Rattn_sPi 2025-03-25

\(a=\dfrac{1}{\sqrt{A}}\)

R10 =

\(R_{10}=\dfrac{\left(a^2-1\right)\cdot R_b\cdot\sqrt{R_k}}{\left(a^2+1\right)\cdot\sqrt{R_k}-2\cdot a\cdot\sqrt{R_b}}\)

R11 =

\(R_{11}=\dfrac{\left(a^2-1\right)\cdot\sqrt{R_b\cdot R_k}}{4\cdot a}\)

R12 =

\(R_{12}=\dfrac{\left(a^2-1\right)\cdot\sqrt{R_b\cdot R_k}}{4\cdot a}\)

R13 =

\(R_{13}=\dfrac{\left(a^2-1\right)\cdot R_b\cdot\sqrt{R_k}}{\left(a^2+1\right)\cdot\sqrt{R_k}-2\cdot a\cdot\sqrt{R_b}}\)


Rb =
Rk =
AP =
☺
			\(a=\dfrac{1}{\sqrt{A}}\)
R10 =			\(R_{10}=\dfrac{\left(a^2-1\right)\cdot R_b\cdot\sqrt{R_k}}{\left(a^2+1\right)\cdot\sqrt{R_k}-2\cdot a\cdot\sqrt{R_b}}\)
R11 =			\(R_{11}=\dfrac{\left(a^2-1\right)\cdot\sqrt{R_b\cdot R_k}}{4\cdot a}\)
R12 =			\(R_{12}=\dfrac{\left(a^2-1\right)\cdot\sqrt{R_b\cdot R_k}}{4\cdot a}\)
R13 =			\(R_{13}=\dfrac{\left(a^2-1\right)\cdot R_b\cdot\sqrt{R_k}}{\left(a^2+1\right)\cdot\sqrt{R_k}-2\cdot a\cdot\sqrt{R_b}}\)

document.write( SzHuEnSzSz( 'Szimmetrikus Π csillapító', 'Symmetric Π attenuator' ) );