# Solving Resistor Circuit Homework: DC Motor, Rf, Rr, E, If, Ir, Pth

• David112234
In summary: The difference between Vdc and the back-emf ε will be dropped across the rotor resistance.In summary, the conversation discusses a shunt-wound DC motor operating from a 140 V DC power line. The motor has a resistance of 248 Ω for the field windings and 4.40 Ω for the rotor. The rotor develops an emf E and draws a current of 4.24 A from the line. Friction losses amount to 41.0 W. The conversation also includes various equations to compute the field and rotor currents, emf, and thermal energy development in both the field windings and rotor. The motor's efficiency is also discussed. The concept of back-emf is also
David112234

## Homework Statement

A shunt-wound DC motor with the field coils and rotor connected in parallel (see the figure) (Figure 1) operates from a 140 V DC power line. The resistance of the field windings, Rf, is 248 Ω . The resistance of the rotor, Rr, is 4.40 Ω . When the motor is running, the rotor develops an emf E. The motor draws a current of 4.24 A from the line. Friction losses amount to 41.0 W .

A
Compute the field current If.
B
Compute the rotor current Ir.
C
Compute the emf E.
D
Compute the rate of development of thermal energy in the field windings.
E
Compute the rate Pth,rotor of development of thermal energy in the rotor.
F
Compute the power input to the motor Pin.
G
Compute the efficiency of the motor.

## Homework Equations

I=v/R

3. The Attempt at a Solution

For Part A it I2=140/248 = .565

For part B since the points A,B, and C have the same potential and the motor has a resistance you can use ohms law again.
I3 = 140/4.40 = 31.81
but that is wrong, why?

Also I do not understand the EMF part, so when the motor becomes active it will develop its own EMF that will lower the batteries voltage. But wouldn't that then change the current trough the battery, and it would no longer be able to create that EMF?
So the motor is dependent on the current caused by the batteries voltage which is dependent on the motor? How does that work?

David112234 said:
when the motor becomes active it will develop its own EMF that will lower the batteries voltage.
The rotor will develop a back-EMF = E.

Thus the voltage that drives the current I3 through Rr = 140V - E

The batteries voltage is not lowered, but constant.

Hesch said:
The rotor will develop a back-EMF = E.

Thus the voltage that drives the current I3 through Rr = 140V - E

The batteries voltage is not lowered, but constant.

What does "back" EMF mean?
the Hint gives me "Think of the back emf from the rotor as reducing the voltage difference from the power supply across the rotor."
But wouldn't the voltage difference across the rotor reach the battery since wires are equipotential? Shouldn't the voltage for A,B, and C be the same?
Where on the diagram would the voltage be different, and wouldn't that still change the current trough the rotor?

David112234 said:
What does "back" EMF mean?
the Hint gives me "Think of the back emf from the rotor as reducing the voltage difference from the power supply across the rotor."
But wouldn't the voltage difference across the rotor reach the battery since wires are equipotential? Shouldn't the voltage for A,B, and C be the same?
Where on the diagram would the voltage be different, and wouldn't that still change the current trough the rotor?

The back-emf appears within the motor, distributed over the rotor coil. The rotor winding also has a resistance though, thanks to the resistivity of the wire that it's made of. A simplifying assumption is to model the resulting distributed system as a single EMF source in series with a single net resistance. So the potential difference between the power supply and the back-emf source is effectively absorbed by the winding resistance.

gneill said:
The back-emf appears within the motor, distributed over the rotor coil. The rotor winding also has a resistance though, thanks to the resistivity of the wire that it's made of. A simplifying assumption is to model the resulting distributed system as a single EMF source in series with a single net resistance. So the potential difference between the power supply and the back-emf source is effectively absorbed by the winding resistance.

I don't understand anything you wrote.
The potential across A,B, and C are are equal, is that correct?
If the battery than develops an EMF where on the diagram would it be?

David112234 said:
I don't understand anything you wrote.
The potential across A,B, and C are are equal, is that correct?
If the battery than develops an EMF where on the diagram would it be?
The potentials at A, B and C (with respect to the bottom node) are all the same. A, B, and C are in fact all one node.

The battery has a constant EMF of Vdc. The battery doesn't develop the back-emf, the motor's rotor coil does. It's inside the blue circle. That's why the blue circle is labeled with the variables ε and Rr, the back-emf and the coil resistance.

gneill said:
The potentials at A, B and C (with respect to the bottom node) are all the same. A, B, and C are in fact all one node.

The battery has a constant EMF of Vdc. The battery doesn't develop the back-emf, the motor's rotor coil does. It's inside the blue circle. That's why the blue circle is labeled with the variables ε and Rr, the back-emf and the coil resistance.

Ok, so current flows from high to lower potential, from top of the rotor to the bottom. EMF does the opposite, it pushes charges to a higher potential. So the motor would create a voltage at point A, since those wires have the same potential than the potential at A,B, and C would now be Vfrom battery - Vfrom motor. Do you see my logic? Is it correct?

No, you seem to think that the node ABC can change potential. It can't.

All three points A, B, and C are fixed at potential Vdc. There is no flexibility with this as the battery is considered to be ideal with a fixed potential difference and the wires are taken to be perfect conductors.

The difference between Vdc and the back-emf ε will be dropped across the rotor resistance.

gneill said:
No, you seem to think that the node ABC can change potential. It can't.

All three points A, B, and C are fixed at potential Vdc. There is no flexibility with this as the battery is considered to be ideal with a fixed potential difference and the wires are taken to be perfect conductors.

The difference between Vdc and the back-emf ε will be dropped across the rotor resistance.

View attachment 109675

Sorry for the late post,
I think my problem stems from not understanding behavior of circuits with more than 1 battery
Lets ignore the middle resistor and use a series circuit for simplicity

So V1 will create a current clockwise, V2 will create a current counter clockwise, which way will the current flow trough the resistor, the battery with the higher Voltage will overpower the other one, is that correct?

So would you be able to just add the Voltages, if battery 1 is 10V and battery 2 is 3V wouldn't the voltage trough the resistor be
10V-7V = 3V ?

I can't imagine what happens physically. If the battery with larger voltage is pushing the current through the other battery opposite direction it would regularly flow what is the EMF doing? How are the charges moving, are they moving up and then back down or are they constantly moving up but with less energy because of the EMF?

David112234 said:
So would you be able to just add the Voltages, if battery 1 is 10V and battery 2 is 3V wouldn't the voltage trough the resistor be
10V-7V = 3V ?

Current flows though a component, voltage is measured across a component. Voltage does not flow through a component.

Potential changes in a circuit sum as you go around the circuit. This is the basis of Kirchhoff's Voltage Law (KVL) wherein if you sum all the potential changes around a closed path the sum will be zero. So in your example circuit, as you do a "KVL walk" clockwise around the circuit the potential differences for the two batteries will sum as +10V - 3V = 7 V. So the net EMF driving the circuit is 7 V, and this 7 V must be dropped across the resistor R if the total sum around the circuit is to be zero.

David112234 said:
I can't imagine what happens physically. If the battery with larger voltage is pushing the current through the other battery opposite direction it would regularly flow what is the EMF doing? How are the charges moving, are they moving up and then back down or are they constantly moving up but with less energy because of the EMF?
The current that enters a battery "in reverse" tends to recharge the battery. The EMF of a battery is typically generated through a chemical reaction that separates charges. When you force current through such a cell in reverse it tends to drive the chemistry the other way. Note that not all batteries are designed to make this process efficient or even workable depending upon how the chemistry sequesters the "spent" components in normal operation. This is why you can by "disposable" and "rechargeable" versions of batteries.

gneill said:
Current flows though a component, voltage is measured across a component. Voltage does not flow through a component.

Potential changes in a circuit sum as you go around the circuit. This is the basis of Kirchhoff's Voltage Law (KVL) wherein if you sum all the potential changes around a closed path the sum will be zero. So in your example circuit, as you do a "KVL walk" clockwise around the circuit the potential differences for the two batteries will sum as +10V - 3V = 7 V. So the net EMF driving the circuit is 7 V, and this 7 V must be dropped across the resistor R if the total sum around the circuit is to be zero.

I made an error there, the lower picture depicts what I meant to write,
So if battery 1 is 10V and battery 2 is 3V wouldn't the current through the resistor be as if there was just 1 (7)V battery instead?

And the recharging of the battery, let's assume they don't discharge and are perfect sources.
The charges are moving up with a U of q(10) but the EMF pointing down is trying to force them downwards and taking away some U of q(3) so the charges by the time reach the resistor they have q(7)J. Is that correct?

David112234 said:
I made an error there, the lower picture depicts what I meant to write,
So if battery 1 is 10V and battery 2 is 3V wouldn't the current through the resistor be as if there was just 1 (7)V battery instead?
Yes.
And the recharging of the battery, let's assume they don't discharge and are perfect sources.
The charges are moving up with a U of q(10) but the EMF pointing down is trying to force them downwards and taking away some U of q(3) so the charges by the time reach the resistor they have q(7)J. Is that correct?
Conceptually that is true, given that a potential difference has units of Joules per Coulomb.

gneill said:
Yes.

Conceptually that is true, given that a potential difference has units of Joules per Coulomb.

So now I have understood how a circuit behaves with two batteries, going back to the original problem, at point C why wouldn't you add the voltages from the battery and from the motor? When the motor is creating an EMF don't you have a circuit with two batteries?

David112234 said:
So now I have understood how a circuit behaves with two batteries, going back to the original problem, at point C why wouldn't you add the voltages from the battery and from the motor? When the motor is creating an EMF don't you have a circuit with two batteries?
The potential at point C is fixed by the battery voltage because there is no resistance between point C and the battery connection. All of the node ABC is fixed by the battery voltage. The emf of the motor is not directly connected between the common node and point C; there is the resistance of the rotor winding in the path. See the diagram in post #8 where the equivalent circuit for the emf and rotor resistance is indicated. The rotor current will create a potential drop across this resistance.

## 1. What is a DC motor?

A DC motor is a type of electric motor that converts direct current (DC) electrical energy into mechanical energy. It typically consists of a stator (stationary part) and a rotor (rotating part), and works on the principle of electromagnetism.

## 2. What do Rf and Rr stand for in resistor circuits?

Rf stands for forward resistance, which is the resistance of a resistor when current flows through it in the forward direction. Rr stands for reverse resistance, which is the resistance of a resistor when current flows through it in the reverse direction.

## 3. What is the difference between E, If, and Ir in a resistor circuit?

E is the voltage or electromotive force applied to a circuit, measured in volts (V). If is the forward current, or the current flowing through a resistor in the forward direction, measured in amperes (A). Ir is the reverse current, or the current flowing through a resistor in the reverse direction, measured in amperes (A).

## 4. How do you solve for the total power (Pth) in a resistor circuit?

The formula for total power (Pth) in a resistor circuit is Pth = E * If, where E is the applied voltage and If is the forward current. Simply multiply the values for E and If to get the total power in watts (W).

## 5. How do you use Kirchhoff's laws to solve resistor circuit homework?

Kirchhoff's laws are used to analyze and solve complex resistor circuits. Kirchhoff's first law, also known as the current law, states that the sum of currents entering a node (or junction) must equal the sum of currents leaving that node. Kirchhoff's second law, also known as the voltage law, states that the sum of voltage drops in a closed loop must equal the sum of voltage sources in that loop. By applying these laws, you can create and solve a system of equations to find the values of unknown resistors or currents in a circuit.

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