Calculating the Flash Frequency in a Neon Lamp Circuit

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The discussion focuses on calculating the flash frequency of a neon lamp circuit with a 4.0 μF capacitor and a resistance of 2.5 x 10^4 ohms. The circuit charges until the voltage reaches 75 V, at which point the neon lamp ionizes and discharges the capacitor, creating a flash of light. Participants clarify the correct values and formulas needed for calculations, emphasizing the importance of using the natural logarithm in the equations. There is some confusion regarding the calculations, but guidance is provided on how to use a calculator or find logarithmic values online. The conversation highlights the learning process involved in understanding the circuit's behavior and calculations.
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Homework Statement


A simple type of blinking light circuit can be constructed using a neon lamp. The
circuit shown here has a 4.0 μF capacitor in parallel with a neon lamp. When the
voltage is low in the RC portion of the circuit, the lamp does not conduct
electricity. Therefore, it is
effectively not there from an
electrical point of view. The RC
circuit will then charge from the
110 V power supply. However,
when the voltage across the
capacitor reaches 75 V, the neon
will ionize very quickly and the
neon lamp will become a very
good conductor, and will immediately discharge the capacitor. The energy stored
in the capacitor will be given off as a flash of orange light, making this a useful
circuit. After the flash, the charging process will start once more since the
voltage will again be low.
a. Determine the flash frequency with the resistance value shown.
b. Make a sketch of the voltage across the capacitor versus time in such a
circuit, showing several periods.

C=4X10^-6 ε=110V Vc=75V R= 2.5 X 10-4

Homework Equations



t=1/f T=RC Vc=ε(1-e^t/RC)



The Attempt at a Solution



So for a) I have T=RC= (2.5X10^-4)(4X10^-6)= 0.1s <-- time constant for the circuit
Vc=ε(1-3^t/RC)
In (1-{Vc/ε))=-t/RC
t= -In (1-V{c/ε})(RC)

What I don't understand is how to calculate that
 
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Lots of jumble here.
R is not 2.5e-4. It's 2.5e4.
Vc is not what you have, which is Vc=ε(1-3^t/RC). What's with the "3"? And is the sign of the exponent correct?

Miraculously, your expression for t is correct if you straighten out your parentheses and write ln instead of In.
FIx that up and go from there.
 
wow, you really are a rude man-lol
yes i see I got a couple typos there-

So for a) I have T=RC= (2.5X10^4)(4X10^-6)= 0.1s <-- time constant for the circuit
Vc=ε(1-e^t/RC)
ln (1-{Vc/ε))=-t/RC
t= -In (1-{Vc/ε})(RC)

What I don't understand is how to calculate that
 
Myr73 said:
wow, you really are a rude man-lol

I calls 'em like I sees 'em - don't take it personally! :smile:

yes i see I got a couple typos there-

So for a) I have T=RC= (2.5X10^4)(4X10^-6)= 0.1s <-- time constant for the circuit
Vc=ε(1-e^t/RC)
ln (1-{Vc/ε))=-t/RC
t= -In (1-{Vc/ε})(RC)

What I don't understand is how to calculate that

What is it you don't understand?

You know Vc, ε, and RC, so why can't you put in the numbers and solve for T = -(RC)ln(1-{Vc/ε})? Got a hand calculator handy?
 
umm I don't know what to do with the " ln ".
 
The ln function is available in the standard Windows calculator (or equivalent if you use a different OS), just set it to "Advanced mode" (alt-2).
 
Oh ok I found it, but if I don't have a calculator, how is that calculated?
 
Myr73 said:
Oh ok I found it, but if I don't have a calculator, how is that calculated?
Google it Write "natural logarithm of N" where N is your number.
In my day we used slide rules. Consider yourself very lucky!
 

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