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

[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.
Express your answer in watts.
E
Compute the rate Pth,rotor of development of thermal energy in the rotor.
Express your answer in watts.
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?

 

[Hesch]

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 R_r = 140V - E

The batteries voltage is not lowered, but constant.

 

[David112234]

The rotor will develop a back-EMF = E.

Thus the voltage that drives the current I₃ through R_r = 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?

 

[gneill]

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.

 

[David112234]

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?

 

[gneill]

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 V_dc. 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 R_r, the back-emf and the coil resistance.

 

[David112234]

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 V_dc. 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 R_r, 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 V_{from battery} - V_{from motor}. Do you see my logic? Is it correct?

 

[gneill]

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 V_dc. 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 V_dc and the back-emf ε will be dropped across the rotor resistance.

 

[David112234]

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 V_dc. 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 V_dc 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?

 

[gneill]

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.

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.

 

[David112234]

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?

 

[gneill]

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.

 

[David112234]

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?

 

[gneill]

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.