Updated greek laters @ latex compiling

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правка #10018 предыдущая #9722 GitHub dbb31ba ← раньше
en/8.3.15.md +7 −7
@@ -80,24 +80,24 @@
<p>
Applying Kirchhoff Second Law (Loop's law). For loop $I$
$$-i_{11}R_1-i_{22}R_2+i_{21}R_2=0$$
− $$-i_{11}R_1+R_2(i_{21}-i_{22})=0 \;(1)$$
+ $$-i_{11}R_1+R_2(i_{21}-i_{22})=0 \quad(1)$$
As $i_{11}R_1 = 10 i_{12}R_1$, that is, $i_{1n} = 10 i_{1(n+1)}$, hence
− $$i_{1n} = \frac{i_{11}}{10^{n-1}} \;\forall~n\geq2 \;(2)$$
+ $$i_{1n} = \frac{i_{11}}{10^{n-1}} \;\forall~n\geq2 \quad(2)$$
For loop $II$ and according $(2)$
$$-i_{N+1}(R_1+r)+i_{2N}R_2=0$$
− $$-\frac{i_{11}}{10^{N-1}}(R_1+r)+i_{2N}R_2=0 \;(3)$$
+ $$-\frac{i_{11}}{10^{N-1}}(R_1+r)+i_{2N}R_2=0 \quad(3)$$
For subsequent bifurcations (applying Kirchhoff First Law)
$$i_{11} = i_{12}+i_{22}$$
$$i_{22} = i_{11}-i_{12}$$
So,
− $$i_{2n} = i_{1(n-1)}-i_{1n} \;\forall~n\geq2 \;(4)$$
+ $$i_{2n} = i_{1(n-1)}-i_{1n} \;\forall~n\geq2 \quad(4)$$
According $(2)$, let's modify $(4)$
$$i_{2n} = \frac{i_{11}}{10^{n-2}}-\frac{i_{11}}{10^{n-1}}$$
− $$i_{2n} = \frac{9i_{11}}{10^{n-1}} \;(5)$$
+ $$i_{2n} = \frac{9i_{11}}{10^{n-1}} \quad(5)$$
According $(5)$ into $(3)$
− $$9R_2 = R_1+r \;(6)$$
+ $$9R_2 = R_1+r \quad(6)$$
According $(5)$, from $(1)$
− $$R_1 = \frac{81}{10} R_2 \;(7)$$
+ $$R_1 = \frac{81}{10} R_2 \quad(7)$$
Finally, solving the equation system formed by $(6)$ and $(7)$, it is obtained
$$\boxed{R_1 = 9r}$$
and,
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