Can someone explain why the voltage here is zero?

Click For Summary
SUMMARY

The discussion centers on a circuit analysis problem involving an 80k ohm resistor where the voltage is calculated to be zero. The user applied Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) but struggled to understand why there is no current through the resistor. The consensus is that if the resistor's leads are shorted, no current flows through it. Additionally, the user inquired about the energy stored in a capacitor, confirming the correct formula as w(t) = 1/2 * C * [v(t)]^2, while also addressing an error in their calculations related to the capacitor's voltage.

PREREQUISITES
  • Understanding of Kirchhoff's Current Law (KCL)
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Basic knowledge of resistor and capacitor circuits
  • Familiarity with the formula for energy stored in a capacitor
NEXT STEPS
  • Study the implications of short-circuiting a resistor in a circuit
  • Learn about the node voltage method for circuit analysis
  • Explore the derivation and applications of the energy formula for capacitors
  • Investigate common mistakes in circuit analysis and how to avoid them
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing resistor-capacitor (RC) circuits will benefit from this discussion.

Jonas E
Messages
15
Reaction score
0

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
 
Physics news on Phys.org
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.
 

Similar threads

Replies
15
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
0
Views
2K