[Vật lí 12] bt điện

T

tukutelove

N

nhat.funsun

Đáp án: C. f = fo

[tex] f_1 = f_2 = f_0 \Leftrightarrow \frac{1}{2\pi \sqrt{L_1 C_1}} = \frac{1}{2\pi \sqrt{L_2 C_2}} \Leftrightarrow L_1 C_1 = L_2 C_2 \Leftrightarrow \frac{L_2}{C_1} = \frac{L_1}{C_2} = \frac{L_1 + L_2 }{C_1 + C_2}[/tex]

L1 nối tiếp L2 =>L
[tex] L = L_1 + L_2 [/tex]

C1 nối tiếp C2 => C
[tex] C = \frac{C_1 C_2}{C1 + C2} [/tex]

[tex] LC = \frac{L_1 + L_2}{C_1 + C_2}C_1 C_2 = \frac{L_2}{C_1}C_1 C_2 = L_2 C_2[/tex]

Vậy
[tex] f = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{L_2 C_2}} = f_0 [/tex]
 
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