Need help with this academic problem

[sidk47]

Hi guys, being a perfectionist, I would like you to tell me how to solve this problem, it is beyond my abilities to solve it by myself.

I would most appreciate any hints.

 

[dorke]

What is "academic" about it?
A high-school grade question.

Hint:
In a series capacitor circuit ,the lower a capacitor the higher the voltage on it,
in a linear fashion regarding the values of the capacitances .

You are given the voltage on the 500uF ,so you can easily find the value of the voltage on the 100uF cap.
Then use KVL to find the voltage on the 1200uF and find Cx.

 

[hevans1944]

Using Q = CV calculate the charge stored in the 500 μF capacitor that has 16 V potential across it. This same charge will also be stored in the 100 μF capacitor and in the parallel-connected capacitance 1200 μF and Cx. Using V = Q/C find the voltage across the 100 μF capacitor. Add this to the 16 V across the 500 μF capacitor and subtract the sum from the 100 V charging potential of the battery. The difference is the voltage across the 1200 μF capacitor and Cx. Using C = Q/V calculate the capacitance of the paralleled capacitors. Using 1/C = 1/Cx + 1/1200 μF solve for Cx. The 40 μF capacitor is irrelevant.

Is this a homework problem?

 

[sidk47]

I'm just trying to study electronics by myself. Thanks for your replies. This is from a text book.

 

[hevans1944]

I'm just trying to study electronics by myself. Thanks for your replies. This is from a text book.

The key to solving this problem is realizing that, since the capacitors to the right of the voltage source are in series, they all have the same charge, Q = CV. The rest is, as you said, academic. Thank you for clarifying your intent.

I noticed that your previous post was in 2013. Don't be a stranger here! We love to help beginners. What do you do in India besides playing video games? We are "construction oriented" here (mostly). Do you have any interest in building DIY projects while you learn?

 

[sidk47]

Hey Hevans,
I managed to find out through calculations based on your advice the proper answer of 800 micro farads.
I don't play video games because my laptop can't really support games. I do try to play guitar every now and then but since I started studying circuit analysis, it's been collecting dust.

I would love to build something, but I fear my lack of knowledge in electronics would be limiting.

 

[hevans1944]

... I would love to build something, but I fear my lack of knowledge in electronics would be limiting.

Well of course it would be limiting, but that's no reason to delay starting the journey.

I am a proponent of "hands on" learning, where experiment and theory go together synergistically to increase knowledge. Start slow and small and build from there. The first thing you need is one of those ubiquitous solderless breadboards... the ones with lots of connected holes you can stick components into to temporarily "wire up" a circuit. Then put together a kit of electronic components and purchase a "wall wart" AC-to-DC low-voltage power supply providing 6 V to 12 V at a few hundred milliamperes of current capability. This will keep you from going broke purchasing batteries. I don't know what your budget is, but for less than US$100 you can buy or build a pretty nice beginners electronics kit.

Most beginners start with LED (light emitting diode) experiments. You should purchase a variety kit with a few dozen LEDs to "play" with since, no doubt, a few will "go up in smoke" as you learn how to use them. Check out the Resources Forum here in Electronics Point for tips. And come back here often with questions when something comes up that you don't understand (yet).

And I wouldn't let that guitar gather too much dust. I don't play guitar, but I get a lot of inspiration from music. There is more to life than electronics.

Hop

 

[hevans1944]

Using 1/C = 1/Cx + 1/1200 μF solve for Cx. The 40 μF capacitor is irrelevant.

Oops! That's the equation for capacitors in series. Since Cx and 1200 μF are in parallel, it's simple addition: C = Cx + 1200 μF. But I see you got the correct answer for Cx = 800 μF despite my "help"