Analog Electronics Magic Mirror

shipeng

Analog Electronics Magic Mirror [Copy link]

 

I have already had several fake hardware engineer friends exposed by this question.

It is known that the initial voltages of capacitors C1, C2, and C3 are all 5V before K1 is closed. What are the potentials at points C2(2) and C3(2) after K1 is closed?

Additional content (2022-11-1 14:13): The voltage directions of the three capacitors are positive at the top and negative at the bottom.

gmchen

The voltage is not marked

maychang

It is OK if the voltage direction of capacitors C2 and C3 is not marked, after all, one end of each is grounded. But it is not OK if the voltage direction of both ends of C1 is not marked.

shipeng

maychang posted on 2022-10-26 16:11 It is OK if the voltage direction of capacitors C2 and C3 is not marked, after all, one end of each is grounded. It is not OK if the voltage direction of both ends of C1 is not marked.

My mistake , the voltage direction of the three capacitors is positive at the top and negative at the bottom.

zhangdaoyu

Wait for the answer and see if my guess is right.

gmchen

It is understandable that analog engineers make mistakes on this question. Analog engineers are usually familiar with circuit theory, such as Kirchhoff's laws. However, this circuit is more physics-oriented, starting from charge, energy conservation, etc.

shipeng

gmchen posted on 2022-10-26 17:34 It is understandable that analog engineers make mistakes on this question. Analog engineers are usually familiar with circuit theory, such as Kirchhoff's laws. But this circuit is more...

According to this, this topic should belong to the category of high school physics.

maychang

Teacher gmchen is right, this problem should start with the conservation of charge and energy.

When switch K is not closed, it is obvious that the potential on the left side of K is higher than the potential on the right side, and the left side is positive and the right side is negative. Therefore, after switch K is closed, the current in K must be from left to right. In other words, after K is closed, capacitor C3 is charged, capacitors C2 and C1 are discharged, and the currents in capacitors C1, C2, and C3 are equal at any time.

maychang

The question requires the voltage across capacitors C2 and C3 when the circuit reaches stability after K is closed. In other words, when will the discharge of C1C2 and the charging of C3 end?

The answer is: when the total energy of the three capacitors is the smallest.

maychang

shipeng posted on 2022-10-26 17:51 According to this, this topic should belong to the category of high school physics

[According to this statement, this topic should belong to the category of high school physics]

High school physics doesn’t go into this much depth.

shipeng

maychang posted on 2022-10-26 18:11 [According to this statement, this topic should fall into the category of high school physics] High school physics cannot be discussed in such depth.

What I mean is that solving this problem will not go beyond the knowledge scope of high school physics?

maychang

shipeng posted on 2022-10-26 19:09 maychang posted on 2022-10-26 18:11 [According to this, this topic should belong to the category of high school physics] High school physics cannot be discussed in such depth. ...

[I mean, answering this question will not exceed the knowledge of high school physics? ]

The system is stable when the total energy is at a minimum. I did learn this in high school, but it was only mentioned briefly.

maychang

shipeng posted on 2022-10-26 19:09 maychang posted on 2022-10-26 18:11 [According to this, this topic should belong to the category of high school physics] High school physics cannot be discussed in such depth. ...

[I mean, answering this question will not exceed the knowledge of high school physics? ]

Sometimes, very important discoveries require only very simple knowledge. Hipparchus of ancient Greece calculated the distance from the Earth to the Moon, and the value measured today is almost the same. The knowledge used is no more than the plane geometry of today's junior high school.

davidzhu210

shipeng posted on 2022-10-26 16:48 Two big guys, my mistake, the voltage direction of the three capacitors is positive at the top and negative at the bottom

Left side = 4.58V, right side = 5.42V.

se7ens

davidzhu210 posted on 2022-10-26 21:10 Left side = 4.58V, right side = 5.42V.

The boy is very brave, I like you

gmchen

Left side = 4.58V, right side = 5.42V.

At first I thought it was simulation, but after a closer look I realized it wasn't.

The exact answer is: 5-(5/12) and 5+(5/12)

hustjtj0806

This is a hard question to answer. Not all college students who have studied circuits can solve it. The exact answer is: 5-(5/12) and 5+(5/12)

shipeng

The above people are all real model electronics masters who will compensate you ten times for the fake ones @

davidzhu210

davidzhu210 posted on 2022-10-26 21:10 Left side = 4.58V, right side = 5.42V.

Here is the solution process:

shipeng

Since the answer has been given by several analog electronics experts, let me talk about my solution: After K1 is closed, the current I flowing through the three capacitors in the loop is always equal. When I=0 after time t, the voltages of the three capacitors no longer change. At this time, the charge changes of the three capacitors are also equal and are dQ = the integral of I in time t. According to the capacitance formula U = Q/C, the voltage changes of the three capacitors dV1, dV2, and dV3 are:

C1 voltage change: dV1 = -dQ/1uF = -dQ/(10E-6);

C2 voltage change: dV2 = -dQ/10uF = -dQ/(10E-5);

C3 voltage change: dV3 = dQ/10uF = dQ/(10E-5);

According to the initial voltage of the three capacitors being 5V, the final voltage of the three capacitors is:

Final voltage of C1: 5 + dV1=5 - dQ/(10E-6);

Final voltage of C2: 5 + dV2=5 - dQ/(10E-5);

Final voltage of C3: 5 + dV3=5 + dQ/(10E-5);

According to the voltage relationship of the three capacitors after the voltage stabilizes, C1+C2=C3, we get:

5 - dQ/(10E-6) + 5 - dQ/(10E-5) = 5 + dQ/(10E-5);

To simplify the calculation, let x = dQ/(10E-5) and substitute it into the above formula:

5 - 10x + 5 - x = 5 + x;

Solve for x=5/12; Final result:

C1 voltage: 5 - 10*5/12;

C2 voltage: 5 - 5/12;

C3 voltage: 5 + 5/12;