Question 7.19: Consider the circuit in Fig. 7.73, and assume that R1=1.5 MΩ......

(a) The smallest value for R_{2}   is Ω and the corresponding time constant for the circuit is

\tau = (R_{1} + R_{2} ) C = ( 1.5 × 10^{6} + 0 ) × 0.1 × 10^{-6} = 0.15 s

The largest value for R_{2}   is 2.5M Ω and the corresponding time constant for the circuit is

\tau = (R_{1} + R_{2} ) C = ( 1.5 + 2.5  ) × 10^{6} × 0.1 × 10^{-6} = 0.4  s

Thus, by proper circuit design, the time constant can be adjusted to introduce a proper time delay in the circuit.

(b) Assuming that the capacitor is initially uncharged v_{C} (0) = 0 , , while v_{C} (∞) = 110 . [/latex]  But

v_{C} (t) = v_{C}(∞) + [ v_{C} (0) – v_{C} (∞) ] e^{-t/\tau} = 110[ 1 – e^{-t/\tau } ]

where \tau = 0.4 s , as calculated in part (a). The lamp glows when v_{C} = 70 V   If  v_{C}(t) = 70 V   at  t = t_{o} , then

70 = 110 [ 1 – e^{-t_{o} / \tau } ]    ⇒       \frac{7}{11} = 1 – e^{-t_{o}/ \tau }

or

e^{-t_{o}/ \tau } = \frac{4}{11}      ⇒    e^{t_{o}/ \tau } = \frac{11}{4}

Taking the natural logarithm of both sides gives

t_{o} = \tau  ln  \frac{-v(∞)}{v(t_{o}) – v(∞) }

The lamp will fire repeatedly every  t_{o} seconds if and only if  v(t_{0}  ) <   v(∞).