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 R1=1.5MΩR_{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 R2R_{2} assume its largest value.

7.73
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(a) The smallest value for R2R_{2} is 0 Ω, and the corresponding time constant for the circuit is

τ=(R1+R2)C=(1.5×106+0)×0.1×106=0.15s\tau=\left(R_{1}+R_{2}\right) C=\left(1.5 \times 10^{6}+0\right) \times 0.1 \times 10^{-6}=0.15 s

The largest value for R2R_{2} is 2.5 MΩ, and the corresponding time constant for the circuit is

τ=(R1+R2)C=(1.5+2.5)×106×0.1×106=0.4s\tau=\left(R_{1}+R_{2}\right) C=(1.5+2.5) \times 10^{6} \times 0.1 \times 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, vC(0)=0v_{C}(0)=0, while vC()=110v_{C}(\infty)=110. But

vC(t)=vC()+[vC(0)vC()]et/τ=110[1et/τ]v_{C}(t)=v_{C}(\infty)+\left[v_{C}(0)-v_{C}(\infty)\right] e^{-t / \tau}=110\left[1-e^{-t / \tau}\right]

where τ = 0.4 s, as calculated in part (a). The lamp glows when vC=70Vv_{C}=70 V . If vC(t)=70V at t=t0v_{C}(t)=70 V \text { at } t=t_{0}, then

70=110[1et0/τ]711=1et0/τ70=110\left[1-e^{-t_{0} / \tau}\right] \quad \Longrightarrow \quad \frac{7}{11}=1-e^{-t_{0} / \tau}

or

et0/τ=411et0/τ=114e^{-t_{0} / \tau}=\frac{4}{11} \quad \Longrightarrow \quad e^{t_{0} / \tau}=\frac{11}{4}

Taking the natural logarithm of both sides gives

t0=τln114=0.4ln2.75=0.4046st_{0}=\tau \ln \frac{11}{4}=0.4 \ln 2.75=0.4046 s

A more general formula for finding t0t_{0} is

t0=τlnv(0)v()v(t0)v()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 t0<τt_{0}<\tau. In this example, that condition is not satisfied.

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