Potential Change in a Circuit Diagram

[daimoku]

[SOLVED] Potential Change in a Circuit Diagram

Homework Statement

http://personalpages.tds.net/~locowise/test/28_29.gif

If the potential at point P is 400 V, what is the potential at point Q?

Homework Equations

Ohm's Law: deltaV=iR
Probably the fact that the sum of changes in potential in a loop of any circuit must be zero.

The Attempt at a Solution

Not real sure on this problem. I tried this:
150V - 2.0i - 3.0i - 50V = 0V and found i to be 20A.

I'm kind of puzzled and can't seem to relate the material in the textbook to this problem. Where does the 400V potential come into play? Thanks for your help!

 

[daimoku]

I'm still not really sure this is the proper way to approach the problem but apparently the solution is correct. Here's what I did:

Used 20A from above.

400V - 50V - 3.0ohms * 20A = 290V

 

[Moetasim]

I understand your confusion with this problem. It appears that the potential at point P is not directly related to the potential at point Q. However, it is important to note that the potential difference between any two points in a circuit is determined by the sum of the changes in potential along the path connecting those two points. In this case, the potential difference between points P and Q is equal to the potential difference between points P and A, plus the potential difference between points A and B, plus the potential difference between points B and C, plus the potential difference between points C and Q.

Using Ohm's Law, we can calculate the potential difference between points P and A as 150 V, and the potential difference between points A and B as 2.0i, where i is the current flowing through the circuit. Similarly, the potential difference between points B and C can be calculated as 3.0i, and the potential difference between points C and Q can be calculated as 50 V.

Therefore, the potential difference between points P and Q can be written as:

deltaV(PQ) = 150 V + 2.0i + 3.0i + 50 V

Since we know that the potential at point P is 400 V, we can set this equal to the potential difference between points P and Q and solve for i:

400 V = 150 V + 2.0i + 3.0i + 50 V

Simplifying, we get:

400 V = 200 V + 5.0i

Solving for i, we get i = 40 A.

Therefore, the potential at point Q is equal to the potential at point C, which can be calculated as:

deltaV(PQ) = 50 V + 3.0(40 A) = 170 V

In conclusion, the potential at point Q is 170 V. I hope this helps clarify the problem for you.