# Homework Help: Capacitive Ciruit

1. Jan 13, 2013

### freshbox

Reference to question 2 part (ii).

I found out that:

Qt =V x Ct
Qt=50 x 1.67 x 10-6
Qt=83.5 x 10-6

I understand that in a series circuit for capacitor Qt=Q1=Q2=Q3

So if that's the case, looking at the circuit, since C2 and C3 are in series, their charge should be 83.5 x 10-6 since Qt=Q1=Q2=Q3

Hence, Qc3=VC
83.5 x 10-6=V x 10 x 10-6
V=8.35V

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Last edited: Jan 13, 2013
2. Jan 13, 2013

### Staff: Mentor

Those 3 are not in series, don't forget C4.

And how do you get that number?

3. Jan 13, 2013

### freshbox

My working:

Qt=83.5μC
C2+C3+C4=10μF

Qc234=83.5μC
Q=VC
83.5μC=V/10μF
Vab=8.35v

Using Voltage Divider Rule:
V3=CtE/C3
=6μF x 8.35/10μF

I am curious is there another way to solve this question besides the above method?

Thanks.

4. Jan 13, 2013

### Staff: Mentor

I don't think your Qt-value is correct.
Apart from that, the solution should be fine.

5. Jan 13, 2013

### freshbox

Ct=1.67μF
Qt=Vt x Ct
=50 x 1.67μF
=83.5μC

May I know which part is wrong?

Thank you.

6. Jan 13, 2013

### Staff: Mentor

I get a different result here.
If you just present some parts of your calculations, it is hard to find the specific location of the error.

7. Jan 13, 2013

### freshbox

C2+C3=6uF
C23+C4=10uF
C1+C5+C234=5/3 (1.666666666666667)

Hence Ct=1.67uF

8. Jan 13, 2013

### Staff: Mentor

Oh sorry, was my error.
Ok, I get the same capacitance now.

9. Jan 13, 2013

### freshbox

Thank you mfb for the help.