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Engineering Circuit problem homework

  • Thread starter Miscing
  • Start date
16
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1. Homework Statement


http://i.imgur.com/ij4eHqM.png?1?5000

2. Homework Equations

KVL, KCL, Ohm's Law, I guess


3. The Attempt at a Solution

Tried node analysis but I get 4 equations in 5 variables, so no specific solution. I don't think the circuit can be simplified either. A friend suggested mesh analysis but I don't think that works here because we have loops but they aren't meshes. I'd like just a hint in the right direction please, perhaps what approach I should try, I'd still like to do most of it myself. Thanks :)
 

gneill

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Consider that no current can flow across the open circuit where Vo is measured. Thus you have a current divider situation for the two branches, and you should be able to determine the currents and thus the individual potentials for Va and Vb:
attachment.php?attachmentid=57295&stc=1&d=1364705207.gif
 

Attachments

16
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Consider that no current can flow across the open circuit where Vo is measured. Thus you have a current divider situation for the two branches, and you should be able to determine the currents and thus the individual potentials for Va and Vb:
attachment.php?attachmentid=57295&stc=1&d=1364705207.gif
Thank you, I managed to get the right answer with that help. Physically, what happens in terms of the resistors at the open terminals + and - of v0 once the current is turned on?

I imagine that initially (for a brief amount of time) we have current flowing one-way into both open terminals, where charge builds up until the terminals reach the same potential as at Va and Vb, respectively. At that point current stops flowing, so there's no drop across the terminal resistors (2k and 4k) and the terminal potentials are simply equal to Va and Vb? Is this right?
 

gneill

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20,486
2,612
Thank you, I managed to get the right answer with that help. Physically, what happens in terms of the resistors at the open terminals + and - of v0 once the current is turned on?

I imagine that initially (for a brief amount of time) we have current flowing one-way into both open terminals, where charge builds up until the terminals reach the same potential as at Va and Vb, respectively. At that point current stops flowing, so there's no drop across the terminal resistors (2k and 4k) and the terminal potentials are simply equal to Va and Vb? Is this right?
Yes, that's it for 'real' components. Ideal components have no physical size, so the transient time would be infinitesimally short.
 
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Yes, that's it for 'real' components. Ideal components have no physical size, so the transient time would be infinitesimally short.
You're a champion, thanks for your help :)
 
16
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Yes, that's it for 'real' components. Ideal components have no physical size, so the transient time would be infinitesimally short.
Actually one more question; what would be the final potential inside each resistor in a realistic and ideal circuit? Are they just at the same potential as the connecting wire?
 

gneill

Mentor
20,486
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Actually one more question; what would be the final potential inside each resistor in a realistic and ideal circuit? Are they just at the same potential as the connecting wire?
Yup. Same potential. No current.
 

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