What is the total capacitance for the RC circuit lab problem with given data?

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SUMMARY

The total capacitance for the RC circuit lab problem can be calculated using the measured capacitances C1 and C2, which are 41μF and 22μF respectively. The half-life measurements indicate that C1 corresponds to 1.200 milliseconds and C2 to 0.600 milliseconds. For series capacitors, the total capacitance is found using the formula 1/C_total = 1/C1 + 1/C2, while for parallel capacitors, the formula is C_total = C1 + C2. The calculations yield C1 as 3.46E-5 F and C2 as 1.73E-5 F, leading to the total capacitance values for both configurations.

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Homework Statement



I did the RC circuit lab
I measure the actual resistance of 1000Ω
Function Generator Resistance: 50 Ω
Function Generator Frequency: 1.4Hz

resistor used was 41uF in C1
Resistor used was 22uf for c2


So i have to find c1, c2 and use c1,c2 to calc. the total capacitance for series/parallel

Data:
Half Life:
C1: 1.200 milli sec
C2: 0.600 milli sec
Series C1, C2: 1.400 millisec
Parallel C1,C2: 2.000 millisec


Homework Equations




C= (t 1/2)/ Rln2



The Attempt at a Solution



C1: ( 1.200e-3)/ 50Ω * ln2 = 3.46E-5
C2: (.600 e-3)/ 50Ω* ln2= 1.73e-5

for parallel/ series capacitors, I'm not sure where to continue.

is it add up the c1 and c2 and that's my series capacitance? and 1/c1 +1/c2= my parallel capacitance?
 
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Use c = c1+c2 if they are in parallel.
Use 1/c = 1/c1 + 1/c2 if they are in series.
 


once i used my solve individual C's to solve in the series/ parallel then I'm done right?
 

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