Một đoạn mạch gồm điện trở R thuần 400 Ω mắc nối t...

D

- Hướng dẫn giải

+ Ta có: 

\(\begin{array}{l}
I = \frac{{N.B.S\omega }}{{\sqrt 2 \sqrt {{R^2} + Z_C^2} }} = \frac{{N.B.S.2\pi .p.n}}{{\sqrt 2 \sqrt {{R^2} + Z_C^2} }} = \frac{{A.n}}{{\sqrt {{R^2} + Z_C^2} }}(A = \frac{{N.B.S.2\pi .p}}{{\sqrt 2 }})\\
 \Rightarrow \frac{{{I_2}}}{{{I_1}}} = \frac{{{n_2}}}{{\sqrt {{R^2} + Z_{C2}^2} }}.\frac{{\sqrt {{R^2} + Z_{C1}^2} }}{{{n_1}}} \Leftrightarrow \sqrt 2  = \frac{{\sqrt {{R^2} + Z_{C1}^2} }}{{\sqrt {{R^2} + Z_{C2}^2} }}
\end{array}\)

+ Khi \(\begin{array}{l}
{n_2} = 2{n_1} \Rightarrow {Z_{C2}} = \frac{{{Z_{C1}}}}{2}\sqrt 2  = \frac{{\sqrt {{R^2} + Z_{C1}^2} }}{{\sqrt {{R^2} + \frac{{Z_{C1}^2}}{4}} }}\\
 \Leftrightarrow 2\left( {{R^2} + \frac{{Z_{C1}^2}}{4}} \right) = {R^2} + Z_{C1}^2 \Rightarrow {Z_{C1}} = 400\sqrt 2 \Omega 
\end{array}\)

+ Khi:

\(\begin{array}{l}
{n_3} = 4{n_1} \Rightarrow {Z_{C3}} = \frac{{{Z_{C1}}}}{4} = 100\sqrt 2  \Rightarrow \frac{{{I_3}}}{{{I_1}}} = \frac{{{n_3}}}{{\sqrt {{R^2} + Z_{C3}^2} }}.\frac{{\sqrt {{R^2} + Z_{C1}^2} }}{{{n_1}}}\\
 \Leftrightarrow \frac{{{I_3}}}{I} = \frac{4}{{\sqrt {{{400}^2} + {{100}^2}.2} }}.\frac{{\sqrt {{{400}^2} + {{400}^2}.2} }}{1} \Rightarrow {I_3} = 6,53I
\end{array}\)