Capacitors in Series and finding voltage

Click For Summary

Homework Help Overview

The discussion revolves around the topic of capacitors in series, specifically focusing on finding the voltage across the first capacitor in a series circuit. Participants are exploring the relationships between charge, capacitance, and voltage in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept that capacitors in series share the same charge but not necessarily the same voltage unless they have equal capacitance. There are attempts to find the equivalent capacitance and relate it to the total voltage. Questions arise about the appropriate formulas to use and the definitions of terms like equivalent voltage.

Discussion Status

There is an ongoing exploration of the relationships between charge, voltage, and capacitance. Some participants have provided guidance on using the formula relating these quantities, while others are clarifying their understanding of the equivalent capacitance and its implications for the voltage across individual capacitors. Multiple interpretations of the problem are being considered.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available or the methods they can use. There is a specific focus on deriving the voltage across the first capacitor based on the total voltage and the equivalent capacitance.

lylos
Messages
77
Reaction score
0
We've been studying capacitors in series in my physics class. I have a problem tonight and I know the answer, I just don't know how to get to it.

The problems is as follows:
37016.jpg

Find the voltage V1 across the first capacitor.
Express your answer in terms of V.
 
Physics news on Phys.org
capacitors in sereis always have the same charge. They do not the same voltage unless they have the same capacitance.

What would be your first step?
 
ranger said:
capacitors in sereis always have the same charge. They do not the same voltage unless they have the same capacitance.

What would be your first step?

The first step is finding that they all have the same charge. Then, I think, I would find the equivalent capacitance which is equal to 11/6 C. From there I don't know where I would get the voltage for each one.
 
Do you know also know of a formula that relates capacitance, voltage, and charge?
 
ranger said:
Do you know also know of a formula that relates capacitance, voltage, and charge?

ofcourse, c=q/v... So Ceq=Q/Veq...
 
lylos said:
ofcourse, c=q/v... So Ceq=Q/Veq...

Good. But instead of Ceq=Q/Veq (not sure what you mean by Veq). It should be Ceq=Q/V, where V is the total source voltage.

We know what Ceq and V is simply V, what's left to be determined? Remember that this quantity would be the same for all caps.
 
ranger said:
Good. But instead of Ceq=Q/Veq (not sure what you mean by Veq). It should be Ceq=Q/V, where V is the total source voltage.

We know what Ceq and V is simply V, what's left to be determined? Remember that this quantity would be the same for all caps.

Ok by the Veq I meant the total voltage... So what we have is that the total voltage is equal to V. The C equivalence is equal to 11/6 C. And that the charge on each capacitor is Q. Now from that information, I am still at a loss as to how to determine the voltage across the first capacitor.

The answer is 6/11 V according to the system my homework is in...
 
Are you sure the Ceq = (11C)/6? Recheck it. If the charge on each cap is Q, then what is Q? Knowing Q, you will be able to find V for any cap.
 
ranger said:
Are you sure the Ceq = (11C)/6? Recheck it. If the charge on each cap is Q, then what is Q? Knowing Q, you will be able to find V for any cap.

Ok, I see that I made the mistake there... It is 6/11 C. So we would then set that 6/11 C = Q(for each capacitor) / (Vtot).

So Q = 6 (Vtot) (C) / 11 for each capacitor.

Now that we have that, we could then find the voltage by knowing the q=v/c equation.

C = (6/11 C Vtot / 11)/V1
C V1 = 6/11 C Vtot
V1 = 6/11 Vtot

And that's the answer! Thank you for your help!
The problem that I had was knowing to get the charge for each capacitor and then work backwards...
 
  • #10
You're very much welcome. And well done on your part :approve:
 

Similar threads

Calculating the voltage across capacitors and resistors with switches
  • · Replies 2 ·
Replies
2
Views
1K
Why Do Capacitors Behave Differently in Series and Parallel Configurations?
  • · Replies 20 ·
Replies
20
Views
2K
Can the Voltage Across a Capacitor Be Greater Than the Source Voltage in a Series RLC Circuit?
  • · Replies 3 ·
Replies
3
Views
2K
Why can we model spherical capacitor with two dielectrics as two capacitors in series?
  • · Replies 4 ·
Replies
4
Views
2K
Two-layer dielectric, four-capacitor model vs three-capacitor model
  • · Replies 5 ·
Replies
5
Views
1K
Finding charge on a capacitor given potential difference across two points
  • · Replies 4 ·
Replies
4
Views
2K
Why Capacitors in Parallels vs. Series: Coaxial Capacitor Case
  • · Replies 2 ·
Replies
2
Views
2K
Capacitor Questions Charging and Discharging
  • · Replies 4 ·
Replies
4
Views
4K
Find the maximum potential difference across a series circuit
  • · Replies 11 ·
Replies
11
Views
4K
Finding Current in Resistors & Charge of Capacitor
  • · Replies 4 ·
Replies
4
Views
2K