Series Resistor Circuit Voltage (2 Power Sources)

[osmanjdt]

if we replace the resistor with led,are the led will light on?

 

[GaryS]

short answer NO you don't have a high enough voltage to push the necessary current to light an LED mot led' require 1.5 v forward bias and about 5ma current with this circuit you could only get about .003 ma and that would be id you connected the led at point B to ground if you put it in series between R2 and R3 you would have even less.
while current would flow in either case it would not be enough that you would see it.

 

[jrwb4gbm]

Where in the original post was point "B" referenced to ground, or referenced to any point for that matter? I think the correct answer needs some reference point but I fail to see it stated.

 

[daba]

The question was "Calculate the voltage at point B in the given circuit."

That is impossible to answer, because the question does not specify what reference for the voltage should be taken.

To be pedantic, it would have been better if the question had been worded "Calculate the potential at point B in the given circuit.". At least that would infer a voltage above ground potential....

I personally hate questions that are given in exams or tests that do not meet one very simple criteria.....

"Is the question ambiguous" - of course in this case it is. You have to make an assumption that the questioner means "with respect to ground potential", but he didn't say so.

Anyway, enough ramblings about the question, the answer is 10.48V "above ground potential"

It is easy to calculate...

Total Potential Difference (wrt ground) across the four resistors = 3V

Total Resistance = 47K + 27K + 56K + 20K = 150K

Current through all resistors = V/R = 3/150000 = 0.00002 A

Volt drop across the 47K resistor = 0.94V
Volt drop across the 27K resistor = 0.54V
Volt drop across the 56K resistor = 1.12V
Volt drop across the 20K resistor = 0.40V

Either way you look at it the potential, above ground, at point B is...

9 + 0.94 + 0.54 = 10.48

or,

12 - 0.40 - 1.12 = 10.48

 

[GaryS]

your math is spot on as far as it goes in the top section and you state that the total voltage across the resisters is 3 v 12V - 9V = 3V that is correct.
but than in the lower statement 9 + 0.94 + 0.54 = 10.48 where did you get the extra 9 V you only have 3V left to work with.
the polarity of the 2 sources is subtractive 12 - 9 = 3 you can't just plug in an extra 9 volts. it has to come from someplace. and you haven't accounted for it. is there a magic power source that is not shown. all voltages and current must be accouted for.

 

[GaryS]

here I a link I think is just interesting don't take it seriously

http://thecompassadjuster.com/education/resources/smoke-theory-of-electricity.html

if I want a good chuckle I read this

 

[daba]

your math is spot on as far as it goes in the top section and you state that the total voltage across the resisters is 3 v 12V - 9V = 3V that is correct.
but than in the lower statement 9 + 0.94 + 0.54 = 10.48 where did you get the extra 9 V you only have 3V left to work with.
the polarity of the 2 sources is subtractive 12 - 9 = 3 you can't just plug in an extra 9 volts. it has to come from someplace. and you haven't accounted for it. is there a magic power source that is not shown. all voltages and current must be accouted for.

Of course I accounted for the "voltages"...

"where did you get the extra 9 V you only have 3V left to work with."

that 9V came from the 9V supply to the left of R1 and R2. there is a 9V potential difference between ground and the connection to the four resistor chain.

perhaps it would help if i stated the potential difference, with respect to ground potential, at all points in the circuit.

Remember that the Potential Difference (PD) across all four resistors is 3V (12V-9V)

a. Positive of the 9V supply, obviously 9.00V
b. junction R1/R2, point A = 9.94V
c. junction R2/R3, point B = 10.48V
d. junction R3/R4, point C = 11.60V
e. Positive of the 12V supply, obviously 12.00V

Volt drop a-b = 0.94V
Volt drop b-c = 0.54V
Volt drop c-d = 1.12V
Volt drop d-e = 0.40V

Total Volt drop = 3.00V

Can I simplify it more ??

 

[GaryS]

the 9V source on the left was used to reduce the 12V to the 3 again 12 -9 = 3
now you have taken away the 9 V source it can't just pop up again it needs to be there
I wish my back account worked that way
take money out and it just refills I don't have to put anything in it just happens you cant get something for nothing
if you can't understand that then I can't explain it. go back to electricity 101 the first thing you should have leraned

 

[daba]

the 9V source on the left was used to reduce the 12V to the 3 again 12 -9 = 3
now you have taken away the 9 V source it can't just pop up again it needs to be there
I wish my back account worked that way
take money out and it just refills I don't have to put anything in it just happens you cant get something for nothing
if you can't understand that then I can't explain it. go back to electricity 101 the first thing you should have leraned

it's a relative thing, perhaps your bank needs to speak to my bank - together they could lift our balances relative to a rich persons.....

please explain what you are having trouble understanding...

 

[Lare]

the 9V source on the left was used to reduce the 12V to the 3 again 12 -9 = 3
now you have taken away the 9 V source it can't just pop up again it needs to be there
I wish my back account worked that way
take money out and it just refills I don't have to put anything in it just happens you cant get something for nothing
if you can't understand that then I can't explain it. go back to electricity 101 the first thing you should have leraned

So Gary you mean that if you take voltage meter and measure voltage between Ground and 9V source plus terminal then voltage show 0Volt?

And again if you measure between ground and plus terminal of 12V source, then again voltage is 0Volts.

Sorry Gary, it don't work that way. Sources negetive terminals are together and between sources there is 3Volts. But 9 and 12 volts are stille there if you measure voltages to ground. (source negative terminals)
If you can't see it, then it is time to go to buy couple resistors and measure with real voltage meter.

 

[osmanjdt]

Voltage drop and Voltage respect to

 

[ebolbol]

The question was "Calculate the voltage at point B in the given circuit."

That is impossible to answer, because the question does not specify what reference for the voltage should be taken.

To be pedantic, it would have been better if the question had been worded "Calculate the potential at point B in the given circuit.". At least that would infer a voltage above ground potential....

I personally hate questions that are given in exams or tests that do not meet one very simple criteria.....

"Is the question ambiguous" - of course in this case it is. You have to make an assumption that the questioner means "with respect to ground potential", but he didn't say so.

Anyway, enough ramblings about the question, the answer is 10.48V "above ground potential"

It is easy to calculate...

Total Potential Difference (wrt ground) across the four resistors = 3V

Total Resistance = 47K + 27K + 56K + 20K = 150K

Current through all resistors = V/R = 3/150000 = 0.00002 A

Volt drop across the 47K resistor = 0.94V
Volt drop across the 27K resistor = 0.54V
Volt drop across the 56K resistor = 1.12V
Volt drop across the 20K resistor = 0.40V

Either way you look at it the potential, above ground, at point B is...

9 + 0.94 + 0.54 = 10.48

or,

12 - 0.40 - 1.12 = 10.48

Thanks to everybody... very educational!

@daba, This post helped me immensely! It really broke it down.

Two questions: Does your breakdown assume the current is flowing one way or the other? Or is that determined by laws of polarity in the circuit just how it's sown?

 

[Doug-P]

 

[daba]

Thanks to everybody... very educational!

@daba, This post helped me immensely! It really broke it down.

Two questions: Does your breakdown assume the current is flowing one way or the other? Or is that determined by laws of polarity in the circuit just how it's sown?

It doesn't matter which way the current is flowing in determining the volt-drop across each resistor.

What it does do is determine whether to add or subtract that volt drop to determine the potential at each node in the circuit.

You have to just look at the circuit to know which way the voltages are distributed - in this case it is obvious that the potential at the positive of the 12V supply is higher than the potential at the positive of the 12V supply. That means that voltages will increase in the resistor network from the 9V to the 12V.

Another way to look at it is to assume a "left-to-right" current flow, subtracting the volt-drops of the resistors at each node. If you do this, the current in the circuit (left-to-right) will be -0.00002 A, and the volt-drop across R1 will be -0.94V .... 9V - -0.94V is 9.94V.

The converse applies also, if you assume current flow is "right-to-left", which is true in this case

I hope that answers both questions

 

[Mickey]

It would be nice to see some student out there with access to a bread board and hardware to actually put this together and measure it. Eliminate all doubt.