Back EMF vs Resistive Voltage Drop: HSC Physics Guide

A DC motor runs from a \(12.0\,\mathrm{V}\) supply. When it drives a heavy load, it draws \(4.0\,\mathrm{A}\). When the load is removed and the motor spins faster, the current falls to \(0.50\,\mathrm{A}\). Yet a voltmeter across its terminals still reads \(12.0\,\mathrm{V}\).

Has the motor’s resistance increased from \(3.0\,\Omega\) to \(24\,\Omega\)?

No. Dividing the terminal voltage by the current gives a misleading answer because a running motor is not just a resistor.

A resistive voltage drop is caused by current flowing through the motor’s windings. Back EMF is an induced voltage produced by the rotating motor, and it opposes the applied voltage.

Both contribute to the motor’s voltage balance, but they come from different physical processes.

For a simple DC motor operating steadily:

\[ V_{\mathrm t}=\mathcal{E}_{\mathrm b}+IR \]

The useful distinction is that \(IR\) depends on current and winding resistance, while back EMF depends on rotational speed and magnetic field strength.

01One terminal voltage, two different effects

Imagine looking inside a small brushed DC motor.

There are copper windings on a rotating armature, magnets that provide a magnetic field, and brushes and a commutator that connect the rotating windings to the electrical supply.

Two things happen when the motor runs.

The copper windings resist the current

Copper conducts electricity well, but it still has electrical resistance.

As current travels through the armature windings, electrical energy is transferred into thermal energy. The windings get warmer.

For winding resistance \(R\), the resistive voltage drop is

\[ V_R=IR \]

where:

• \(V_R\) is the resistive voltage drop, in volts (V).
• \(I\) is the armature current, in amperes (A).
• \(R\) is the effective armature winding resistance, in ohms (\(\Omega\)).

For example, if the windings have a resistance of \(1.0\,\Omega\) and carry \(3.0\,\mathrm{A}\), the resistive drop is \(3.0\,\mathrm{V}\).

The important point is that the resistance does not disappear when the motor stops rotating. If current flows through the stationary windings, they still produce a resistive voltage drop.

The rotating windings produce an opposing EMF

Now consider what happens when the armature turns.

The conducting windings move through the magnetic field. Electromagnetic induction produces an EMF in those windings.

During normal motoring operation, this induced EMF acts against the applied voltage. It is called back electromotive force, or back EMF.

You may also see it called counter-EMF.

Suppose the supply provides \(12\,\mathrm{V}\) and the rotating motor produces \(9\,\mathrm{V}\) of opposing back EMF.

Only \(3\,\mathrm{V}\) remains to drive current through the winding resistance:

\[ V_R=12-9=3\,\mathrm{V} \]

If the resistance is \(1.0\,\Omega\), the current is \(3.0\,\mathrm{A}\).

This does not mean that back EMF is another resistance. The motor is generating an opposing voltage because its conductors are moving through a magnetic field.

The equation and its sign convention

For steady motoring operation, the applied terminal voltage must balance the back EMF and the resistive drop:

\[ V_{\mathrm t}=\mathcal{E}_{\mathrm b}+IR \]

Here:

• \(V_{\mathrm t}\) is the voltage measured across the motor terminals, in volts.
• \(\mathcal{E}_{\mathrm b}\) is the back EMF, in volts.
• \(I\) is the armature current, in amperes.
• \(R\) is the winding resistance, in ohms.

We take current entering the motor’s positive terminal as positive. Back EMF is positive for the motor’s normal direction of rotation, where it opposes the applied voltage.

Rearranging gives two useful forms:

\[ \mathcal{E}_{\mathrm b}=V_{\mathrm t}-IR \]

\[ I=\frac{V_{\mathrm t}-\mathcal{E}_{\mathrm b}}{R} \]

These equations use a simplified steady-state DC motor model. We initially assume a constant magnetic field, approximately constant winding temperature, and negligible brush voltage drops and inductive effects.

A real motor is more complicated, particularly during switching and rapid changes in current. We will return to those limitations.

What happens when the shaft rotates faster?

Before reading further, predict what happens if the motor speeds up while its terminal voltage and winding resistance stay constant.

Does the current increase because the motor is spinning faster?

It actually decreases.

For a motor with a constant magnetic field, back EMF is approximately proportional to angular speed:

\[ \mathcal{E}_{\mathrm b}=k\omega \]

Here \(\omega\) is angular speed in radians per second (\(\mathrm{rad\,s^{-1}}\)), and \(k\) is the motor’s back-EMF constant, in \(\mathrm{V\,s\,rad^{-1}}\).

As speed increases, back EMF increases. That leaves less voltage across the winding resistance, reducing the current.

This gives us a way to separate the two effects.

The most revealing row is the one about zero current.

A rotating motor can produce an EMF even when its electrical circuit is open. But with no current flowing, there is no resistive voltage drop.

That observation suggests an experimental way to distinguish them.

For a fuller explanation of why the induced EMF opposes the supply, see Back emf in a DC motor: HSC Physics explained. Here, our focus is how to identify and measure the two voltage contributions separately.

02How an experiment can distinguish them

Suppose you have a small permanent-magnet DC motor and want to answer a specific question:

When the current changes as the motor speeds up or slows down, is that because the winding resistance has changed, or because the motor is generating a different back EMF?

A useful investigation needs measurements of voltage, current, and rotational speed.

The first arrangement: measure a running motor

Start with an adjustable, current-limited DC supply.

Connect its positive terminal to an ammeter, then connect the ammeter to the motor’s positive terminal. Complete the circuit from the other motor terminal back to the supply’s negative terminal.

Connect a voltmeter directly across the motor terminals.

Attach a controllable mechanical load to the shaft. An optical tachometer measures the shaft’s rotational speed without needing to make electrical contact.

What should this arrangement show?

If the motor behaved as a fixed resistor, changing the mechanical load would not change the current at a constant terminal voltage.

For example, a \(1.0\,\Omega\) resistor connected across \(12.0\,\mathrm{V}\) would draw

\[ I=\frac{12.0}{1.0}=12.0\,\mathrm{A} \]

regardless of whether anything nearby was rotating.

A real running motor behaves differently.

Illustrative measurements

Consider the following hypothetical data for a permanent-magnet DC motor.

All readings are invented for this explanation, not taken from a particular experiment. Assume the winding temperature and magnetic field remain constant, and that the motor has an effective winding resistance of \(1.0\,\Omega\).

A brief, current-limited test with the shaft stationary gives a terminal voltage of \(2.0\,\mathrm{V}\) and a settled current of \(2.0\,\mathrm{A}\).

With back EMF zero at rest, the simplified model gives

\[ R=\frac{2.0}{2.0}=1.0\,\Omega \]

Such measurements need care: holding a powered motor stationary can produce substantial current and heating. In practice, resistance measurements must also account for brush contact effects and temperature.

Now suppose the motor is supplied with \(12.0\,\mathrm{V}\) and operated under different mechanical loads.

Follow the moderate-load reading.

The current is \(2.0\,\mathrm{A}\), so the voltage drop across the windings is

\[ V_R=(2.0)(1.0)=2.0\,\mathrm{V} \]

The remaining terminal voltage must balance the back EMF:

\[ \mathcal{E}_{\mathrm b}=12.0-2.0=10.0\,\mathrm{V} \]

Now compare this with the heavy-load reading.

The motor slows from \(200\) to \(160\,\mathrm{rad\,s^{-1}}\). Its back EMF decreases from \(10.0\) to \(8.0\,\mathrm{V}\).

This leaves an additional \(2.0\,\mathrm{V}\) across the winding resistance, so the current doubles from \(2.0\) to \(4.0\,\mathrm{A}\).

The winding resistance has not changed.

There is another useful test of the data.

At \(200\,\mathrm{rad\,s^{-1}}\), the inferred back EMF is \(10.0\,\mathrm{V}\). At \(160\,\mathrm{rad\,s^{-1}}\), it is \(8.0\,\mathrm{V}\).

In both cases,

\[ \frac{\mathcal{E}_{\mathrm b}}{\omega} =0.050\,\mathrm{V\,s\,rad^{-1}} \]

The constant ratio is what the fixed-field induction model predicts.

That is stronger evidence than merely observing that the current has changed.

The problem with the first design

There is still a limitation.

The voltmeter connected across the running motor reads the total terminal voltage. It does not directly display the back EMF.

We calculated back EMF by subtracting an assumed resistive drop:

\[ \mathcal{E}_{\mathrm b}=V_{\mathrm t}-IR \]

That calculation is only as reliable as our estimate of \(R\).

If the windings get hotter, their resistance generally increases. If brush voltage drops are significant, or the magnetic field changes, our simple model may become less accurate.

How could we test the induced EMF more independently?

The revised arrangement: spin the motor with its supply disconnected

Disconnect the motor from its electrical supply.

Keep a high-resistance voltmeter connected across its terminals, but use a separate mechanical drive to rotate its shaft at a controlled speed.

The permanent magnets still provide a magnetic field. The rotating windings still produce an induced EMF.

However, the open electrical circuit means that almost no current flows.

Therefore,

\[ IR\approx0 \]

and the terminal voltmeter reads approximately the induced EMF:

\[ V_{\mathrm{oc}}\approx\mathcal{E}_{\mathrm b} \]

Here \(V_{\mathrm{oc}}\) is the open-circuit terminal voltage in volts.

Using our illustrative motor, externally rotating the shaft at \(200\,\mathrm{rad\,s^{-1}}\) should produce an open-circuit voltage of approximately \(10.0\,\mathrm{V}\).

At \(160\,\mathrm{rad\,s^{-1}}\), the expected reading is \(8.0\,\mathrm{V}\).

This revised test is valuable because the voltage can be observed without a substantial resistive drop.

Notice the change in the experiment’s logic.

The first arrangement inferred back EMF from the voltage balance while the motor was powered. The second arrangement tests the existence and size of the induced voltage while almost no current flows.

Both arrangements describe the same physical effect.

For a meaningful comparison, the shaft speed and magnetic field must be the same in the powered and open-circuit tests. A separately excited motor would also need its field maintained when the armature supply was disconnected.

03Worked method: separate the voltages before calculating power

Most HSC-style questions supply the winding resistance rather than asking you to measure it.

The reliable method is to calculate \(IR\) first, subtract it from the motor’s terminal voltage, and then interpret the two contributions.

04Three mistakes that make motor voltage questions go wrong

Mistake 1: Using terminal voltage divided by current to find winding resistance

For an ordinary resistor,

\[ R=\frac{V}{I} \]

works because the entire voltage is a resistive drop.

For a running DC motor, some of the terminal voltage balances back EMF.

So the correct calculation is

\[ R=\frac{V_{\mathrm t}-\mathcal{E}_{\mathrm b}}{I} \]

provided the back EMF is known.

A motor’s apparent \(V_{\mathrm t}/I\) ratio can change dramatically with speed even when its winding resistance stays almost constant.

Mistake 2: Assuming the back EMF is whatever the voltmeter reads

A voltmeter across a powered motor measures the motor’s terminal voltage.

It does not separately measure the back EMF.

Under steady operation, you must subtract the resistive drop to infer the back EMF.

An open-circuit generator test can measure the induced voltage more directly, but only if the shaft is externally driven at the relevant speed and the magnetic field is maintained.

Mistake 3: Forgetting where the supply voltage is measured

Suppose a motor is powered through a resistor, long connecting wires, or a source with noticeable internal resistance.

The motor terminal voltage may be smaller than the source’s stated voltage.

You must then distinguish:

• Voltage lost elsewhere in the circuit.
• Resistive voltage drop inside the motor.
• Back EMF generated by the motor.

The equation

\[ V_{\mathrm t}=\mathcal{E}_{\mathrm b}+IR \]

uses the voltage actually across the motor, not necessarily the ideal source voltage.

Also remember the model’s limits. During rapid current changes, winding inductance contributes an additional voltage. Brush drops, magnetic-field changes, and temperature-dependent resistance can also matter. Do not use a steady-state voltage balance to describe every instant after a switch is closed.

05Questions and solutions

The questions increase in difficulty. The first checks the basic distinction. The later problems ask you to interpret experiments, deal with changing supply voltage, and decide what can actually be inferred from incomplete measurements.

Question 1

A DC motor operates steadily from a \(12.0\,\mathrm{V}\) supply.

Its armature resistance is \(0.75\,\Omega\), and its current is \(4.0\,\mathrm{A}\).

Calculate:

a. The resistive voltage drop across its windings.

b. The back EMF.

c. The apparent resistance obtained from \(V/I\), and explain why this is not the winding resistance.

Solution 1

The resistive drop is \(3.0\,\mathrm{V}\), the back EMF is \(9.0\,\mathrm{V}\), and the apparent resistance is \(3.0\,\Omega\), not the actual \(0.75\,\Omega\) winding resistance.

For part a, Ohm’s law gives

\[ \begin{aligned} V_R&=IR\\ &=(4.0)(0.75)\\ &=3.0\,\mathrm{V} \end{aligned} \]

For part b, the terminal voltage balances the back EMF and the winding drop:

\[ \begin{aligned} \mathcal{E}_{\mathrm b} &=V_{\mathrm t}-IR\\ &=12.0-3.0\\ &=9.0\,\mathrm{V} \end{aligned} \]

For part c,

\[ \frac{V_{\mathrm t}}{I} =\frac{12.0}{4.0} =3.0\,\Omega \]

This is not the winding resistance because the entire \(12.0\,\mathrm{V}\) does not act across the resistance.

Only \(3.0\,\mathrm{V}\) is the resistive drop. The remaining \(9.0\,\mathrm{V}\) balances the induced back EMF.

The tempting mistake is to treat a spinning motor as if it were a stationary resistor.

Question 2

A permanent-magnet DC motor has a winding resistance of \(0.60\,\Omega\). Its terminal voltage remains constant at \(18.0\,\mathrm{V}\).

Initially, it rotates at \(2700\,\mathrm{rpm}\) and draws \(3.0\,\mathrm{A}\).

A heavier mechanical load is applied. After the motor settles, its current has increased to \(5.0\,\mathrm{A}\).

Assume the magnetic field and winding resistance remain unchanged.

a. Calculate the back EMF before and after the load increases.

b. Determine the new rotational speed.

c. Calculate the winding heating power in both conditions.

d. A student claims the winding resistance has decreased because \(V/I\) has decreased. Evaluate this claim.

Solution 2

The back EMF falls from \(16.2\,\mathrm{V}\) to \(15.0\,\mathrm{V}\), the speed falls to \(2500\,\mathrm{rpm}\), and winding heating increases from \(5.4\,\mathrm{W}\) to \(15.0\,\mathrm{W}\). The winding resistance has not decreased.

For part a, the initial back EMF is

\[ \begin{aligned} \mathcal{E}_{\mathrm{b1}} &=V_{\mathrm t}-I_1R\\ &=18.0-(3.0)(0.60)\\ &=16.2\,\mathrm{V} \end{aligned} \]

After the load increases,

\[ \begin{aligned} \mathcal{E}_{\mathrm{b2}} &=18.0-(5.0)(0.60)\\ &=15.0\,\mathrm{V} \end{aligned} \]

For part b, back EMF is proportional to rotational speed because the magnetic field is constant.

Let \(n\) represent speed in rpm.

\[ \frac{n_2}{n_1} = \frac{\mathcal{E}_{\mathrm{b2}}} {\mathcal{E}_{\mathrm{b1}}} \]

Substituting gives

\[ \begin{aligned} n_2&=2700\left(\frac{15.0}{16.2}\right)\\ &=2500\,\mathrm{rpm} \end{aligned} \]

The motor slows by only about \(7.4\%\), yet its current increases substantially.

This is possible because the winding resistance is small. A relatively small change in opposing EMF can therefore cause a large change in current.

For part c, initial winding heating is

\[ \begin{aligned} P_{\mathrm{heat1}} &=(3.0)^2(0.60)\\ &=5.4\,\mathrm{W} \end{aligned} \]

After the load increases,

\[ \begin{aligned} P_{\mathrm{heat2}} &=(5.0)^2(0.60)\\ &=15.0\,\mathrm{W} \end{aligned} \]

For part d, the apparent voltage-to-current ratios are

\[ \frac{18.0}{3.0}=6.0\,\Omega \]

and

\[ \frac{18.0}{5.0}=3.6\,\Omega \]

Neither value represents the motor’s actual winding resistance.

The student has incorrectly assigned the entire terminal voltage to the resistance.

The actual resistance remains \(0.60\,\Omega\). The change in current is explained by the smaller back EMF produced at the lower speed.

Question 3

A technician wants to determine the winding resistance of a permanent-magnet DC motor without holding its shaft stationary.

During steady motoring operation, the motor’s angular speed is \(180\,\mathrm{rad\,s^{-1}}\). The terminal voltage is \(13.0\,\mathrm{V}\), and the current entering the motor is \(2.5\,\mathrm{A}\).

The technician then disconnects the electrical supply and uses another machine to drive the shaft at exactly \(180\,\mathrm{rad\,s^{-1}}\).

A high-resistance voltmeter connected across the open-circuit motor reads \(11.0\,\mathrm{V}\).

Assume the magnetic field, winding temperature, and direction of rotation remain unchanged. Neglect brush voltage drops.

a. Determine the armature winding resistance.

b. Explain why the two terminal voltage readings differ, even though the motor rotates at the same speed.

c. Suppose the technician accidentally performed the open-circuit test at \(150\,\mathrm{rad\,s^{-1}}\). Predict the open-circuit voltage and calculate the incorrect resistance that would be inferred by treating it as a same-speed measurement.

Solution 3

The winding resistance is \(0.80\,\Omega\). The different terminal voltages arise because current produces a resistive drop during motoring, but almost no current flows during the open-circuit test. Using an unmatched shaft speed would incorrectly suggest a resistance of approximately \(1.53\,\Omega\).

For part a, at the same speed and magnetic field, the induced EMF is the same in both tests.

In the open-circuit test, current is approximately zero, so

\[ \mathcal{E}_{\mathrm b}\approx11.0\,\mathrm{V} \]

During powered operation,

\[ V_{\mathrm t}=\mathcal{E}_{\mathrm b}+IR \]

Therefore,

\[ \begin{aligned} R&=\frac{V_{\mathrm t}-\mathcal{E}_{\mathrm b}}{I}\\ &=\frac{13.0-11.0}{2.5}\\ &=0.80\,\Omega \end{aligned} \]

For part b, the powered motor has two contributions to its terminal voltage:

\[ 13.0=11.0+2.0 \]

The back EMF is \(11.0\,\mathrm{V}\), while the winding drop is \(2.0\,\mathrm{V}\).

During the open-circuit test, the motor is still rotating and generating an EMF, but almost no current flows.

Therefore, the resistive drop is negligible and the terminal voltage is approximately \(11.0\,\mathrm{V}\).

For part c, with the magnetic field unchanged, induced EMF is proportional to angular speed.

At \(150\,\mathrm{rad\,s^{-1}}\),

\[ \begin{aligned} \mathcal{E}_{150} &=11.0\left(\frac{150}{180}\right)\\ &\approx9.17\,\mathrm{V} \end{aligned} \]

Here \(\mathcal{E}_{150}\) is the open-circuit EMF at the lower speed.

If this value were incorrectly paired with the original powered reading, the calculated resistance would be

\[ \begin{aligned} R_{\mathrm{wrong}} &=\frac{13.0-9.17}{2.5}\\ &\approx1.53\,\Omega \end{aligned} \]

That is almost twice the correct resistance.

The error comes from assuming the induced EMF is unchanged when the shaft speed has changed.

The essential experimental control is not simply switching off the supply. It is maintaining the same shaft speed, magnetic field, and winding condition between measurements.

Question 4

An ideal \(18.0\,\mathrm{V}\) voltage source is connected in series with a resistance of \(0.50\,\Omega\), representing the internal resistance of a practical power supply.

This combination drives a permanent-magnet DC motor with an armature resistance of \(0.50\,\Omega\).

Initially, the motor draws \(2.0\,\mathrm{A}\) and rotates steadily at \(300\,\mathrm{rad\,s^{-1}}\).

A heavier mechanical load slows the shaft by \(25\%\).

Assume the magnetic field and both resistances remain constant, and neglect all other electrical losses.

a. Find the initial voltage across the motor terminals and the initial back EMF.

b. Determine the new current after the motor reaches its lower steady speed.

c. Find the new motor terminal voltage, winding voltage drop, and back EMF.

d. Calculate the heating power in the supply’s internal resistance and the motor windings under the new conditions.

e. Explain why using \(18.0\,\mathrm{V}\) as the motor terminal voltage would give an incorrect answer.

Solution 4

The initial motor terminal voltage is \(17.0\,\mathrm{V}\) and the initial back EMF is \(16.0\,\mathrm{V}\). After the motor slows, the current increases to \(6.0\,\mathrm{A}\), the motor terminal voltage falls to \(15.0\,\mathrm{V}\), and each resistance dissipates \(18.0\,\mathrm{W}\).

The crucial point is that the supply’s internal resistance and the motor’s winding resistance are separate.

Let \(R_{\mathrm s}=0.50\,\Omega\) be the supply resistance and \(R_{\mathrm m}=0.50\,\Omega\) be the motor winding resistance.

For part a, the supply resistance initially drops

\[ \begin{aligned} V_{\mathrm s,drop} &=I_1R_{\mathrm s}\\ &=(2.0)(0.50)\\ &=1.0\,\mathrm{V} \end{aligned} \]

The initial motor terminal voltage is therefore

\[ V_{\mathrm{t1}}=18.0-1.0=17.0\,\mathrm{V} \]

The motor winding drop is

\[ I_1R_{\mathrm m}=(2.0)(0.50)=1.0\,\mathrm{V} \]

So the initial back EMF is

\[ \mathcal{E}_{\mathrm{b1}} =17.0-1.0 =16.0\,\mathrm{V} \]

For part b, the speed falls to \(75\%\) of its original value.

The new angular speed is

\[ \omega_2=(0.75)(300)=225\,\mathrm{rad\,s^{-1}} \]

Because the magnetic field is constant, the back EMF also falls to \(75\%\):

\[ \mathcal{E}_{\mathrm{b2}} =(0.75)(16.0) =12.0\,\mathrm{V} \]

The ideal source voltage balances three contributions: back EMF, motor winding drop, and supply resistance drop.

\[ 18.0=\mathcal{E}_{\mathrm{b2}} +I_2R_{\mathrm m} +I_2R_{\mathrm s} \]

Rearranging,

\[ \begin{aligned} I_2&=\frac{18.0-12.0}{0.50+0.50}\\ &=\frac{6.0}{1.00}\\ &=6.0\,\mathrm{A} \end{aligned} \]

For part c, the motor terminal voltage is

\[ \begin{aligned} V_{\mathrm{t2}} &=18.0-I_2R_{\mathrm s}\\ &=18.0-(6.0)(0.50)\\ &=15.0\,\mathrm{V} \end{aligned} \]

The motor winding drop is

\[ I_2R_{\mathrm m} =(6.0)(0.50) =3.0\,\mathrm{V} \]

The motor’s voltage balance is therefore

\[ 15.0=12.0+3.0 \]

The back EMF is \(12.0\,\mathrm{V}\), as predicted from the change in speed.

For part d, each resistance has the same value and carries the same current, so both produce equal heating.

For the supply resistance,

\[ \begin{aligned} P_{\mathrm{s,heat}} &=I_2^2R_{\mathrm s}\\ &=(6.0)^2(0.50)\\ &=18.0\,\mathrm{W} \end{aligned} \]

For the motor windings,

\[ \begin{aligned} P_{\mathrm{m,heat}} &=I_2^2R_{\mathrm m}\\ &=(6.0)^2(0.50)\\ &=18.0\,\mathrm{W} \end{aligned} \]

The motor’s electromagnetic conversion power is

\[ P_{\mathrm{em}} =(12.0)(6.0) =72.0\,\mathrm{W} \]

The ideal source supplies

\[ P_{\mathrm{source}} =(18.0)(6.0) =108\,\mathrm{W} \]

The energy balance is consistent:

\[ 108=18+18+72 \]

For part e, using \(18.0\,\mathrm{V}\) as the motor terminal voltage would ignore the voltage dropped across the supply’s internal resistance.

The actual motor terminal voltage falls as current increases.

The tempting shortcut is to assume that a supply labelled \(18.0\,\mathrm{V}\) always places \(18.0\,\mathrm{V}\) directly across the motor. That is only justified when voltage drops elsewhere in the circuit are negligible.

This problem contains two separate resistive voltage drops, neither of which is back EMF.

Question 5

A permanent-magnet DC machine is connected to an adjustable electrical source that can both supply and absorb current.

An external shaft controller can hold the machine at a constant angular speed, even when the electromagnetic torque changes direction.

During two initial tests, the shaft speed, magnetic field, and winding temperature are identical. The winding resistance and back EMF are unknown.

Current entering the machine from the electrical source is defined as positive.

The following readings are obtained:

The direction of shaft rotation is unchanged between tests.

A student argues that the reversal of current means the back EMF must also have reversed, so the two readings cannot be used to determine a single winding resistance.

a. Evaluate the student’s claim. Determine the winding resistance and the back EMF from the two tests.

b. Identify the direction of electromagnetic energy conversion during test B. Calculate the terminal electrical power, resistive heating power, and electromagnetic conversion power, and show that energy is conserved.

The machine is then allowed to become warmer. In test C, the external controller maintains exactly the original shaft speed and direction, but the terminal readings are now \(10.6\,\mathrm{V}\) and \(+3.0\,\mathrm{A}\).

Assume the permanent magnets may have weakened with heating but cannot have become stronger. The machine’s geometry remains unchanged.

c. Can test C uniquely determine the new winding resistance and induced EMF? Find the minimum possible winding resistance, and give two different sets of winding resistance and back EMF that agree with the reading.

d. State an additional electrical measurement that would distinguish those possibilities. If that measurement gives \(6.7\,\mathrm{V}\), determine the new winding resistance.

Neglect brush drops, inductive effects, and other electrical losses. Assume the simplified steady-state motor equation remains valid in both directions of current.

Solution 5

The student’s claim is incorrect. The winding resistance during tests A and B is \(1.0\,\Omega\), and the induced EMF is \(7.0\,\mathrm{V}\) in both tests. Reversing current does not reverse the back EMF when the shaft continues rotating in the same direction through the same magnetic field.

During test B, the machine converts mechanical energy into electrical energy. Test C cannot uniquely separate resistance and induced EMF, although it establishes that the winding resistance is at least \(1.2\,\Omega\) under the stated assumptions.

The key is to distinguish what determines each quantity.

For part a, during steady operation,

\[ V_{\mathrm t}=\mathcal{E}_{\mathrm b}+IR \]

During test A,

\[ 11.0=\mathcal{E}_{\mathrm b}+4.0R \]

During test B,

\[ 5.0=\mathcal{E}_{\mathrm b}-2.0R \]

The same induced EMF appears in both equations because the shaft speed, field strength, and direction of rotation are unchanged.

Subtracting the second equation from the first eliminates the EMF:

\[ \begin{aligned} 11.0-5.0&=6.0R\\ R&=\frac{6.0}{6.0}\\ &=1.0\,\Omega \end{aligned} \]

Substituting into test A gives

\[ \begin{aligned} \mathcal{E}_{\mathrm b} &=11.0-(4.0)(1.0)\\ &=7.0\,\mathrm{V} \end{aligned} \]

Test B confirms the same induced EMF:

\[ \mathcal{E}_{\mathrm b} =5.0-(-2.0)(1.0) =7.0\,\mathrm{V} \]

The negative current changes the direction of the resistive voltage contribution, not the direction of the induced EMF.

A negative current does not imply negative resistance or reverse shaft rotation.

For part b, test B has current

\[ I=-2.0\,\mathrm{A} \]

Since positive current was defined as entering the machine, this means current actually flows out of the machine towards the electrical source.

The terminal electrical power, with power entering the machine taken as positive, is

\[ \begin{aligned} P_{\mathrm t} &=V_{\mathrm t}I\\ &=(5.0)(-2.0)\\ &=-10.0\,\mathrm{W} \end{aligned} \]

The negative sign means \(10.0\,\mathrm{W}\) is delivered electrically from the machine to the source.

Resistive heating is always non-negative:

\[ \begin{aligned} P_{\mathrm{heat}} &=I^2R\\ &=(-2.0)^2(1.0)\\ &=4.0\,\mathrm{W} \end{aligned} \]

The electromagnetic conversion power, using the motor sign convention, is

\[ \begin{aligned} P_{\mathrm{em}} &=\mathcal{E}_{\mathrm b}I\\ &=(7.0)(-2.0)\\ &=-14.0\,\mathrm{W} \end{aligned} \]

The negative result means energy conversion is occurring from the mechanical system into the electrical system.

The external shaft controller supplies \(14.0\,\mathrm{W}\) of mechanical power for electromagnetic conversion, in addition to any mechanical losses excluded from the model.

Of this, \(10.0\,\mathrm{W}\) goes to the electrical source and \(4.0\,\mathrm{W}\) becomes winding heat.

The signed power balance is

\[ -10.0=-14.0+4.0 \]

This explains how the machine can generate electrical power while still heating its windings.

The student might expect the induced EMF to reverse when the current reverses because, during ordinary motoring, current and rotation are connected.

But the shaft is externally controlled in this experiment. Its rotation determines the induced EMF, while the difference between terminal voltage and induced EMF determines the current.

For part c, test C occurs at the same speed but at a different temperature.

Its voltage equation is

\[ 10.6=\mathcal{E}_{\mathrm C}+3.0R_{\mathrm C} \]

Here \(\mathcal{E}_{\mathrm C}\) and \(R_{\mathrm C}\) are the induced EMF and winding resistance during test C.

We have only one equation but two unknowns.

It would be tempting to assume that the back EMF is still exactly \(7.0\,\mathrm{V}\) because the speed is unchanged.

However, the question explicitly allows the magnetic field to weaken. Since induced EMF depends on both speed and field strength, the EMF might be smaller.

The magnets cannot have become stronger, so

\[ \mathcal{E}_{\mathrm C}\leq7.0\,\mathrm{V} \]

For the same direction of rotation and magnetic field, the induced EMF remains non-negative.

Rearranging the voltage equation gives

\[ R_{\mathrm C} =\frac{10.6-\mathcal{E}_{\mathrm C}}{3.0} \]

The smallest possible resistance occurs when the induced EMF is as large as permitted, namely \(7.0\,\mathrm{V}\).

Therefore,

\[ \begin{aligned} R_{\mathrm{C,min}} &=\frac{10.6-7.0}{3.0}\\ &=1.2\,\Omega \end{aligned} \]

So the warm winding resistance must be at least \(1.2\,\Omega\).

We can construct two consistent explanations.

Possibility 1: The magnetic field is unchanged.

If

\[ \mathcal{E}_{\mathrm C}=7.0\,\mathrm{V} \]

then

\[ R_{\mathrm C} =\frac{10.6-7.0}{3.0} =1.2\,\Omega \]

The higher resistance accounts for the new reading.

Possibility 2: The magnetic field has weakened.

Suppose

\[ \mathcal{E}_{\mathrm C}=6.7\,\mathrm{V} \]

Then

\[ R_{\mathrm C} =\frac{10.6-6.7}{3.0} =1.3\,\Omega \]

This also agrees exactly with the measured terminal voltage and current:

\[ 10.6=6.7+(3.0)(1.3) \]

Both possibilities are consistent with the stated physics.

Therefore, we can conclude that the winding resistance has increased from its original \(1.0\,\Omega\), but we cannot determine its exact new value from the single loaded reading.

This is the important logical distinction: incomplete information does not always mean we know nothing. Sometimes it allows a definite bound without allowing a unique numerical answer.

For part d, the additional measurement should isolate the induced EMF from the resistive drop.

Disconnect the electrical source and have the external controller continue rotating the warm machine at the same shaft speed.

Measure the open-circuit voltage across the machine using a high-resistance voltmeter.

With almost zero current, the winding drop is negligible. The open-circuit reading therefore gives the induced EMF under the new conditions.

If that reading is \(6.7\,\mathrm{V}\), then

\[ \begin{aligned} R_{\mathrm C} &=\frac{10.6-6.7}{3.0}\\ &=1.3\,\Omega \end{aligned} \]

The warm winding resistance is \(1.3\,\Omega\), and the back EMF has fallen to \(6.7\,\mathrm{V}\).

The extra measurement distinguishes two competing explanations that the original terminal readings could not separate.

That is the deeper lesson behind distinguishing back EMF from a resistive voltage drop. A terminal voltage and current do not automatically tell you how much voltage belongs to each physical effect. You must know something independently about the winding resistance, shaft motion, or magnetic field.

This same method becomes useful when investigating motor efficiency, generator operation, and regenerative braking. In each case, the next question is not simply how much current flows, but whether the measured electrical power is becoming heat, mechanical energy, or electrical energy returned to a source.