The circuit in Fig. 8.123 is the electrical analog of body functions used in medical schools to study convulsions. The analog is as follows: C1= Volume of fluid in a drug C2 = Volume of blood stream in a specified region R1 = Resistance in the passage of the drug from the input to the blood stream

Question

The circuit in Fig. 8.123 is the electrical analog of body functions used in medical schools to study convulsions. The analog is as follows:

C_{1}= Volume of fluid in a drug

C_{2}= Volume of blood stream in a specified region

R_{1}= Resistance in the passage of the drug from the input to the blood stream

R_{2}= Resistance of the excretion mechanism, such as kidney, etc.

v_{0} = Initial concentration of the drug dosage

v(t) = Percentage of the drug in the blood stream

Find v(t) for t > 0 given that C_{1} =0.5 \mu F, C_{2} =5 \mu F, R_{1}= 5 M\Omega , and v_{0}= 60 u(t) V.

Accepted Answer

For t = 0-, v(0) = 0.

For t > 0, the circuit is as shown below.

At node a,

[latex]\begin{array}{l}\left(\mathrm{v}_{\mathrm{o}}-\mathrm{v} / \mathrm{R}_{1}=\left(\mathrm{v} / \mathrm{R}_{2} ight)+\mathrm{C}_{2} \mathrm{d} \mathrm{v} / \mathrm{dt} ight. \\\mathrm{v}_{\mathrm{o}}=\mathrm{v}\left(1+\mathrm{R}_{1} / \mathrm{R}_{2} ight)+\mathrm{R}_{1} \mathrm{C}_{2} \mathrm{dv} / \mathrm{dt} \\60=(1+5 / 2.5)+\left(5 imes 10^{6} imes 5 imes 10^{-6} ight) \mathrm{dv} / \mathrm{dt} \\60=3 \mathrm{v}+25 \mathrm{dv} / \mathrm{dt} \\\mathrm{v}(\mathrm{t})=\mathrm{V}_{\mathrm{s}}+\left[\mathrm{Ae}^{-3 \mathrm{t} / 25} ight] \\ ext{where} 3 \mathrm{V}_{\mathrm{s}}=60 ext { yields } \mathrm{V}_{\mathrm{s}}=20 \\\mathrm{v}(0)=0=20+\mathrm{A} ext { or } \mathrm{A}=-20 \\\mathrm{v}(\mathrm{t})={20}\left(1-\mathrm{e}^{-3 t / 25} ight) \mathrm{V}\end{array}[/latex]

Question 8.82: The circuit in Fig. 8.123 is the electrical analog of body f...

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