Question

A 2.00 µF and a 5.50 µF capacitor can be connected in series or parallel, as can a 30.0 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 in series, capacitors in parallel

-resistors in parallel, capacitors in series

-capacitors and resistors in parallel

Please show how to solve! I'm really confused.

Answer 1

time constant in RC circuit is given by:

= time constant = R*C

Case A.

When both capacitors are in series, then

1/Ceq = 1/C1 + 1/C2

Ceq = C1*C2/(C1 + C2)

Ceq = 2.00*5.50/(2.00 + 5.50) = 1.47 F

When Both resistors are in series, then

Req = R1 + R2

Req = 30.0 + 100 = 130.0 k-ohm

Now time constant of this circuit will be:

= Req*Ceq

= 130.0*10^3*1.47*10^-6

= 0.191 sec

Case B.

When both capacitors are in parallel, then

Ceq = C1 + C2

Ceq = 2.00 + 5.50 = 7.50 F

When Both resistors are in series, then

Req = R1 + R2

Req = 30.0 + 100 = 130.0 k-ohm

Now time constant of this circuit will be:

= Req*Ceq

= 130.0*10^3*7.50*10^-6

= 0.975 sec

Case C.

When both capacitors are in series, then

1/Ceq = 1/C1 + 1/C2

Ceq = C1*C2/(C1 + C2)

Ceq = 2.00*5.50/(2.00 + 5.50) = 1.47 F

When Both resistors are in parallel, then

1/Req = 1/R1 + 1/R2

Req = R1*R2/(R1 + R2)

Req = 30.0*100/(30.0 + 100) = 23.08 k-ohm

Now time constant of this circuit will be:

= Req*Ceq

= 23.08*10^3*1.47*10^-6

= 0.034 sec

Case D.

When both capacitors are in parallel, then

Ceq = C1 + C2

Ceq = 2.00 + 5.50 = 7.50 F

When Both resistors are in parallel, then

1/Req = 1/R1 + 1/R2

Req = R1*R2/(R1 + R2)

Req = 30.0*100/(30.0 + 100) = 23.08 k-ohm

Now time constant of this circuit will be:

= Req*Ceq

= 23.08*10^3*7.50*10^-6

= 0.173 sec

Let me know if you've any query.