Question

3. Three capacitors (3 uF, 4uF, and 6 uF) are connected in both series and parallel. The 3 uF and 6 uF are in series, and the 4 uF is in parallel with the other 2 capacitors. These capacitors are connected in series with two 10K ohm resistors in parallel. If 500. volts is applied to the RC circuit, how long will it take for the voltage to drop to 100. volts after the initial voltage is turned off?

Answer 1

null

total capacitance = 4 + 3||6 = 4 +3*6/ ( 3+6) = 6 μF = 6*10^-6 F

total resistance = 10 K || 10 || = 10*10/ ( 10+10) = 5 kΩ =5*10^3 Ω

the value of the voltage across the capacitor ( Vc ) at any instant in time during the charging period is given as:

100 = 500 * ( 1- exp( -t / ( 5*10³*6*10^-6 ) )

1/5 = 1- exp( -t / ( 5*10³*6*10^-6 ) )

exp( -t / ( 5*10³*6*10^-6 ) = 4/5

-t / ( 5*10³*6*10^-6 = ln ( 4/5)

t= - 5*10^3*6*10^-6*ln(4/5) =0.00669430 sec

t =6.694*10^-3 s