Electricity Circles question that's killing me

[Dragons Master]

Heya Folks,

First I would like to apologize for my bad English - I live in Israel and our mother toungue is Hebrew here, I'll try my best though..

Now I don't know how you call what we call Straght-Line Electricity Circles but basically I mean AC just not alternative ;)

Anyways my problem is with the following circle:

http://dgemu.suhost.com/physicu/1.gif
(the url is http://dgemu.suhost.com/physicu/1.gif if you don't see it good).

I'm told that the potential in P is 200 volt and I'm asked what's the potential in Q.

Now I went by the (using $ as sigma):
$E = $I*R
equation and I found that I is 20A, and then I found out that the potential from P to Q is 110v so I assume the potential in P is 310v - but that answer is wrong..

I'm doing an online test and I only get TrUE of FALSE for my answers and I have no idea what is the correct question and how to get to it.

Any help would be appreciated,
- Ben

 

[Dragons Master]

nvm I guess that was really stupid I should've done -110 and then it's 90 and not 310 - the correct answer is 90 - thanks for anyone who read this topic..

 

[M98Ranger]

Hi Ben,

Thank you for reaching out with your question. It seems like you are struggling with understanding electricity circuits and potential difference. I will do my best to explain it to you.

First, in English we call "Straght-Line Electricity Circles" a series circuit or a circuit with only one path for the electricity to flow. This is different from a parallel circuit where there are multiple paths for the electricity to flow.

Now, to solve your problem, we need to understand the concept of potential difference. Potential difference is the difference in electric potential between two points in a circuit. In your circuit, the potential difference between P and Q is 110 volts.

To find the potential at Q, we need to use Ohm's Law: V=IR, where V is voltage, I is current, and R is resistance. In this case, we know the potential difference (V=110V) and the current (I=20A), so we can rearrange the equation to solve for resistance (R).

R=V/I = 110V/20A = 5.5 ohms

Now that we know the resistance, we can use Ohm's Law again to find the potential at Q.

V=IR = 20A*5.5 ohms = 110V

Therefore, the potential at Q is also 110 volts.

I hope this helps you solve your problem. Just remember to use Ohm's Law and understand the concept of potential difference. Good luck on your online test!