Can someone explain why the voltage here is zero?

AI Thread Summary
The discussion revolves around a circuit analysis where the voltage across an 80k resistor is found to be zero, leading to confusion about the absence of current through it. The user applied Kirchhoff's laws but misinterpreted the circuit's configuration, as there is no current flow when the resistor's leads are shorted. Additionally, a separate question about capacitor energy calculations was raised, but it was noted that the original circuit did not include a capacitor. The conversation highlights the importance of providing complete circuit details for accurate analysis. Overall, clarity in circuit representation and understanding of electrical principles is essential for resolving such queries.
Jonas E
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Homework Statement


I have the circuit in the picture below, and using KCL and KVL I found that the voltage over the 80k resistor is 0. However, I don't understand why this is correct. Can someone explain why there is no current through it?

Homework Equations


None

The Attempt at a Solution


I used the node voltage method to get 25ib + 1/(16k)-i0 = 0. After that I used KVL through the 80k resistor, 16k resistor and 1V source, which gave me 80k * ib - 1 + 1 = 0. So ib = 0.
circuit2.png
 
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Jonas E said:
Can someone explain why there is no current through it?
Imagine you took a resistor and wired its leads together. Why would there be any current flowing in it?

Jonas E said:
After that I used KVL through the 80k resistor, 16k resistor and 1V source, which gave me 80k * ib - 1 + 1 = 0. So ib = 0.
I guess you mean you used KVL for the two loops and "added" the equations together. There's no loop that takes you through each of those components.

You could have stopped at 80k*ib = 0.
 
milesyoung said:
Imagine you took a resistor and wired its leads together. Why would there be any current flowing in it?I guess you mean you used KVL for the two loops and "added" the equations together. There's no loop that takes you through each of those components.

You could have stopped at 80k*ib = 0.

Thanks for the reply. Could you also tell if the energy in a capacitor at time t equals half of its capacitance times its voltage at time t? So w(t) = 1/2 * C * v(t) ?
 
milesyoung said:

Sorry, I forgot to square the voltage. But is this correct then: w(t) = 1/2 * C * [v(t)] ^2 ? I couldn't understand the wikipedia article. You see, I used this formula, but it doesn't give me the right answer
 
Jonas E said:
You see, I used this formula, but it doesn't give me the right answer
Answer for what? There's no capacitor in your schematic.
 
milesyoung said:
Answer for what? There's no capacitor in your schematic.

EDIT: Ok, sorry again, but I just found out what I did wrong: When taking the square root, I forgot that there would be two answers. The negative one turned out to be the correct one. Thanks for helping me out figure out my original problem!

Sorry, the original post was just part of the problem.

I used that schematic to find Rth = 16k ohms and Vth = -19.8 V. I used this to make an RC circuit (C = 0.25 * 10^-6 F), which I solved to find that the capacitor's voltage is given by v(t) = [ 19.8 * (e^(-250t) - 1) ]. Everything up until this point was correct according to the solutions manual. The final question was to find how long it takes for the energy stored in the capacitor to reach 36 % of its final value. This is where I set up the equation w(t) = 0.36 * w(infinity). But for some reason this doesn't give me the correct answer. I used the following values:

w(t) = 0.5 * (0.25 *10^-6) * [ 19.8 * (e^(-250t) - 1) ]^2

w(infinity) = 0.5 * (0.25 *10^-6) * [ 19.8 * ( 0 - 1) ]^2 = 4.9005 * 10^-5 J

Could you help me find the mistake in my reasoning?
 
Last edited:
Jonas E. You are introducing a second topic unrelated to the original post. For this RC network you must start a new thread.

I'm not satisfied that your original circuit has been correctly dispensed with yet. I think you should post the whole exercise, not just a piece, so that we can see the full picture, rather than dealing with your uncertain interpretation of a piece out of context.
 

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