Question

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an RC circuit with a half-life of t1/2 = 3.00 s. Your supervisor is concerned that the initiation of the process is occurring too soon and that the half-life needs to be extended. He asks you to change the resistance of the circuit to make the half-life longer. All you can find in the supply room is a single 48.0 Ω resistor. You look at the RC circuit and see that the resistance is 40.0 Ω. You combine the new resistor with the old to extend the half-life of the circuit. Determine the new half-life (in s).

Answer 1

`t'_(1\2) = 6.6 sec`

Step-by-step explanation:

the half life of the given circuit is given by

`t_(1\2) =\tau ln2`

where [/tex]\tau = RC[/tex]

`t_(1\2) = RCln2`

Given
`t_(1\2) = 3 sec`

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

`t'_(1\2) =R'Cln2`

Divide equation 2 by 1

`(t'_(1\2))/(t_(1\2)) = (R'Cln2)/(RCln2) = (R')/(R)`

`t'_(1\2) = t'_(1\2)(R')/(R)`

putting all value we get new half life

`t'_(1\2) = 3 * (88)/(40)  = 6.6 sec`

`t'_(1\2) = 6.6 sec`