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Bond [772]
3 years ago
12

g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],

and 푅푅=17 [Ω/m]. Find 훾훾,훼훼,훽훽,휆휆, and 푍푍0.
Engineering
1 answer:
yKpoI14uk [10]3 years ago
5 0

Explanation:

\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}

Therefore,

-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}

Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}

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bulgar [2K]

Answer:

YES

Explanation:

If we connect batteries in series then the output voltage is the sum of the individual voltage of each battery i.e if you connect three 12 volts batteries in series then their output voltage will be 12+12+12=36 volts, but the current rating of the batteries in series will be same of the individual battery rating in 'mah'.

But when we connect the batteries in parallel their voltage is not added  but their current rating in mah is addition of their individual rating.

So, If you want 24 volts from three 12 volts battery then you can connect two of them in series and the other one in parallel with them this will give 24 volts and the current will be addition of the two series batteries and the third which is in parallel with them. You can use this configuration if current value is not a big factor.

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