How a Split-Ring Commutator Changes a Generator’s Waveform

Imagine turning a generator coil while watching its voltage on an oscilloscope. With two slip rings, the trace rises above zero for half a revolution, then falls below zero for the next half. Replace those slip rings with a split-ring commutator. What happens to the negative half of the waveform?

It flips above zero.

The coil still generates an alternating emf. The split-ring commutator reverses the coil’s connections to the external circuit every half-turn, just as the induced emf changes direction. The external output becomes a series of positive pulses instead of a positive-negative sine wave.

For an ideal single-coil generator, the change is:

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

Here, \(E_0\) is the peak induced emf in volts, and \(\theta\) is the coil’s rotation angle. We’ve chosen the output polarity so the split-ring pulses are positive.

This is called pulsating DC. The voltage keeps the same polarity, but it still falls to zero twice per revolution. A split-ring commutator doesn’t turn a sine wave into a steady horizontal line.

01The experiment: change the contacts, not the coil

To understand why the waveform changes, keep the generator itself the same and change only the device connecting its rotating coil to the stationary circuit.

Picture a rectangular coil between two magnetic poles. The north pole is on the left, the south pole is on the right, and the magnetic field points from north to south.

A shaft runs through the centre of the coil, perpendicular to the magnetic field. An external motor turns the shaft at constant speed.

The coil has two ends, which we’ll call X and Y. They must connect to an external resistor and an oscilloscope without twisting the wires as the coil rotates.

That’s the job of the sliding contacts.

First arrangement: two continuous slip rings

Connect coil end X to one complete metal ring mounted on the rotating shaft. Connect coil end Y to a second, separate metal ring.

A stationary carbon brush presses against each ring. Call the brushes P and Q.

Brush P connects to the positive input of the oscilloscope. Brush Q connects to its negative input. A resistor can be connected between the same brushes to provide a load.

As the shaft turns, each ring rotates beneath its brush, maintaining electrical contact.

Crucially, X always connects to P, and Y always connects to Q.

What should the oscilloscope show?

Because the induced emf in the coil reverses every half-turn, the voltage between P and Q must alternate between positive and negative. The expected ideal waveform is a sine wave.

This arrangement solves the problem of connecting a rotating coil to stationary wires. It doesn’t solve the problem of obtaining a single-polarity output.

Revised arrangement: one split-ring commutator

Now replace the two complete rings with two conducting half-rings separated by insulating gaps.

The half-rings are attached to the same rotating shaft. End X connects to one segment, and end Y connects to the other.

The stationary brushes remain opposite one another, connected to the same external circuit.

During the first half-turn:

• X contacts P.
• Y contacts Q.

During the second half-turn:

• X contacts Q.
• Y contacts P.

The connections have exchanged places.

Suppose the coil initially makes X positive relative to Y. Current can flow from P through the external resistor to Q.

Half a turn later, the induced emf makes Y positive relative to X. But Y now contacts P.

So P remains the positive external brush.

Two changes have happened together: the coil’s emf has reversed, and its connection to the stationary circuit has reversed.

The expected oscilloscope trace is therefore positive during both half-turns.

This comparison isolates the commutator’s role. The magnetic field, coil area, number of turns, and rotation speed haven’t changed.

One qualification matters: these are ideal predictions, not reported experimental measurements. A real oscilloscope measures the voltage between the brushes, which can also be affected by internal resistance, brush contact, and rotation speed. To compare the two arrangements fairly, those conditions must be controlled.

02Why the commutator switches when the emf is zero

The split ring has to reverse the connections at the right moment. To find that moment, we need to follow the changing magnetic flux through the coil.

Start with the coil’s orientation

Imagine an arrow pointing straight out of one face of the coil. That arrow is the coil’s area normal.

Let \(\theta\) be the signed rotation angle between this chosen normal and the magnetic field, measured in the direction of rotation.

At \(\theta=0^\circ\), the normal points along the magnetic field. The plane of the coil is perpendicular to the field, so the magnetic flux is maximum and positive.

At \(\theta=90^\circ\), the plane of the coil is parallel to the field. The flux is zero.

At \(\theta=180^\circ\), the normal points opposite the field. The flux has its maximum negative value.

For a uniform magnetic field, the flux through one turn is

\[ \Phi=BA\cos\theta \]

where:

• \(\Phi\) is magnetic flux in webers (\(\mathrm{Wb}\)).
• \(B\) is magnetic flux density in tesla (\(\mathrm{T}\)).
• \(A\) is the area of one turn in square metres (\(\mathrm{m^2}\)).
• \(\theta\) is the orientation angle.

Now consider what happens as the coil rotates from \(0^\circ\) towards \(90^\circ\).

The flux decreases from its maximum towards zero. Does the induced emf decrease as well?

No. It increases from zero towards its maximum.

The reason is that induced emf depends on how quickly the flux changes, not the amount of flux present.

The flux derivative gives the internal emf

For a coil containing \(N\) identical turns, Faraday’s law gives

\[ e_c=-N\frac{d\Phi}{dt} \]

Here, \(e_c\) is the signed emf induced in the rotating coil, in volts (\(\mathrm{V}\)), and \(t\) is time in seconds.

The minus sign represents Lenz’s law: the induced effect opposes the change in magnetic flux producing it.

If the coil rotates at a constant angular speed \(\omega\), measured in radians per second (\(\mathrm{rad\,s^{-1}}\)), then \(\theta=\omega t\), taking \(\theta=0\) at \(t=0\).

Substituting the flux equation gives

\[ \begin{aligned} e_c&=-N\frac{d\Phi}{dt}\\ &=NBA\omega\sin\theta. \end{aligned} \]

Define the peak emf as

\[ E_0=NBA\omega. \]

Therefore,

\[ e_c=E_0\sin\theta. \]

We’ve chosen the winding direction and signs so that the coil emf is positive during the first half-turn. Reversing the chosen coil terminals would reverse every internal emf sign.

The important geometry is now visible:

Notice what happens at \(180^\circ\).

The flux stops decreasing and begins increasing. Its derivative changes sign, so the induced emf changes sign.

At that same position, a correctly timed split-ring commutator reverses the connections.

Because the emf is momentarily zero, the ideal commutator can switch without producing an abrupt reversal of the external voltage.

This is the precise connection between the geometry and the commutator.

The commutator doesn’t create the positive voltage peaks. The changing magnetic flux creates the emf. The commutator determines which polarity reaches the external circuit.

If you’d like to revisit how the rotating coil produces a sinusoidal emf in the first place, see AC Generators: Sinusoidal EMF and Slip Rings Explained.

03How the split ring changes the waveform

Let’s write the comparison in a way you can use directly in an HSC question.

With slip rings

The external voltage follows the internal coil emf:

\[ V_{\text{slip}}=E_0\sin\theta. \]

During the first half-turn, the voltage is positive. During the second half-turn, it is negative.

The external circuit therefore receives alternating voltage.

With a correctly timed split-ring commutator

The output voltage is

\[ V_{\text{split}}=E_0|\sin\theta|. \]

The absolute-value signs mean that both positive and negative internal emf values appear as positive external voltages.

A useful analogy is swapping the red and black oscilloscope leads every time the coil’s emf is about to reverse. What would have been a negative reading becomes positive.

The split ring performs that switching mechanically.

The analogy has a limit: a real commutator carries current through physical sliding contacts. It isn’t modifying a recorded signal after the event.

The waveform still reaches zero

Here is a prediction that catches a common mistake.

If the generator rotates at 20 revolutions per second, does its split-ring output contain 20 positive pulses or 40?

It contains 40.

One full revolution produces two positive half-sine pulses.

Let \(f_r\) be the rotational frequency in hertz, \(f_p\) the frequency at which the positive pulses repeat, and \(T_p\) the time between successive positive peaks.

Then

\[ f_p=2f_r \]

and

\[ T_p=\frac{1}{2f_r}. \]

At \(f_r=20\ \mathrm{Hz}\),

\[ \begin{aligned} f_p&=2(20)=40\ \mathrm{Hz},\\ T_p&=\frac{1}{40}=0.025\ \mathrm{s}. \end{aligned} \]

The positive peaks are separated by 25 milliseconds.

The distinction is worth keeping precise:

Changing from slip rings to a split ring doesn’t double the peak emf. It doubles the number of positive peaks appearing per revolution.

04What if the output isn’t a clean positive waveform?

The ideal model assumes the commutator reverses connections exactly when the induced emf passes through zero.

That timing matters.

Suppose the brushes are shifted so the connections reverse slightly after the coil emf changes sign.

For a short angular interval, the emf has already reversed, but the contacts haven’t. The external voltage can become negative before the commutator finally switches.

In an ideal mathematical model, the delayed switch produces a sudden voltage jump. In a real generator, contact resistance, brief interruptions, or short circuits during switching can complicate the trace further.

An oscilloscope trace can therefore help distinguish different causes.

A negative output section doesn’t automatically mean the generator contains slip rings. A split ring connected or timed incorrectly can also produce one.

And a perfectly positive waveform doesn’t prove the voltage is constant.

These distinctions apply to the idealised single-coil generator model used in NSW Physics Stage 6 (2017), the syllabus assessed in the 2026 HSC. Real generators may contain multiple coils, many commutator segments, and additional electrical effects.

05Questions and solutions

Question 1

An ideal single-coil generator rotates at \(5.0\ \mathrm{Hz}\). Its internal emf is

\[ e_c=10.0\sin(10\pi t) \]

where \(e_c\) is in volts and \(t\) is in seconds.

The generator uses a correctly timed split-ring commutator, with the external polarity chosen to produce positive pulses.

a) Calculate the external voltage at \(t=0.150\ \mathrm{s}\).

b) Determine the number of positive voltage peaks produced each second.

c) State the external voltage at the same instant if the split ring were replaced with two slip rings.

Solution 1

The split-ring voltage is \(+10.0\ \mathrm{V}\) at \(0.150\ \mathrm{s}\), the generator produces 10 positive peaks per second, and the corresponding slip-ring voltage is \(-10.0\ \mathrm{V}\).

For part (a), calculate the internal emf first:

\[ \begin{aligned} e_c&=10.0\sin[10\pi(0.150)]\\ &=10.0\sin(1.5\pi)\\ &=-10.0\ \mathrm{V}. \end{aligned} \]

The split-ring commutator reverses this polarity externally:

\[ V_{\text{split}}=|-10.0|=10.0\ \mathrm{V}. \]

For part (b), each revolution produces two positive pulses:

\[ \begin{aligned} f_p&=2f_r\\ &=2(5.0)\\ &=10\ \mathrm{Hz}. \end{aligned} \]

There are 10 positive peaks per second.

For part (c), slip rings preserve the coil’s terminal connections, so the external voltage follows the signed internal emf:

\[ V_{\text{slip}}=-10.0\ \mathrm{V}. \]

The tempting error is treating the negative internal emf as impossible in a DC generator. It is only the external brush polarity that has been rectified.

Question 2

A 50-turn coil rotates uniformly in a constant magnetic field.

During the first quarter-turn, from \(\theta=0^\circ\) to \(90^\circ\), the magnetic flux through each turn changes from \(+0.0120\ \mathrm{Wb}\) to zero in \(10.0\ \mathrm{ms}\).

The flux follows the usual cosine relationship, and the coil’s induced emf is positive during this first quarter-turn.

a) Calculate the average internal emf during this interval.

b) Determine the instantaneous peak internal emf, including its value at \(90^\circ\).

c) If the generator uses a correctly timed split-ring commutator, calculate its average external voltage during the third quarter-turn, from \(180^\circ\) to \(270^\circ\).

d) A student argues that the answer to part (a) must be the maximum emf and that installing a split ring doubles that maximum. Explain both errors.

Assume an ideal generator.

Solution 2

The average emf in the first quarter-turn is \(+60.0\ \mathrm{V}\), the instantaneous peak emf is \(+94.2\ \mathrm{V}\), and the average split-ring output during the third quarter-turn is \(+60.0\ \mathrm{V}\). The split ring does not double the maximum emf.

For part (a), Faraday’s law gives the average emf:

\[ \overline{e_c}=-N\frac{\Delta\Phi}{\Delta t}. \]

Convert the elapsed time:

\[ 10.0\ \mathrm{ms}=0.0100\ \mathrm{s}. \]

The flux change is

\[ \Delta\Phi=0-0.0120=-0.0120\ \mathrm{Wb}. \]

Therefore,

\[ \begin{aligned} \overline{e_c} &=-50\frac{-0.0120}{0.0100}\\ &=+60.0\ \mathrm{V}. \end{aligned} \]

This is the average emf over an interval, not the instantaneous emf at the end.

For part (b), the coil completes one quarter of a revolution in \(0.0100\ \mathrm{s}\).

Its full period is

\[ T_r=4(0.0100)=0.0400\ \mathrm{s}. \]

Hence,

\[ \begin{aligned} \omega&=\frac{2\pi}{0.0400}\\ &=157.1\ \mathrm{rad\,s^{-1}}. \end{aligned} \]

The maximum flux magnitude per turn is \(BA=0.0120\ \mathrm{Wb}\).

So

\[ \begin{aligned} E_0&=NBA\omega\\ &=50(0.0120)(157.1)\\ &=94.2\ \mathrm{V}. \end{aligned} \]

At \(90^\circ\),

\[ \begin{aligned} e_c&=94.2\sin90^\circ\\ &=+94.2\ \mathrm{V}. \end{aligned} \]

The emf has increased from zero to its maximum during the quarter-turn. Its average must therefore be smaller than its peak.

For part (c), the flux during the third quarter-turn changes from \(-0.0120\ \mathrm{Wb}\) to zero.

The average internal emf is

\[ \begin{aligned} \overline{e_c} &=-50\frac{0-(-0.0120)}{0.0100}\\ &=-60.0\ \mathrm{V}. \end{aligned} \]

The internal emf is negative throughout the interior of this quarter-turn.

A correctly timed split ring reverses its external connection throughout that interval, giving an average output of

\[ \overline{V_{\text{split}}}=+60.0\ \mathrm{V}. \]

For part (d), the student’s first mistake is confusing an average rate of change with an instantaneous rate of change. Faraday’s law using \(\Delta\Phi/\Delta t\) calculates the average emf over a finite interval.

The second mistake is confusing polarity switching with stronger electromagnetic induction. The commutator changes which terminal is connected to each brush; it doesn’t change \(N\), \(B\), \(A\), or \(\omega\).

The peak remains \(94.2\ \mathrm{V}\).

Question 3

A simple split-ring generator rotates at \(12\ \mathrm{Hz}\) in a uniform magnetic field. In its original arrangement, the external voltage is positive at brush P relative to brush Q, with a peak magnitude of \(6.0\ \mathrm{V}\).

A \(3.0\ \Omega\) resistor is connected between P and Q.

The shaft has a fixed positive rotation direction, and its angle \(\theta\) is always measured from the original field direction. The commutator remains mechanically attached to the coil.

Consider three changes to the original arrangement, made at the same rotation speed magnitude:

a) Reverse the direction of shaft rotation, but leave the magnetic field unchanged.

b) Restore the original rotation direction, but reverse the north and south magnetic poles.

c) Reverse both the shaft rotation direction and the magnetic field direction.

For each case, determine the polarity of the external voltage, the signed peak current through the resistor, and the pulse repetition rate. Explain why the results differ.

Take current from P through the resistor to Q as positive. Ignore the brief transient while the generator is stopped or reversed.

Solution 3

Reversing either the shaft direction or the magnetic field alone reverses the polarity of the external pulses. Reversing both restores the original polarity. In every case, the peak current magnitude is \(2.0\ \mathrm{A}\), and the pulse repetition rate is \(24\ \mathrm{Hz}\).

The results are:

For the current magnitude, Ohm’s law gives

\[ \begin{aligned} I_0&=\frac{V_0}{R}\\ &=\frac{6.0\ \mathrm{V}}{3.0\ \Omega}\\ &=2.0\ \mathrm{A}. \end{aligned} \]

The sign is determined by the external voltage polarity.

For part (a), reversing the shaft reverses the direction in which the flux changes at any given physical coil orientation.

Since

\[ e_c=-N\frac{d\Phi}{dt}, \]

the internal emf reverses sign.

However, the commutator’s connection pattern at each physical coil orientation hasn’t changed. It still connects the same segment to the same brush at that orientation.

Consequently, the external pulse polarity also reverses.

For part (b), reversing the magnetic field changes the sign of the flux through the coil at every fixed orientation.

For unchanged rotation, this also reverses the flux derivative and induced emf.

Again, the commutator geometry is unchanged, so the external voltage polarity reverses.

For part (c), reversing both the magnetic field and the shaft direction produces two sign reversals in the flux derivative.

They cancel, restoring the original internal emf sign at each position.

The commutator then produces the same external polarity as before.

Finally, the speed magnitude is unchanged in all three cases, so the peak emf magnitude remains \(6.0\ \mathrm{V}\).

The pulse repetition rate is

\[ \begin{aligned} f_p&=2f_r\\ &=2(12)\\ &=24\ \mathrm{Hz}. \end{aligned} \]

The tempting prediction is that a split-ring generator always makes brush P positive regardless of how the shaft rotates. It doesn’t.

The commutator corrects the alternating polarity associated with its intended field and rotation arrangement. It doesn’t automatically compensate for an externally imposed reversal of shaft rotation or magnetic field.

Question 4

An ideal generator has the internal coil emf

\[ e_c=10.0\sin\theta \]

where \(\theta\) is the rotation angle in degrees, measured in the normal forward direction.

The split-ring commutator has been incorrectly aligned.

Instead of reversing its connections at \(0^\circ\) and \(180^\circ\), it reverses them at \(30^\circ\) and \(210^\circ\).

The external connections are such that:

• For \(30^\circ<\theta<210^\circ\), the brush voltage equals \(e_c\).
• For the remaining part of the revolution, the brush voltage equals \(-e_c\).

Assume the switching is instantaneous in the ideal model.

a) Identify every angular interval during which the external voltage is negative.

b) Calculate the voltage immediately before and immediately after each commutator reversal.

c) A technician proposes increasing the generator’s rotation speed without adjusting the brushes. Would this remove the negative sections? Explain.

Solution 4

The external voltage is negative from \(0^\circ\) to \(30^\circ\), and from \(180^\circ\) to \(210^\circ\). At both switching positions, the ideal voltage jumps from \(-5.0\ \mathrm{V}\) to \(+5.0\ \mathrm{V}\). Increasing the rotation speed alone cannot remove the negative sections.

For part (a), examine the sign of the internal emf and the connection arrangement separately.

During \(0^\circ<\theta<30^\circ\), the internal emf is positive because \(\sin\theta>0\).

However, the commutator is in the arrangement that makes the external voltage equal to \(-e_c\).

Therefore, the external voltage is negative.

From \(30^\circ\) to \(180^\circ\), the internal emf and external voltage are both positive.

At \(180^\circ\), the internal emf crosses zero and becomes negative.

But the commutator doesn’t switch until \(210^\circ\).

So for \(180^\circ<\theta<210^\circ\), the external voltage is negative again.

After \(210^\circ\), the contacts reverse, and the remaining negative internal emf appears positive externally.

For part (b), calculate the internal emf at the two switching angles.

At \(30^\circ\),

\[ \begin{aligned} e_c&=10.0\sin30^\circ\\ &=+5.0\ \mathrm{V}. \end{aligned} \]

Immediately before switching, the external voltage is \(-e_c\), giving \(-5.0\ \mathrm{V}\).

Immediately after switching, it becomes \(+e_c\), giving \(+5.0\ \mathrm{V}\).

The jump is

\[ \begin{aligned} \Delta V&=5.0-(-5.0)\\ &=+10.0\ \mathrm{V}. \end{aligned} \]

At \(210^\circ\),

\[ \begin{aligned} e_c&=10.0\sin210^\circ\\ &=-5.0\ \mathrm{V}. \end{aligned} \]

Immediately before switching, the external voltage equals \(e_c=-5.0\ \mathrm{V}\).

Immediately after switching, it equals \(-e_c=+5.0\ \mathrm{V}\).

Again, the ideal jump is \(+10.0\ \mathrm{V}\).

At the exact switching instants, the idealised model does not specify a unique voltage during contact transfer. The calculated values are the limits immediately before and after.

For part (c), increasing the speed increases the induced emf magnitude through

\[ E_0=NBA\omega. \]

But the incorrect switching angles remain \(30^\circ\) and \(210^\circ\).

The coil emf still changes sign at \(0^\circ\) and \(180^\circ\).

The negative intervals therefore remain, although their voltage magnitudes and durations in seconds change.

The tempting idea is that the presence of a split-ring commutator guarantees non-negative output. That only works when its switching is correctly timed to the sign reversals of the induced emf.

A correctly aligned commutator would switch at the emf zero crossings instead.

Question 5

Extension challenge – two coils, two possible connections

Two identical generator coils rotate on the same shaft in a uniform magnetic field. Their planes are perpendicular, so their induced emfs are \(90^\circ\) out of phase.

The signed internal emfs are

\[ e_1=6.0\sin\theta \]

and

\[ e_2=6.0\cos\theta \]

where both emfs are measured in volts.

A designer proposes two arrangements.

Arrangement A: Each coil has its own correctly timed split-ring commutator. The two stationary outputs are then connected in series with their positive polarities adding.

Arrangement B: The two rotating coils are first connected in series, with their signed emfs adding according to the equations above. Their combined emf is then passed through one shared split-ring commutator, timed to reverse at the zero crossings of the combined emf.

Assume ideal contacts, negligible internal resistance, constant angular speed, and no other source of voltage.

a) Determine whether each arrangement can produce zero external voltage while rotating.

b) Find the minimum and maximum external voltage for each arrangement.

c) Calculate the external voltage of both arrangements at \(\theta=135^\circ\).

d) A student argues that because neither coil has zero induced emf at \(135^\circ\), the combined generator cannot produce zero voltage there. Explain precisely when that argument works and when it fails.

e) Could changing the switching positions of the shared split-ring commutator in Arrangement B eliminate every zero in its output? Justify your answer.

Solution 5

Arrangement A never reaches zero voltage and varies between \(6.0\ \mathrm{V}\) and approximately \(8.49\ \mathrm{V}\). Arrangement B reaches zero twice per revolution and varies between \(0\ \mathrm{V}\) and approximately \(8.49\ \mathrm{V}\).

The surprising difference is caused by the order in which the voltages are added and rectified.

For Arrangement A, the two emfs are rectified separately:

\[ V_A=6.0|\sin\theta|+6.0|\cos\theta|. \]

For Arrangement B, the signed emfs are added before rectification:

\[ V_B=6.0|\sin\theta+\cos\theta|. \]

These expressions are not equivalent.

For part (a), in Arrangement A, the output could be zero only if both \(|\sin\theta|\) and \(|\cos\theta|\) were zero simultaneously.

But

\[ \sin^2\theta+\cos^2\theta=1. \]

They cannot both be zero.

Therefore, Arrangement A always produces a positive voltage while rotating.

Arrangement B can produce zero whenever

\[ \sin\theta+\cos\theta=0. \]

This occurs at

\[ \theta=135^\circ,\ 315^\circ. \]

So Arrangement B has two zero-voltage positions per revolution.

For part (b), Arrangement A reaches its minimum when one of the trigonometric terms has magnitude 1 and the other has magnitude 0.

For example, at \(0^\circ\),

\[ \begin{aligned} V_A&=6.0(0)+6.0(1)\\ &=6.0\ \mathrm{V}. \end{aligned} \]

Its maximum occurs when the two magnitudes are equal, each being \(\sqrt{2}/2\):

\[ \begin{aligned} V_{A,\max} &=6.0\frac{\sqrt2}{2} +6.0\frac{\sqrt2}{2}\\ &=6.0\sqrt2\\ &\approx8.49\ \mathrm{V}. \end{aligned} \]

For Arrangement B, use the identity

\[ \sin\theta+\cos\theta =\sqrt2\sin(\theta+45^\circ). \]

Its maximum output is therefore

\[ V_{B,\max}=6.0\sqrt2 \approx8.49\ \mathrm{V}. \]

Its minimum is zero, because the combined signed emf can cancel completely.

For part (c), at \(135^\circ\),

\[ \begin{aligned} e_1&=6.0\sin135^\circ\\ &=+4.24\ \mathrm{V},\\ e_2&=6.0\cos135^\circ\\ &=-4.24\ \mathrm{V}. \end{aligned} \]

In Arrangement A, each commutator has already made its coil’s external voltage positive before the two outputs are added:

\[ \begin{aligned} V_A&=|4.24|+|-4.24|\\ &\approx8.49\ \mathrm{V}. \end{aligned} \]

Using unrounded values gives the stated \(8.49\ \mathrm{V}\).

In Arrangement B, the two signed voltages are added inside the rotor first:

\[ \begin{aligned} e_{\text{total}}&=4.24-4.24\\ &=0\ \mathrm{V}. \end{aligned} \]

The shared commutator receives zero emf, so

\[ V_B=0\ \mathrm{V}. \]

For part (d), the student’s argument works only when the two coils are rectified independently and their positive outputs are then added.

It fails when the induced emfs are connected in series before rectification.

Two nonzero voltages can cancel if they have equal magnitudes and opposite signs. The fact that neither coil has zero emf does not prevent their signed sum from being zero.

For part (e), changing the shared commutator’s switching positions cannot eliminate these zeros.

At \(135^\circ\), for example, the combined input to that commutator is already zero.

Changing its connections can reverse the sign of a nonzero voltage, but it cannot turn a zero input into a nonzero output without another source or an energy-storage mechanism.

In fact, moving the switching positions away from the combined emf’s zero crossings would risk producing negative external sections, as in Question 4.

The deeper principle is that rectifying two emfs separately and then adding them is not the same as adding signed emfs and rectifying the result.

This is the bridge to understanding why practical generators use carefully arranged coils and commutator connections. Multiple coils can reduce voltage ripple, but simply adding more coils is not enough. Their relative phases, electrical connections, and switching positions determine whether their contributions reinforce or cancel.

For the wider generator design, including how current is delivered to a load and why multiple coils can produce a smoother output, see DC generators: how rotating coils produce direct current.