Question

A 2.00 µF and a 4.50 µF capacitor can be connected in series or parallel, as can a 45.0 kΩ and a 100 kΩ resistor. Calculate the four RC time constants (in s) possible from connecting the resulting capacitance and resistance in series.

resistors and capacitors in series

resistors in series, capacitors in parallel

resistors in parallel, capacitors in series

capacitors and resistors in parallel

Answer 1

Given,

C1 = 2 uF ; C2 = 4.5 uF ;

R1 = 45 kOhm ; R2 = 100 kOhm

resistors and capacitors in series

We know that,

time constant = tau = Req Ceq

Req = 45 + 100 = 145 kOhm

Ceq = 2 x 4.5/(2 + 4.5) = 1.38 uF

Tau = 145 x 10^3 x 1.38 x 10^-6 = 0.201 s

Hence, Tau = 0.201 s = 201 milli sec

resistors in series, capacitors in parallel

Req = 45 + 100 = 145 kOhm

Ceq = 2 + 4.5 = 6.5 uF

Tau = 145 x 10^3 x 6.5 x 10^-6 = 0.943 s

Hence, tau = 0.943 s = 943 milli sec

resistors in parallel, capacitors in series

Req = 100 x 45/(100 + 45) = 31.03 kOhm

Ceq = 2 x 4.5/(2 + 4.5) = 1.38 uF

Tau = 31.03 x 10^3 x 1.38 x 10^-6 = 0.043 s

Hence, tau = 0.043 s = 43 milli sec

capacitors and resistors in parallel

Req = 100 x 45/(100 + 45) = 31.03 kOhm

Ceq = 2 + 4.5 = 6.5 uF

tau = 31.03 x 10^3 x 6.5 x 10^-6 = 0.202 s

Hence, tau = 0.202 s = 202 milli sec