Question 7.19: Consider the circuit in Fig. 7.73, and assume that R1 = 1.5M...
Consider the circuit in Fig. 7.73, and assume that R_{1}=1.5 M \Omega, 0 < R < 2.5 MΩ. (a) Calculate the extreme limits of the time constant of the circuit. (b) How long does it take for the lamp to glow for the first time after the switch is closed? Let R_{2} assume its largest value.
(a) The smallest value for R_{2} is 0 Ω, and the corresponding time constant for the circuit is
\tau=\left(R_{1}+R_{2}\right) C=\left(1.5 \times 10^{6}+0\right) \times 0.1 \times 10^{-6}=0.15 sThe largest value for R_{2} is 2.5 MΩ, and the corresponding time constant for the circuit is
\tau=\left(R_{1}+R_{2}\right) C=(1.5+2.5) \times 10^{6} \times 0.1 \times 10^{-6}=0.4 sThus, 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}(\infty)=110. But
where τ = 0.4 s, as calculated in part (a). The lamp glows when v_{C}=70 V . If v_{C}(t)=70 V \text { at } t=t_{0}, then
70=110\left[1-e^{-t_{0} / \tau}\right] \quad \Longrightarrow \quad \frac{7}{11}=1-e^{-t_{0} / \tau}or
e^{-t_{0} / \tau}=\frac{4}{11} \quad \Longrightarrow \quad e^{t_{0} / \tau}=\frac{11}{4}Taking the natural logarithm of both sides gives
t_{0}=\tau \ln \frac{11}{4}=0.4 \ln 2.75=0.4046 sA more general formula for finding t_{0} is
t_{0}=\tau \ln \frac{v(0)-v(\infty)}{v\left(t_{0}\right)-v(\infty)}The lamp will fire repeatedly every τ seconds if and only if t_{0}<\tau. In this example, that condition is not satisfied.