Basic ohm's law problem, 2 loads wired in series, 120 volt supply, and many questions.

Thread Starter

Gailileo

Joined Jun 8, 2020
10
The problem is, "There are two 120 motors wired in series, one is 60 watts and one is 4 watts. The result is; "

A. Both motors operate normally. B. Neither motor operates normally. C. Only the 60 watt motor operates. D. Only the 4 watt motor operates.

I'm more interested in the math behind this problem. To try and figure this out I made the following calculations:

using P1 = (E^2)/R1 ----> R1 = E^2/P1 -----> R1 = (120^2)/60 = 240 ohms.

Then R2 = (120^2)/4 = 3600 ohms

I didn't use voltage drops in this calculation because I figured the motors were designed for these specs.

Next, I wanted to find the total amps on the circuit (now imagining them wired in series).
I used three different formulas from the power wheel.
IT = PT/ET ----> IT = 64/120 = .53333A
IT = sqrt(PT/RT) -----> IT = sqrt(64/3840) = 0.1289A.
IT = ET/RT ----> IT = 120/3840 = .03125A

I expected these values to be equal, but, they are not. I tested out all of these values trying to calculate the voltage drop across each resistor.
Voltage was conserved only when I used the third value.
E1 = 240(.03125) = 7.5V
E2 = 3600(.03125) = 112.5V
E1 + E2 = 120V

Then I calculated the amp draw for both motors.

I1 = sqrt(60/240) = 0.5A
I2 = sqrt(4/3600) = .033333A

My conclusion, based on these calculations, is that neither motor is operating normally because there aren't enough amps on the circuit (.03125A) to satisfy their needs.

My questions are:
Am I wrong? If so what have I not considered?
If I am correct, is my reasoning sound, or did I come to the correct conclusion through faulty reasoning? (which is just as bad as getting it wrong)
On the power wheel it seems as if there are only two equations, Watt's Law and Ohm's Law, and they are used as a system of equations. Is this why the other two formulas I used yielded different results? (EDIT: I think I made a mistake by assuming the motors were dissipating the power they were rated for. I re-calculated the watts produced at each load based on the total amp value that I moved forward with, and the total watts produced was 3.872W, not 64W.)

As you can tell, I have very little experience modeling these types of problems so please bare with me.
 
Last edited:

ericgibbs

Joined Jan 29, 2010
12,942
hi G.
Welcome to AAC.
The question wording is a little ambiguous

E

A. Both motors operate normally.
B. Neither motor operates normally.
C. Only the 60 watt motor operates.
D. Only the 4 watt motor operates.

'My conclusion, based on these calculations, is that neither motor is operating normally '
 

Sensacell

Joined Jun 19, 2012
2,833
Your numbers look fine.

The question is really terrible.
In the real world, motors are very nonlinear loads, if this was about lightbulbs, it would be more logical.

You passed the test, but whoever created this question fails.
 

BobTPH

Joined Jun 5, 2013
3,318
@MrChips, Because those are the currents with them powered alone.

I think they will mark you wrong on this, though, in reality you are right.

They are going to claim the 4W motor is operating normally because it is getting nearly the voltage and current that it is specified for.

Bob
 

MrChips

Joined Oct 2, 2009
23,541
@MrChips, Because those are the currents with them powered alone.

I think they will mark you wrong on this, though, in reality you are right.

They are going to claim the 4W motor is operating normally because it is getting nearly the voltage and current that it is specified for.

Bob
Are you sure that you are addressing the correct poster?
 

BobaMosfet

Joined Jul 1, 2009
1,776
The problem is, "There are two 120 motors wired in series, one is 60 watts and one is 4 watts. The result is; "

A. Both motors operate normally. B. Neither motor operates normally. C. Only the 60 watt motor operates. D. Only the 4 watt motor operates.

I'm more interested in the math behind this problem. To try and figure this out I made the following calculations:

using P1 = (E^2)/R1 ----> R1 = E^2/P1 -----> R1 = (120^2)/60 = 240 ohms.

Then R2 = (120^2)/4 = 3600 ohms

I didn't use voltage drops in this calculation because I figured the motors were designed for these specs.

Next, I wanted to find the total amps on the circuit (now imagining them wired in series).
I used three different formulas from the power wheel.
IT = PT/ET ----> IT = 64/120 = .53333A
IT = sqrt(PT/RT) -----> IT = sqrt(64/3840) = 0.1289A.
IT = ET/RT ----> IT = 120/3840 = .03125A

I expected these values to be equal, but, they are not. I tested out all of these values trying to calculate the voltage drop across each resistor.
Voltage was conserved only when I used the third value.
E1 = 240(.03125) = 7.5V
E2 = 3600(.03125) = 112.5V
E1 + E2 = 120V

Then I calculated the amp draw for both motors.

I1 = sqrt(60/240) = 0.5A
I2 = sqrt(4/3600) = .033333A

My conclusion, based on these calculations, is that neither motor is operating normally because there aren't enough amps on the circuit (.03125A) to satisfy their needs.

My questions are:
Am I wrong? If so what have I not considered?
If I am correct, is my reasoning sound, or did I come to the correct conclusion through faulty reasoning? (which is just as bad as getting it wrong)
On the power wheel it seems as if there are only two equations, Watt's Law and Ohm's Law, and they are used as a system of equations. Is this why the other two formulas I used yielded different results? (EDIT: I think I made a mistake by assuming the motors were dissipating the power they were rated for. I re-calculated the watts produced at each load based on the total amp value that I moved forward with, and the total watts produced was 3.872W, not 64W.)

As you can tell, I have very little experience modeling these types of problems so please bare with me.
Watt's Law is more applicative.

Neither motor will run. Because you only have 120V potential, and both motors want the full potential, the smaller motor acts as a higher inductive impedance- which means it will get most of the voltage dropped across it (about 113V), and the larger motor which is a current sink, will only get < 8V or so. The smaller motor acts as the bottle neck and will limit the current to < 33mA. That is the reason neither will run.
 

Thread Starter

Gailileo

Joined Jun 8, 2020
10
hi G.
Welcome to AAC.
The question wording is a little ambiguous

E

A. Both motors operate normally.
B. Neither motor operates normally.
C. Only the 60 watt motor operates.
D. Only the 4 watt motor operates.

'My conclusion, based on these calculations, is that neither motor is operating normally '
If the motors are in series, how can their currents be different?
The amps they are receiving are the same (.03125A). The "amp draw", if I am using the phrase correctly, is the amount of current a load needs to produce the watts it was designed for.
 

Thread Starter

Gailileo

Joined Jun 8, 2020
10
Watt's Law is more applicative.

Neither motor will run. Because you only have 120V potential, and both motors want the full potential, the smaller motor acts as a higher inductive impedance- which means it will get most of the voltage dropped across it (about 113V), and the larger motor which is a current sink, will only get < 8V or so. The smaller motor acts as the bottle neck and will limit the current to < 33mA. That is the reason neither will run.
Thank you. This gives me more insight into what's happening on the circuit.
 

Thread Starter

Gailileo

Joined Jun 8, 2020
10
@MrChips, Because those are the currents with them powered alone.

I think they will mark you wrong on this, though, in reality you are right.

They are going to claim the 4W motor is operating normally because it is getting nearly the voltage and current that it is specified for.

Bob
I am definitely worried about this. It's not going to be graded by a human who might give me partial credit for sound reasoning. The next question is exactly the same as this one but instead of motors, light bulbs are wired in series. I imagine the 60 watt bulb would be so dim it would not be detectable, but the 4 watt bulb would produce a noticeable amount of light.
 
Last edited:

Thread Starter

Gailileo

Joined Jun 8, 2020
10
Thank you Eric. How did you produce this sim? Is it a free app? I was looking for something to help me simulate these problems. I have a multimeter and both the light bulbs, but no receptacle for the 4 watt bulb :(
 

crutschow

Joined Mar 14, 2008
27,231
Neither motor will run. Because you only have 120V potential, and both motors want the full potential,
I disagree.
Since one motor is much smaller than the other, the small one will drop most of the voltage when they are in series, so the small one will run.
Whether that's "normally" or not is the question.
 

Thread Starter

Gailileo

Joined Jun 8, 2020
10
Yes, I forgot to ask about that. I am assuming that at 120v, the circuit is residential (1 hot and a neutral from a 240vac single phase supply). I think there are two versions of Ohm's Law/Watts Law, one for AC and one for DC. I have been wondering if I was expected to apply DC mathematics to AC circuits. If so, why? But no, I don't know if the motors are AC or DC.
 

Thread Starter

Gailileo

Joined Jun 8, 2020
10
I should probably mention that the test is for basic electricity, a kind of employment readiness certification. Perhaps certain details are being ignored (like in first semester physics, when we ignore drag).
 

ericgibbs

Joined Jan 29, 2010
12,942
hi G,
I would say unless clearly stated as AC at120Vrms, frequency 50Hz etc... regarding the motor type and supply, consider it as a DC supply.

As stated earlier a poorly worded question.
E
 
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