RC Circuit ; why RC=(VT/2 delta V)

AI Thread Summary
The discussion focuses on proving the equation RC = (VT/2 delta V) in the context of RC circuits involving capacitor charging and discharging. Participants emphasize measuring resistance accurately and experimenting with different frequencies to observe changes in cycle time. The equations governing charge (Q) and voltage (V) across the capacitor are provided, with attempts to manipulate these equations to derive the desired relationship. Despite various approaches, including Taylor expansion, the original poster struggles to arrive at the correct proof. The conversation highlights the complexities involved in understanding the relationships within RC circuits.
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Homework Statement


I need to prove that RC=(VT/2 delta V)


Select the circle components of Charging and discharging of a capacitor
Measure the resistance of resistor with a multimeter
When you run a different frequency the cycle time change
run different frequency that T<<tao, T=tao, T>>tao
Choose components So that T<<tao,
V - the voltage of Function Generator
delta V= Voltage amplitude


Homework Equations



when -T/2<t<0
Q(a)=-cv+Ae^(-t/tao)

when T/2>t>0
Q(b)=cv+Be^(-t/tao)

A=(2cv)/(e^(T/2tao)+1)
B=(-2cv)/(e^(-T/2tao)+1)

The Attempt at a Solution



Q=VC

Va=Qa/C

Vb=Qb/C

delta V= Vb-Va

RC=tao

I place Vb-Va on RC=(VT/2 delta V) and I played with the equation

but I didn't get RC=(VT/2 delta V)
 
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I tried another solution but I got stuck:

Va=Qa/C

I used Va:

delta V= V(0)-V(-T/2)

VT/2 delta V

e^(T/2tao)=1+T/2tao -- taylor

and I get RC+T/4

what wrong?
 
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