole pair, its frequency equals the shaft’s rotational frequency.

Revised arrangement: a split-ring commutator

The slip-ring design maintains electrical contact, but it cannot keep the terminal voltage at one polarity.

To produce a unidirectional output, replace the two continuous rings with a split-ring commutator.

A simple split-ring commutator contains two conducting semicircular segments separated by insulating gaps. Each segment connects to one end of the rotating coil.

Two stationary brushes press against the segments.

During the first half-turn, X connects to P and Y connects to Q.

During the second half-turn, X connects to Q and Y connects to P.

The connections exchange as the induced emf reverses. With the brushes correctly positioned, this switching occurs around the emf’s zero crossings.

As a result, the voltage across the external terminals keeps the same polarity during forward rotation.

The induced emf in the coil has not suddenly doubled its frequency. The commutator has changed which coil end is connected to each external terminal.

For an ideal one-pole-pair generator rotating in our chosen forward direction:

\[
\begin{aligned}
V_{\mathrm{slip}}&=E_0\sin\theta\\
V_{\mathrm{split}}&=E_0|\sin\theta|
\end{aligned}
\]

Here, \(V_{\mathrm{slip}}\) and \(V_{\mathrm{split}}\) are terminal voltages in volts (V), and \(E_0\) is the peak induced emf in volts.

The absolute-value operation represents the reversal of the external connections. It is valid for this ideal arrangement, with the brushes correctly timed and terminal polarity chosen so that forward rotation produces positive pulses.

The split-ring output is called pulsating DC. Its polarity stays the same, but its magnitude repeatedly falls to zero.

For one pole pair, there are two positive pulses per revolution.

Illustrative data: the same coil with different contacts

Suppose the shaft rotates at 300 revolutions per minute (rpm), with one magnetic pole pair.

Its mechanical frequency is:

\[
f_{\mathrm m}=\frac{300}{60}=5.0\ \mathrm{Hz}
\]

One revolution takes:

\[
T_{\mathrm m}=\frac{1}{5.0}=0.200\ \mathrm{s}
\]

The symbol \(T_{\mathrm m}\) represents the mechanical period in seconds.

Assume the generated emf has a peak magnitude of 6.0 V, with negligible internal resistance and ideal electrical contacts.

The following are calculated illustrative readings, not measurements from a real experiment.

One millisecond (ms) is \(0.001\) seconds.

With slip rings, the voltage completes one positive-negative cycle in 200 ms. Its electrical frequency is 5 Hz.

With a split ring, positive peaks occur at 50 ms and 150 ms. They are separated by 100 ms.

The external pulse frequency is therefore:

\[
f_{\mathrm{ripple}}
=\frac{1}{0.100}
=10\ \mathrm{Hz}
\]

Here, \(f_{\mathrm{ripple}}\) is the rate at which the same-polarity voltage pulses repeat.

It is twice the induced AC frequency, even though the shaft speed and magnetic geometry have not changed.

04Why electrical frequency can exceed mechanical frequency

So far, the generator has contained one north-south pole pair.

Now imagine a machine with four magnetic poles arranged around its rotor: north, south, north, south.

That is two pole pairs.

During one mechanical revolution, a suitable winding experiences two complete repetitions of the magnetic field pattern.

The induced emf can therefore complete two AC cycles during one shaft revolution.

For an idealised winding in a symmetrical multipole generator, the changing flux linkage can be modelled as:

\[
\lambda=\lambda_0\cos(p\theta)
\]

The symbol \(\lambda_0\) represents the maximum magnitude of flux linkage in weber-turns.

Applying Faraday’s law gives:

\[
\mathcal E
=p\lambda_0\omega\sin(p\theta)
\]

This is a sinusoidal multipole model, not the exact uniform-field formula for every real coil geometry. A real winding’s flux linkage depends on its shape, the distribution of magnetic poles, and the magnetic field.

The important feature is the angle \(p\theta\).

When the shaft advances through \(360^\circ\), the electrical waveform advances through \(p\) complete cycles.

Thus:

\[
f_{\mathrm e}=p f_{\mathrm m}
\]

If the shaft speed is supplied as \(n\) in rpm:

\[
f_{\mathrm e}=p\frac{n}{60}
\]

For example, a four-pole generator at 300 rpm has two pole pairs.

Its shaft frequency is 5 Hz, but the induced emf frequency is:

\[
f_{\mathrm e}=2(5)=10\ \mathrm{Hz}
\]

This frequency increase comes from the magnetic pole arrangement, not from the electrical contacts.

An important limit of the split-ring shortcut

You might now multiply by two again and predict 20 positive pulses per second from a split ring.

That can be correct for a properly rectified 10 Hz emf. But there is a condition.

A basic two-segment split-ring commutator reverses its connections every \(180^\circ\) of mechanical rotation.

In a two-pole-pair winding, the induced emf reverses every \(90^\circ\) of mechanical rotation.

Those events no longer line up.

Simply fitting the original two-segment commutator to a four-pole winding does not guarantee a unidirectional output. A multipole DC generator needs commutation designed around its winding and magnetic geometry.

With properly timed rectification, an ideal sinusoidal emf of frequency \(f_{\mathrm e}\) produces a pulse rate of \(2f_{\mathrm e}\). Without that condition, the shortcut is unsafe.

This becomes especially important when interpreting unusual oscilloscope traces.

05Frequency is about timing; emf magnitude is about the flux derivative

Two generators can produce the same electrical frequency but very different peak voltages.

For a single coil in the uniform-field model, the maximum induced emf is:

\[
E_0=NBA\omega
\]

The angular speed is related to mechanical frequency by:

\[
\omega=2\pi f_{\mathrm m}
\]

Increasing the shaft speed increases both the induced frequency and the maximum emf, provided the magnetic field, coil dimensions, and number of turns stay unchanged.

But changing the ring type alters neither the shaft speed nor the emf generated inside the coil.

06How to identify which frequency a question is asking for

Before using a frequency formula, identify what is actually being counted.

The last relationship assumes ideal rectification of a sinusoidal voltage.

The first two relationships concern the rotating magnetic geometry. The others concern what is measured at the terminals.

There are three useful checks when a question gives an unexpected frequency.

First, count magnetic pole pairs, not individual magnetic poles. Four poles means two pairs.

Second, distinguish a complete positive-negative AC cycle from a single positive peak. Two peaks do not necessarily mean two AC cycles.

Third, determine whether the output has been rectified. A split-ring commutator is one way to obtain a pulsating unidirectional output, but an external rectifier can produce a similar waveform from slip-ring AC.

There is also a subtle sign issue. A commutator does not guarantee that a particular labelled terminal will always be positive under every condition. Reversing the rotation direction or magnetic field can reverse the terminal polarity.

The familiar positive absolute-value waveform assumes a specified rotation direction, field direction, and brush polarity.

Real generators add further complications. Brush switching is not instantaneous, internal resistance affects terminal voltage, magnetic fields may not produce perfect sine waves, and multiple windings may smooth the DC output. These effects change the waveform shape without invalidating the basic pole-pair relationship for the fundamental induced frequency.

For a detailed explanation of the switching mechanism, see How a Split-Ring Commutator Changes a Generator’s Waveform.

07Questions and solutions

Question 1

A simple AC generator has one north-south magnetic pole pair and uses slip rings.

Its shaft rotates at 900 rpm.

Calculate:

a) The mechanical frequency.

b) The electrical frequency.

c) The number of complete AC cycles produced in \(0.40\ \mathrm s\).

d) The period of the induced voltage.

Solution 1

The mechanical and electrical frequencies are both 15 Hz. The generator produces six complete AC cycles in 0.40 s, and each cycle lasts approximately 0.0667 s.

Part a: Mechanical frequency

Convert revolutions per minute into revolutions per second:

\[
f_{\mathrm m}=\frac{900}{60}=15\ \mathrm{Hz}
\]

Part b: Electrical frequency

There is one pole pair, so \(p=1\).

\[
f_{\mathrm e}=1(15)=15\ \mathrm{Hz}
\]

One mechanical revolution produces one complete AC cycle.

Part c: Number of cycles

Multiply frequency by time:

\[
N_{\mathrm{cycles}}
=15(0.40)=6.0
\]

Here, \(N_{\mathrm{cycles}}\) is the number of complete voltage cycles.

Part d: Period

\[
T_{\mathrm e}=\frac{1}{15}
=0.0667\ \mathrm s
\]

The slip rings allow the positive and negative half-cycles to reach the external circuit without reversing the connections.

Counting both the positive and negative extrema as separate AC cycles would incorrectly double the frequency.

Question 2

An ideal single-pole-pair generator uses slip rings. Its coil has 50 turns, each enclosing an area of \(0.020\ \mathrm{m^2}\), in a uniform \(0.20\ \mathrm T\) magnetic field.

At \(t=0\), the coil’s normal points along the magnetic field, and the shaft begins rotating at a constant 300 rpm in the positive angular direction.

At \(t=25\ \mathrm{ms}\), the motor speed changes instantaneously to 600 rpm and remains constant afterwards. The angular position is continuous.

Treat the instantaneous speed change as an idealisation, and ignore internal electrical effects.

Determine:

a) The coil’s angle at the instant of the speed change.

b) The magnetic flux linkage immediately before and after the change.

c) The induced emf immediately before and after the change.

d) The electrical frequency before and after the change, and the time from \(t=0\) until the next positive emf maximum.

Explain why the voltage may change suddenly even though the flux linkage does not.

Solution 2

The coil reaches \(45^\circ\) at 25 ms. Its flux linkage remains \(0.141\ \mathrm{Wb\,turns}\), while its induced emf jumps from approximately +4.44 V to +8.89 V. The next positive maximum occurs at \(t=37.5\ \mathrm{ms}\).

The key is that emf depends on the rate of change of flux linkage, not its instantaneous value.

Part a: Find the angle.

Before the change:

\[
\begin{aligned}
f_{\mathrm m,1}&=\frac{300}{60}
=5.0\ \mathrm{Hz}\\
\omega_1&=2\pi(5.0)
=10\pi\ \mathrm{rad/s}
\end{aligned}
\]

The angle reached after \(0.025\ \mathrm s\) is:

\[
\begin{aligned}
\theta&=\omega_1t\\
&=10\pi(0.025)\\
&=\frac{\pi}{4}=45^\circ
\end{aligned}
\]

Part b: Calculate the flux linkage.

The maximum flux linkage is:

\[
\begin{aligned}
\lambda_0&=NBA\\
&=50(0.20)(0.020)\\
&=0.200\ \mathrm{Wb\,turns}
\end{aligned}
\]

At \(45^\circ\):

\[
\begin{aligned}
\lambda&=0.200\cos45^\circ\\
&=0.141\ \mathrm{Wb\,turns}
\end{aligned}
\]

The coil’s physical orientation cannot jump simply because its angular speed changes.

Therefore, the flux linkage has the same value immediately before and immediately after the speed change.

Part c: Calculate the emf on each side of the change.

Before the speed increases:

\[
\begin{aligned}
\mathcal E_1
&=\lambda_0\omega_1\sin\theta\\
&=0.200(10\pi)\sin45^\circ\\
&=+4.44\ \mathrm V
\end{aligned}
\]

After the speed increases:

\[
\omega_2=2\pi(10)=20\pi\ \mathrm{rad/s}
\]

\[
\begin{aligned}
\mathcal E_2
&=0.200(20\pi)\sin45^\circ\\
&=+8.89\ \mathrm V
\end{aligned}
\]

The induced emf doubles because the flux linkage is changing twice as rapidly at the same orientation.

Part d: Determine frequencies and timing.

With one pole pair, the electrical frequency matches the mechanical frequency.

Before the change:

\[
f_{\mathrm e,1}=5.0\ \mathrm{Hz}
\]

After the change:

\[
f_{\mathrm e,2}=10\ \mathrm{Hz}
\]

These are the frequencies for the two constant-speed intervals. The entire motion does not have one fixed frequency because its speed changes.

The next positive maximum occurs at \(90^\circ\).

The remaining angle after the speed change is:

\[
\Delta\theta
=\frac{\pi}{2}-\frac{\pi}{4}
=\frac{\pi}{4}
\]

Using the new angular speed:

\[
\begin{aligned}
\Delta t
&=\frac{\pi/4}{20\pi}\\
&=0.0125\ \mathrm s
\end{aligned}
\]

Therefore:

\[
t_{\mathrm{peak}}
=0.025+0.0125
=0.0375\ \mathrm s
\]

The tempting mistake is to use the new 10 Hz frequency as though it had applied since \(t=0\). That would ignore the angle already covered at the slower speed.

The other mistake is to assume continuous flux linkage requires continuous emf. It does not: a continuous quantity can have an abruptly changing rate of change.

Question 3

Two sealed generator modules are driven by identical shafts rotating at 600 rpm.

They have been constructed to produce sinusoidal induced emf waveforms and adjusted to have equal peak voltage magnitudes.

Their internal designs are known to be one of the following:

• Module A: One magnetic pole pair, with a simple correctly timed split-ring commutator.
• Module B: Two magnetic pole pairs, with slip rings.

A sensor counts only positive voltage maxima. It ignores negative minima.

Both modules produce exactly 20 positive maxima per second.

a) Determine the induced electrical frequency of each module.

b) Explain why the sensor cannot identify which contact arrangement is being used.

c) Predict the number of positive and negative voltage extrema observed during one shaft revolution on a signed oscilloscope trace for each module.

d) An unknown module displays alternating positive and negative voltage half-cycles. Identify its design. Could changing the shaft speed alone make the other design produce the same alternating polarity pattern during steady rotation?

Solution 3

Module A generates 10 Hz AC internally and delivers 20 positive pulses per second. Module B generates 20 Hz AC and delivers an alternating voltage with 20 positive maxima and 20 negative minima per second.

The identical positive-peak readings do not imply identical induced electrical frequencies.

Part a: Calculate the induced frequencies.

Both shafts have mechanical frequency:

\[
f_{\mathrm m}=\frac{600}{60}
=10\ \mathrm{Hz}
\]

Module A has one pole pair:

\[
f_{\mathrm e,A}=1(10)
=10\ \mathrm{Hz}
\]

Module B has two pole pairs:

\[
f_{\mathrm e,B}=2(10)
=20\ \mathrm{Hz}
\]

Part b: Explain the matching sensor readings.

Module A’s split-ring commutator converts each 10 Hz AC cycle into two positive voltage pulses.

\[
f_{\mathrm{ripple,A}}
=2(10)=20\ \mathrm{Hz}
\]

Module B uses slip rings. Its output remains a 20 Hz AC waveform.

Every AC cycle contains one positive maximum, so it produces 20 positive maxima per second.

The sensor sees only positive maxima. It cannot tell whether they came from rectification of a lower-frequency emf or from unrectified higher-frequency AC.

Part c: Count extrema per shaft revolution.

For Module A, each mechanical revolution produces one induced AC cycle, which becomes two positive pulses.

Therefore, per revolution:

• Positive peaks: 2
• Negative minima: 0

For Module B, each mechanical revolution produces two complete AC cycles.

Therefore, per revolution:

• Positive peaks: 2
• Negative minima: 2

A signed oscilloscope distinguishes the arrangements because it retains the polarity information discarded by the positive-peak counter.

Part d: Identify the unknown module.

The unknown module is Module B.

Its slip rings maintain the same connection between each end of the coil and the external terminals. The induced emf therefore appears with alternating polarity.

Changing Module A’s steady rotational speed changes its pulse spacing and amplitude. Reversing its rotation direction can reverse the polarity of all its pulses, but it does not make the properly commutated output alternate between positive and negative every half-cycle.

The deeper lesson is that peak counts alone do not uniquely determine the underlying physical process. We need to know what the measuring instrument discards.

Question 4

A generator contains one magnetic pole pair, a single rectangular coil, and a correctly timed split-ring commutator.

Initially, its shaft rotates clockwise at 600 rpm. The magnetic field points from the north pole on the left towards the south pole on the right.

With a fixed choice of oscilloscope terminals P and Q, the generator produces positive voltage pulses with peaks of +12 V, separated by 0.050 s.

After the apparatus is modified, the following readings are recorded:

• The shaft rotates anticlockwise at 300 rpm.
• The voltage peaks are still positive, reaching +6.0 V.
• The peaks are separated by 0.100 s.

The coil, brush positions, brush wiring, and magnitude of the magnetic field are unchanged. The only permitted changes are the shaft’s speed and direction, and a possible reversal of the magnetic field.

a) Determine the induced electrical frequency and output pulse frequency before and after the modifications.

b) Explain the change in peak voltage.

c) Must the magnetic poles have been reversed? Justify your conclusion using both the direction of rotation and the observed voltage polarity.

Solution 4

Yes. The magnetic field must also have been reversed to keep the measured voltage pulses positive after the shaft’s rotation direction reversed.

The observed changes in pulse spacing and peak magnitude are consistent with the shaft speed being halved.

Part a: Calculate the frequencies.

Initially:

\[
f_{\mathrm m,1}=\frac{600}{60}
=10\ \mathrm{Hz}
\]

With one pole pair:

\[
f_{\mathrm e,1}=10\ \mathrm{Hz}
\]

The correctly timed commutator produces two pulses per revolution:

\[
f_{\mathrm{ripple,1}}
=2(10)=20\ \mathrm{Hz}
\]

The interval between pulses is:

\[
T_{\mathrm{ripple,1}}
=\frac{1}{20}
=0.050\ \mathrm s
\]

After modification:

\[
f_{\mathrm m,2}=\frac{300}{60}
=5.0\ \mathrm{Hz}
\]

Therefore:

\[
\begin{aligned}
f_{\mathrm e,2}&=5.0\ \mathrm{Hz}\\
f_{\mathrm{ripple,2}}&=10\ \mathrm{Hz}
\end{aligned}
\]

The new pulse interval is:

\[
T_{\mathrm{ripple,2}}
=\frac{1}{10}
=0.100\ \mathrm s
\]

Both measured pulse intervals match the predicted values.

Part b: Explain the peak-voltage change.

For the same coil and magnetic-field magnitude:

\[
E_0=NBA|\omega|
\]

The peak emf is directly proportional to the magnitude of angular speed.

Halving the speed halves the peak emf:

\[
E_{0,2}
=12\left(\frac{300}{600}\right)
=6.0\ \mathrm V
\]

This agrees with the recorded peak magnitude.

Part c: Determine whether the field reversed.

For fixed coil connections and brush positions, the sign of the generated emf at a particular shaft orientation depends on both the magnetic-field direction and the signed angular velocity.

Reversing only the rotation direction reverses the sign of the emf at that orientation.

The commutator still exchanges the coil connections at the same geometric positions, so the external pulse polarity would also reverse.

If the field remained unchanged, the new pulses would be negative relative to the original P-Q terminal convention.

But the measured pulses are positive.

Reversing the magnetic field as well reverses the emf sign a second time. The two reversals cancel.

The modifications therefore require both anticlockwise rotation and reversed magnetic poles.

A tempting argument is that a split-ring commutator always produces positive voltage. More precisely, it produces a consistent polarity for a specified field direction, rotation direction, and brush arrangement.

Change one of those directions, and the polarity can change.

Question 5

A generator contains four magnetic poles, arranged as two north-south pole pairs around a rotor.

A particular winding is designed so that its flux linkage varies sinusoidally with shaft angle according to:

\[
\lambda=\lambda_0\cos(2\theta)
\]

Here, \(\theta\) is mechanical shaft angle, measured from \(0^\circ\) to \(360^\circ\) over one revolution, and \(\lambda_0\) is the maximum flux-linkage magnitude.

The shaft rotates steadily in the direction of increasing \(\theta\) at 300 rpm.

The winding’s induced emf, measured with a fixed reference polarity between its ends X and Y, is:

\[
\mathcal E=E_0\sin(2\theta)
\]

where \(E_0\) is positive.

An engineer connects this winding to stationary brushes P and Q using a basic two-segment split-ring commutator.

The commutator is arranged so that:

• From \(0^\circ\) to \(180^\circ\), X connects to P, and Y connects to Q.
• From \(180^\circ\) to \(360^\circ\), Y connects to P, and X connects to Q.

The contact changes occur at \(0^\circ\) and \(180^\circ\). Ignore the brief switching intervals.

A student argues:

“Four poles give two electrical cycles per revolution. A split ring turns every negative half-cycle positive. Therefore, the external circuit must receive four positive pulses per revolution.”

Determine whether the student is correct.

a) Calculate the mechanical frequency and the induced electrical frequency.

b) Determine the polarity of the voltage at P relative to Q at shaft angles \(45^\circ\), \(135^\circ\), \(225^\circ\), and \(315^\circ\).

c) Find the number of positive and negative voltage lobes produced in one mechanical revolution, and the corresponding rates per second.

d) Explain exactly why the student’s prediction succeeds for the familiar two-pole generator but fails for this arrangement.

e) Describe what must change in the commutation or rectification process to produce four positive pulses per mechanical revolution. Determine the resulting ideal pulse frequency.

Solution 5

The student is incorrect. The winding produces two AC cycles per revolution, but the specified two-segment commutator does not rectify all four half-cycles correctly.

Instead, the external output has two positive voltage lobes and two negative voltage lobes per mechanical revolution.

The difficulty is not the frequency calculation. It is the assumption that the commutator switches whenever the induced emf reverses.

Part a: Calculate both frequencies.

The mechanical frequency is:

\[
f_{\mathrm m}=\frac{300}{60}
=5.0\ \mathrm{Hz}
\]

There are two pole pairs.

\[
f_{\mathrm e}=2(5.0)=10\ \mathrm{Hz}
\]

Thus, the winding generates two complete AC cycles per mechanical revolution.

Part b: Track the voltage polarity.

The induced emf follows:

\[
\mathcal E=E_0\sin(2\theta)
\]

Evaluate the sign at each specified angle.

At \(45^\circ\):

\[
\mathcal E=E_0\sin90^\circ=+E_0
\]

The connections have not reversed, so P is positive relative to Q.

At \(135^\circ\):

\[
\mathcal E=E_0\sin270^\circ=-E_0
\]

The connections still have not reversed. P is therefore negative relative to Q.

At \(225^\circ\):

\[
\mathcal E=E_0\sin450^\circ=+E_0
\]

The winding’s emf is positive again, but the commutator has now exchanged X and Y between the brushes.

Therefore, P is negative relative to Q.

At \(315^\circ\):

\[
\mathcal E=E_0\sin630^\circ=-E_0
\]

The winding’s emf is negative, and the commutator connections are reversed.

The two sign changes cancel, giving positive voltage at P relative to Q.

The results are:

Part c: Count the voltage lobes.

Across one mechanical revolution, the terminal-voltage signs follow this pattern:

\[
+,\quad -,\quad -,\quad +
\]

Each lobe occupies a \(90^\circ\) interval.

Therefore, one revolution produces two positive lobes and two negative lobes.

The shaft completes five revolutions per second.

The rates are:

\[
\begin{aligned}
f_{\mathrm{positive}}&=2(5)=10\ \mathrm{s^{-1}}\\
f_{\mathrm{negative}}&=2(5)=10\ \mathrm{s^{-1}}
\end{aligned}
\]

The quantities \(f_{\mathrm{positive}}\) and \(f_{\mathrm{negative}}\) here represent lobe counts per second, not the frequency of complete sinusoidal AC cycles.

There are 20 voltage lobes per second in total, but they are not all positive.

This output is not pulsating DC because its terminal polarity changes.

Part d: Explain the failure of the familiar rule.

In a simple generator with one pole pair, the induced emf reverses every \(180^\circ\) of mechanical rotation.

A two-segment commutator also exchanges its connections every \(180^\circ\).

The two events coincide.

That is why a simple split ring can turn the negative half of each induced AC cycle into a positive pulse.

Now consider the four-pole arrangement.

The induced emf is proportional to \(\sin(2\theta)\).

Its polarity changes at:

\[
0^\circ,\ 90^\circ,\ 180^\circ,\
270^\circ,\ 360^\circ
\]

But the specified commutator switches only at:

\[
0^\circ,\ 180^\circ,\ 360^\circ
\]

The winding’s emf reverses at \(90^\circ\) and \(270^\circ\) without a corresponding reversal of the external connections.

Those missed switching points cause negative voltage lobes at the terminals.

The student’s original argument treats a split-ring commutator as an automatic full-wave rectifier, regardless of the number of magnetic pole pairs.

It is not. The switching geometry must match the winding’s emf reversals.

Part e: Produce correctly rectified output.

To obtain four positive lobes per mechanical revolution from this winding, the external connection polarity must reverse whenever the raw emf changes sign.

In this ideal model, that means properly timed switching at every \(90^\circ\) of mechanical rotation rather than only every \(180^\circ\).

This would require an appropriate commutation design, using suitably arranged segments and windings, or an external full-wave rectifier.

The resulting ideal terminal waveform would be:

\[
V_{\mathrm{rectified}}
=E_0|\sin(2\theta)|
\]

It would contain four positive pulses per revolution.

At five revolutions per second:

\[
f_{\mathrm{ripple}}
=4(5)=20\ \mathrm{Hz}
\]

The induced AC frequency would remain 10 Hz.

This final comparison exposes the central distinction.

Magnetic geometry determines how often the winding’s flux linkage repeats. The flux derivative determines the induced emf. The contact arrangement determines how that emf appears at the external terminals.

To solve a generator-frequency question, keep those three stages separate.

Once the frequency is known, a different question follows: how large will the induced voltage become? That depends on the magnetic field, the winding, and the rotation speed. The next useful calculation is developed in How to Calculate Peak EMF in a Rotating Coil.