Finding Potential Difference between two points homework

[harshakantha]

R1 = 120ohms
R2 = 820ohms
R3 = 2200ohms
E1 = 15v
E2 = 9v

What is the potential difference between points "B" and "C"?

(I've attached the circuit diagram)

 

[ehild]

What have you learned about electric current and electric potential so far?

ehild

 

[Femme_physics]

Hmm. I'm not a hmework helper and not entirely sure myself, but I believe E2 won't play a factor since its current leads nowhere. Is that right, helpers?

 

[harshakantha]

I know the basic electronic theories, Ohms law, Kirchhoff's Theorem and etc,

 

[ehild]

I know the basic electronic theories, Ohms law, Kirchhoff's Theorem and etc,

You should write them under the title "Relevant equations".
And then show some attempt to solve the problem.
As a first hint: Does any current flow through resistor R2?

ehild

 

[harshakantha]

I'm sorry this will not happen again, I'm new to this forum :)
ok I think there is no current flowing through the R2

 

[ehild]

Hmm. I'm not a hmework helper and not entirely sure myself, but I believe E2 won't play a factor since its current leads nowhere. Is that right, helpers?

No, it is wrong, it influences the potential difference between C and B.
And "its current" has no sense. It is a voltage source, characterized by its emf (electromotive force).

ehild

 

[ehild]

ok I think there is no current flowing through the R2

Very well, so what do you know about the potential drop across R2?

ehild

 

[harshakantha]

before that I think we should calculate the current through the whole circuit

 

[harshakantha]

ya it's wrong, we need E2 when we calculate potential difference of point "B"

 

[ehild]

No, you just said that zero current flows through R2. And you know Ohm's law. What does it say? ehild

 

[harshakantha]

according to the method I've studied in school, we should follow that step, if I'm wrong please show me how to get the potential difference of point B and C

 

[harshakantha]

Oh I've found a old post from this forum, could you please check this, it has the same steps that I've mentioned
https://www.physicsforums.com/showthread.php?t=158460

 

[ehild]

ya it's wrong, we need E2 when we calculate potential difference of point "B"

A point has a potential, and potential difference is established between two points. So it is "potential at B, U_B" and "potential difference between B and C, U_BC =U_C-U_B".
The zero of the potential can be assigned to any point of a circuit. So potential of B means potential with respect to the zero point.

ehild

 

[harshakantha]

Thanks that is another problem i had on my head :),it's mean we can assign zero point to any point, thanks for telling me that before i ask, so you still didn't tell me how to fine 'Uc" or "Ub"

 

[I like Serena]

Hi harshakantha! Welcome to PF!

Just a friendly bit of (unasked) advice.
Ehild is helping you.
See the "Homework Helper" medal?
That means ehild know this stuff quite well and knows how to help people!
As far as I'm concerned ehild has already shown remarkable patience...

Cheers!

 

[harshakantha]

ya Serena, he knows how to help people very well :-)

 

[ehild]

Harshakantha, I want to help you, so answer my questions, please. I need to know what your problems are.
What does Ohm's law state?

What is the potential difference between A and B? We will assign the zero of potential to point A. ehild

 

[harshakantha]

there is substantially constant ratio between the applied potential difference and the current flowing through a resistor, this relation ship known as Ohm's law.
can u give me a hint to get potential of point B :)

 

[harshakantha]

ehild now I'm going to leave computer, I hope you will leave a better answer for my question when next time i log on to The PF, till then bye... :)

 

[ehild]

As others might be interested in this thread, I continue.

According to Ohm's law, zero current means zero potential difference across a resistor. So the potential at B is the same as at the positive terminal of the battery E2.

There is current in the closed loop at the LHS of the drawing: KVL can be applied to get it. The potential of B with respect to A is obtained from Ohm's law again. If you know the potential both at B and C the potential difference is just a subtraction.

ehild

 

[harshakantha]

Hi ehild, I solve the question,

Total Resistance = R1+R3 = (120+2200)ohms
consider current through Resister R1 and R3 is "I"
then I=E1/(R1+R3) = 15/(120+2200) = 0.0065A

At point "C" let's assume there to be zero potential

1) by going over R3 we lose -14.3V (0.0065*R3)
2) The E2 power supply adds 9V, leaving us with -5.3V (-14.3V + 9V)
3) no current flows through R2
4) therefor the potential different between point "B" and "C" is 5.3V

is this correct??

 

[ehild]

Yes, but add for clarity that B is negative with respect to C.

ehild

 

[harshakantha]

you mean to the final answer?? can you show me how to write the final answer :)

 

[ehild]

you mean to the final answer?? can you show me how to write the final answer :)

The answer depends how your teacher defined the potential difference between two points, say B and C. It might be Ub-Uc or Uc-Ub. But it is always correct if you give the absolute value and indicate which point is more positive or more negative than the other one.

ehild

 

[harshakantha]

Thank you so much ehild, I appreciate your help on this matter very much, once again thanks for helping me

 

[Femme_physics]

I took on solving it myself and got 5.22V (I realized drawing "I2" was a mistake)

http://img32.imageshack.us/img32/3595/voltagedrop.jpg

 

[ehild]

You are welcome.. And one more little hint: When doing calculations, do not round the data too early, too much. You rounded the current to two significant digits so the error of the result is 0.1 V. During calculations, keep at least one more significant digits as needed in the final result.

 

[ehild]

I took on solving it myself and got 5.22V (I realized drawing "I2" was a mistake)

Good job! Congratulations!

Just a little hint, the same I said to harshakantha: Do not round off too much during the calculations. That current was 0.00647 with 3 digits. Both of you rounded it to 0.0065 which caused an error of 0.066 V in the voltage in harshakantha's calculation as he multiplied it with the bigger resistance. You used the smaller resistance, so your error is one magnitude less, 0.004 V.

ehild

 

[Femme_physics]

Thank you ehild, so encouraging! :)

You're right I had a feeling I was rounding too much.