Capacitors - Voltage

1. Dec 20, 2008

FiskiranZeka

Capacitors - Voltage [ Solved ]

Can you explain me the steps of solving this problem and write the process/calculation ?
Thank you very much.

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Last edited: Dec 20, 2008
2. Dec 20, 2008

Hootenanny

Staff Emeritus
Welcome to PF,

That's not the way it works here. We will help you through your homework, but we won't do it for you.

What have you tried thus far?

3. Dec 20, 2008

FiskiranZeka

You are asking how i treid to solve this right ?
So, i explain;

q=V.C

3C + C --> 4C
q = 4C. V1

On the Upper Wire, there is 10 Voltage
On the Upper Wire, there is Tottal 4C + 1C --> 4/5 C
On the Upper Wire,
Q= 4C.V1 or 4/5C.Vtotal
( Vtotal = 10 Volt )
So, V1= 2 Volt,
V1/V2 = 1/4
So V2=8 Volt,
On the downside Wire,
above Cx is 8 Voltage,
So above 2C must be 2 Voltage, because 2+8 = 10 Volt ( Total Voltage )
q on the downside wire is;
q= V2.Cx or q = 2C. ( Voltage of 2C )
So, V2.Cx = 2C. ( Voltage of 2C )
V2 = 8 Volt,
8.Cx = 2C . 2Volt
Cx = 1/2 is the wrong answer, what i have found...

4. Dec 20, 2008

Hootenanny

Staff Emeritus
Everything looks good up until this point. Your next statement is incorrect:
Picking up from your previous like you have:

$$8C_x = 4C$$

Hence,

$$C_x = \frac{1}{2}{\color{red}\bold{C}}$$

You still need to eliminate the unknown Capacitance (C). Do you follow?

5. Dec 20, 2008

FiskiranZeka

Yeah sorry,
my calculations say -->
Cx = 1/2 C

But correct answer is Cx = 8C

As the answer has Unknown Capacitance (C) too...
We dont need to eliminate C...
No problem... Just need to find how many C is equal to Cx.

6. Dec 20, 2008

Hootenanny

Staff Emeritus
After discussing this problem with another Homework Helper, we both agree that the correct answer is Cx = 1/2*C as you have.

Wherever you got the 'correct' answer of Cx = 8*C is incorrect.

7. Dec 20, 2008

FiskiranZeka

Yeayy !! Thank you !!
I have also doubt that 8C Could be wrong, but i couldn't be sure, 'cos it's Physic Book's answer. Anyway, now im glad

// Problem Solved.

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