91Ó°ÊÓ

Consider the following infinite resistors network of resistances$R_1$ and$R_2$ extends to infinity toward the right Since the network is infinite, the resistance of the network to the right of points c and d is also equal to $R_T$, as shown in the lower figure, so simply we have two parallel resistors$R_2$ and$R_1$ which are connected in series with two resistors of , so we can write,

$\begin{array}{r}{R}_T=2R_1+{\left(\frac{1}{R_2}+\frac{1}{R_T}\right)}^{−1}\\ R_T=2R_1+\frac{R_2{R}_T}{R_2+R_T}\\ \end{array}$

Or

$R_T^2−2R_1{R}_T−2R_1{R}_2=0$

this is a quadratic equation that has a solution of,

$R_T=R_1Â\pm \sqrt{{R_1}^2+2R_1{R}_2}$

but since $R_T$> 0, then,

$R_T=R_1Â\pm \sqrt{{R_1}^2+2R_1{R}_2}$

let the parallel combination of $R_2$and $R_T$be $R_{eq}$, that is,

$R_{eq}={\left(\frac{1}{R_2}+\frac{1}{R_T}\right)}^{−1}=\frac{R_2{R}_T}{R_2+R_T}$

the potential difference between the points c and d can be written in terms of the potential difference between a and b by applying the loop rule, note that the same current is passing through$R_1$ and$R_{eq}$ since they are connected in series, so we get

role="math" localid="1668322955028" $V_{ab}=2I{R}_1-V_{cd}=0$

where the potential difference$V_{cd}$ is the potential across $R_{eq}$. The current following in the circuit is,

$I=\frac{V_{ab}}{2R_1+R_{eq}}$

thus,

$V_{ab}−2\frac{V_{ab}{R}_1}{2R_1+R_{eq}}−V_{cd}=0$

solve this equation for $V_{cd}$, we get,

$V_{cd}=V_{ab}\frac{R_{eq}}{2R_1+R_{eq}}=V_{ab}\frac{1}{2R_1/R_{eq}+1}$

Using (2) we can write the factor$\mathrm{β}$ given by the problem as,

$\mathrm{β}=\frac{2R_1\left(R_2+R_T\right)}{R_2{R}_T}=\frac{2R_1}{R_{eq}}$

thus

$V_{cd}=V_{ab}\frac{1}{1+\mathrm{β}}$

Now let the potential difference between terminals a and b at the left end of the infinite network is V, using the result of step 1, the potential difference between the upper and lower wires one segment from the left end is,

$V_1=\frac{V_{\mathrm{ο}}}{\left(1+\mathrm{β}\right)}$

and for two segments is,

role="math" localid="1668323423579" $V_2=\frac{V_1}{\left(1+\mathrm{β}\right)}=\frac{V_{\mathrm{ο}}}{{\left(1+\mathrm{β}\right)}^2}$

therefore, we can generalize the formula after n segments, to be,

$V_n=\frac{V_{n-1}}{\left(1+\mathrm{β}\right)}=\frac{V_{\mathrm{ο}}}{{\left(1+\mathrm{β}\right)}^n}$

now we need to find how many segments are needed to decrease the potential difference$V_n$ to less than 10% of$V_o$ for $R_1$=$R_2$ . In this case, we have,

$R_T=R_1Â\pm \sqrt{{R_1}^2+2R_1{R}_2}=R_1(1+\sqrt{3})$

hence,

$\mathrm{β}=\frac{2(2+\sqrt{3})}{(1+\sqrt{3})}=2.73$

we require $V_n/V_o$< 0.01, that is,

$\frac{1}{(1+\mathrm{β}{)}^n}=\frac{1}{(1+2.73{)}^n}≤0.01$

or,

$100≤(1+2.73{)}^n$

take the natural logarithm for both sides we get,

$$\begin{gathered} 4.605≤nIn(3.73) \\ 3.5≤n \end{gathered}$$

but since n must be an integer, then,

n = 4

(c) In this part we need to find$R_1$ and$\mathrm{β}$ for$R_1=6.3×{10}^3\mathrm{Î\copyright }$ and $R_1=6.4×{10}^3\mathrm{Î\copyright }$. Using the formulas above we get

$\begin{array}{r}{R}_T=6400\mathrm{Î\copyright }+\sqrt{(6400{)}^2+2(6400\mathrm{Î\copyright })\left(8.00×{10}^8\mathrm{Î\copyright }\right)}=3.2×{10}^6\mathrm{Î\copyright }\\ R_T=3.2×{10}^6\mathrm{Î\copyright }\end{array}$

and,

$\begin{array}{r}\mathrm{β}=\frac{2\left(6000\mathrm{Î\copyright }\right)\left(3.2×{10}^6\mathrm{Î\copyright }+8.0×{10}^6\mathrm{Î\copyright }\right)}{\left(3.2×{10}^6\mathrm{Î\copyright }\right)\left(8.0×{10}^8\mathrm{Î\copyright }\right)}=4.0×{10}^{−3}\\ \mathrm{β}=4.0×{10}^{−3}\end{array}$

For an axon segment of length$△x=1.0\mathrm{μ}m$ and total length x = 2.0 mm, we have

$n=\frac{2.0×{10}^{−3}\mathrm{m}}{1.0×{10}^{−6}\mathrm{m}}=2000\text{Segments}$

then the fraction of the potential difference between the inside and outside of the axon decrease over a distance of x by,

$V_{2000}=\frac{V_{\mathrm{ο}}}{(1+\mathrm{β}{)}^{2000}}$

or,

$\begin{array}{r}\frac{V_{2000}}{V_o}=\frac{1}{{\left(1+4.0×{10}^{−3}\right)}^{2000}}=3.4×{10}^{−4}\\ \frac{V_{2000}}{V_o}=3.4×{10}^{−4}\end{array}$

(e) Finally, $R_2=3.3×{10}^{12}$, we get

$R_T=6400\mathrm{Î\copyright }+\sqrt{(6400{)}^2+2(6400\mathrm{Î\copyright })\left(3.3×{10}^{12}\mathrm{Î\copyright }\right)}=2.1×{10}^{12}\mathrm{Î\copyright }$

and,

$\mathrm{β}=\frac{2\left(6000\mathrm{Î\copyright }\right)\left(2.1×{10}^{12}\mathrm{Î\copyright }+3.3×{10}^{12}\mathrm{Î\copyright }\right)}{\left(2.1×{10}^{12}\mathrm{Î\copyright }\right)\left(3.3×{10}^{12}\mathrm{Î\copyright }\right)}=6.2×{10}^{−5}$

thus,

$\begin{array}{r}\frac{V_{2000}}{V_o}=\frac{1}{{\left(1+6.2×{10}^{−5}\right)}^{2000}}=0.88\\ \frac{V_{2000}}{V_o}=0.88\end{array}$