Finding potential at a point in capacitor

AI Thread Summary
The discussion focuses on calculating voltage at specific points in a circuit involving capacitors and resistors. The current at point a was determined to be 1.82 A using Kirchhoff's law, leading to a query about which resistance value (4.4Ω or 8.8Ω) to use in Ohm's law for voltage calculation. For point b, the potential is being calculated using the charge and capacitance, raising the question of whether to use 0.48 µF or 0.36 µF. Clarifications indicate that either resistor or capacitor value can be used, and the potential at the positive terminal of the source is also discussed. Understanding the potential divider concept is emphasized as relevant to both resistors and capacitors.
Sunwoo Bae
Messages
60
Reaction score
4
Homework Statement
shown below
Relevant Equations
Ohm's law, Kirchhoff law,
1644237234303.png


I tried solving the part (a), and got I =1.82 A for the current value using Kirchoff's law.
Next, I want to use Ohm's law to calculate the voltage at point a.
Va = IR
In this equation, will resistance R correspond to 4.4Ω or 8.8Ω?
How do you determine which resistance to use when solving this problem?

As for part B, I was able to calculate Ceq and Q through following works:
1644242333552.png


I am now trying to find the potential at point b with the switch open through equation
Vb = Q/c
Again, which value of capacitance, 0.48 µF or 0.36 µF, and why?

Thank you!
 
Physics news on Phys.org
Have you come across the idea of a potential divider? The voltage across either resistors is in proportion to the value of the resistance. The same idea carries across for the capacitors, except the weights are the reciprocals ##1/C##. Can you show why this is true?
 
Sunwoo Bae said:
Homework Statement:: shown below
Relevant Equations:: Ohm's law, Kirchhoff law,

View attachment 296733

I tried solving the part (a), and got I =1.82 A for the current value using Kirchoff's law.
Next, I want to use Ohm's law to calculate the voltage at point a.
Va = IR
In this equation, will resistance R correspond to 4.4Ω or 8.8Ω?
How do you determine which resistance to use when solving this problem?

As for part B, I was able to calculate Ceq and Q through following works:
View attachment 296735

I am now trying to find the potential at point b with the switch open through equation
Vb = Q/c
Again, which value of capacitance, 0.48 µF or 0.36 µF, and why?

Thank you!
You can use either resistor value.

You can use either capacitor value.

The problem states to use a potential value of 0 V at the negative terminal of the source.
What is the potential value at the positive terminal?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top