Question

Use the graph to determine the time to 'half-max'.
t_1/2 =  s

Calculate the capacitance (C) of the capacitor.
C =  µF

Determine the percent error between the calculated capacitance and the value on the capacitor (330 µF).

% error =

calculated ? 330 µF

330 µF

× 100% =  %

What is the maximum charge for the capacitor in this experiment (approximately)?

Charge equals capacitance multiplied by voltage, or 0.000330 F × 4.00 V, or 0.00132 C or 1320 µC.Charge equals capacitance multiplied by voltage, or 330 F × 4.00 V, or 1.32 C.     Charge equals capacitance divided by voltage, or 0.000330 F / 4.00 V, or 0.00132 C or 1320 µC.Charge equals capacitance multiplied by the inverse of the voltage, or 0.000330 F × 1/4.00 V, or 0.0000825 C or 82.5 µC.

Answer 1

1) Thalf =10.14 - 9.99 = 0.26sec

2)

C = t1/2/ ln 2/ R

RC = 0.26/0.693=0.375

C = 0.375/R

consider R = 1000 ohm

C = 0.375/1000 = 375 microfarad

3)% error = (calculated- 330)/calculated ) 100%

=( 375-330/375 ) 100%

= (0.12)100%

= 12%

(4)

Q = CV = 0.000330( 4 V) = 1320 µC