What is the mistake in my series capacitors homework?

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Homework Help Overview

The discussion revolves around a problem involving series capacitors, specifically focusing on the voltage across capacitors and the effective capacitance in a circuit. Participants are examining the relationships between voltage, charge, and capacitance in the context of the original poster's calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to express the total voltage in terms of individual voltages across capacitors but questions the accuracy of their approach. Some participants provide alternative expressions for voltage and discuss the effective capacitance in series configurations.

Discussion Status

Participants are actively engaging with the original poster's calculations, offering different perspectives on how to derive the total voltage and effective capacitance. There is a mix of interpretations regarding the relationships between voltage and charge, with some guidance provided on the effective capacitance formula.

Contextual Notes

There appears to be confusion regarding the definitions and relationships of voltage across capacitors in series, as well as the implications of the original poster's calculations. The discussion is constrained by the need for clarity on these concepts without reaching a definitive conclusion.

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V = q/c
so v = q/c + q/2c + q/3c
q/c(1+1/2+1/3)
q/c (11/6)
hope this helps
 
So total V is (11/6)q/c ? Can you explain how that is V1?
 
Instinctlol said:
So total V is (11/6)q/c ? Can you explain how that is V1?
I can't get you. Actually is 11/6 times the V
 
Capacitors are connected in series. So the effective capacitance is
1/Cs = 1/C + 1/2C + 1/3C
= 11/6C
Cs = 6C/11
Now substitute this value for C for the equation V = 11/6(Q/C) you ll get Q/C which is nothing but V or V1
Hope you got it now
 

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