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

Answer:

$t'_{1\2} = 6.6 sec$

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

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

putting all value we get new half life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$