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A digital voltmeter charged a 1-uF capacitor to the level of an unknown voltage using a...
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The time period (time for one cycle) T is by definition the reverse of the frequency F
`T =1/F =1/(100*10^3) =10^-5 seconds = 10 "microseconds"`
The total time taken by 141 cycles is
`t = 141*T =141*10^-5 =1.41*10^-3 seconds =1.41 miliseconds`
This is the total time of charging therefore the total charge on the capacitor is
`Q =I*t = 10^-3*1.41*10^-3 =1.41*10^-6 C`
Now, by definition again the capacitance is
`C = Q/U`
Therefore the voltage on the capacitor is
`U =Q/C = (1.41*10^-6)/10^-6 =1.41 "Volt"`
Posted by valentin68 on September 11, 2013 at 3:30 PM (Answer #1)
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