Solving Capacitive Circuit: Why is 8.35V Wrong?

  • Thread starter freshbox
  • Start date
  • Tags
    Ciruit
In summary, there was a question about the calculation of Qt where the correct answer was 83.5μC. It was suggested that in a series circuit, Qt should equal Q1, Q2, and Q3. However, it was pointed out that C4 was also in the circuit and not in series with the other capacitors. The correct calculation was shown, and it was mentioned that there may be other methods to solve the problem.
  • #1
freshbox
290
0
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

But how come my answer is wrong? Please advise, thank you.
 

Attachments

  • Picture 16.jpg
    Picture 16.jpg
    30.6 KB · Views: 394
  • Picture 17.jpg
    Picture 17.jpg
    27.1 KB · Views: 448
Last edited:
Physics news on Phys.org
  • #2
freshbox said:
I understand that in a series circuit for capacitor Qt=Q1=Q2=Q3
Those 3 are not in series, don't forget C4.

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
And how do you get that number?
 
  • #3
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
= 5.01V Answer

I am curious is there another way to solve this question besides the above method?Thanks.
 
  • #4
I don't think your Qt-value is correct.
Apart from that, the solution should be fine.
 
  • #5
Ct=1.67μF
Qt=Vt x Ct
=50 x 1.67μF
=83.5μC

May I know which part is wrong?


Thank you.
 
  • #6
freshbox said:
Ct=1.67μF
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
C2+C3=6uF
C23+C4=10uF
C1+C5+C234=5/3 (1.666666666666667)

Hence Ct=1.67uF
 
  • #8
Oh sorry, was my error.
Ok, I get the same capacitance now.
 
  • #9
Thank you mfb for the help.
 

1. Why is the voltage reading of 8.35V considered wrong in a capacitive circuit?

The voltage reading of 8.35V is considered wrong in a capacitive circuit because it is likely a result of an error in measurement or calculation. In a capacitive circuit, the voltage should be continuously changing due to the charging and discharging of the capacitor. Therefore, a steady voltage reading of 8.35V is not expected and may indicate a mistake in the circuit or an incorrect assumption.

2. What is the correct voltage reading in a capacitive circuit?

The correct voltage reading in a capacitive circuit will depend on the specific circuit and its components. However, in a steady-state condition, the voltage across a fully charged capacitor should be equal to the applied voltage. In an AC circuit, the voltage reading will vary depending on the frequency and capacitance of the circuit.

3. Could a voltage reading of 8.35V be possible in a capacitive circuit?

In theory, a voltage reading of 8.35V could be possible in a capacitive circuit. However, it would likely require specific circuit components and conditions, such as a very low frequency and a high capacitance value, to achieve this voltage reading. In most practical cases, a steady voltage reading of 8.35V is unlikely and may indicate an error in the circuit.

4. What are some common mistakes that can lead to a wrong voltage reading in a capacitive circuit?

Some common mistakes that can lead to a wrong voltage reading in a capacitive circuit include incorrect calculation of the voltage or capacitance, using the wrong units, or not taking into account the effects of resistance in the circuit. Additionally, errors in the circuit connections or malfunctioning equipment can also result in incorrect voltage readings.

5. How can I accurately measure the voltage in a capacitive circuit?

To accurately measure the voltage in a capacitive circuit, it is important to use the correct equipment and techniques. This may include using a digital multimeter with a capacitance measurement function, ensuring proper connections in the circuit, and taking multiple readings to account for any fluctuations. It is also crucial to have a thorough understanding of the circuit and its components to accurately interpret the voltage readings.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Back
Top