Verified answer

tau 1=9 ms

tau 2=?

c1=3 µF

c2=3*6/6+3=2

tau 1=rc1 ...............1

tau2=rc2.................2

divide i and 2,

9/tau2=3/2

tau2=6 ms

therefore resulting time constant If an additional 6.0-µF is added in series to the 3.0 µF capacitor

is 6 ms..

time constant with one resistor R and one capacitor C1 in series: tau1 = R C1

equivalent capacitor C when C1 ans C2 are in series: 1/C = 1/C1+1/C2 ---> C=C1C2/(C1+C2)

the new time constant is : tau = RC = RC1C2/(C1+C2)

then tau = tau1 C2/(C1+C2) = tau1 * 1/(C1/C2 + 1)

then if C2 = 2 C1 ; tau = tau1 * 1/(1+1/2) = 2/3 tau1