Potential of Point A to Point B in Difficult RC Circuit

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Homework Help Overview

The discussion revolves around determining the potential difference between two points in a circuit, specifically in a scenario described as a difficult RC circuit, although it is noted that there are no capacitors present. Participants are exploring the implications of circuit connections and current flow in the context of resistors and a voltage source.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss methods to find the potential difference and current using various circuit analysis techniques, including Millman's Theorem and Kirchhoff's Current Law (KCL). There are questions about how to simplify the circuit and the implications of having an open circuit between points a and b.

Discussion Status

Some participants have offered hints and suggestions for approaching the problem, such as using KCL and recognizing the lack of current in certain branches of the circuit. There is an acknowledgment of confusion regarding the circuit's configuration and the terminology used.

Contextual Notes

There is a noted misunderstanding regarding the classification of the circuit as an RC circuit, as it lacks capacitors. Participants are also grappling with the implications of the circuit being open at certain points and the resulting current flow.

AGGENGR
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Warrning: Posting template must be used for homework questions
Find the potential of point a with respect to point b in the figure (Fighttps://s.yimg.com/hd/answers/i/1ffe0f8aeaf145679cde38f7812b1c7f_A.png?a=answers&mr=0&x=1413484693&s=0943630e47eb74ae023909f7d39f74e3 ure 1) .

If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 10.0V battery.

V=IR, Req=R1+R2, 1/Re = 1/R1+1/R2

So basically added up the resistors in series on the side getting 4 ohm and 3 ohm respectively. I am stuck because the a-b is disconnected which is something i haven't seen until this problem. also i can't seem to simplify the circuit into a single "line" if you know what i mean.
 
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AGGENGR said:
Find the potential of point a with respect to point b in the figure (Fighttps://s.yimg.com/hd/answers/i/1ffe0f8aeaf145679cde38f7812b1c7f_A.png?a=answers&mr=0&x=1413484693&s=0943630e47eb74ae023909f7d39f74e3 ure 1) .

If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 10.0V battery.

V=IR, Req=R1+R2, 1/Re = 1/R1+1/R2

So basically added up the resistors in series on the side getting 4 ohm and 3 ohm respectively. I am stuck because the a-b is disconnected which is something i haven't seen until this problem. also i can't seem to simplify the circuit into a single "line" if you know what i mean.

Or just write the KCL equations for the circuit. I would put ground at the middle right of the circuit, between the 3 resistors. For the open-circuit case, there is no current flowing in the middle branch (obviously).

Can you show us the KCL equations?
 
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Incidentally, this is not an RC circuit because there are no capacitors.

When you do have to find the transient voltages and currents in a RC circuit in DC by using differential equations, you will then realize a new level of "difficult".
Finding voltages and currents in an RC circuit in AC has challenges as well, but is not as difficult as the dreaded DC transient RC circuits.
Both of these will be coming to you soon assuming you are in an engineering program.

A hint to solve the problem...there is ZERO current thru that middle branch as suggested above.
 
berkeman said:
Or just write the KCL equations for the circuit. I would put ground at the middle right of the circuit, between the 3 resistors. For the open-circuit case, there is no current flowing in the middle branch (obviously).

KCL?
 
AGGENGR said:
KCL?

Kirchoff's Current Law (KCL). I'm guessing by your response that you haven't covered that yet in class. It's a very easy and useful technique to learn -- maybe check it out at Wikipedia, and you will be ahead of the rest of your class! :-)
 
Ahh i see now. Thanks
 

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