Предмет: Физика, автор: 0020490p0wsws

Частота коливань у коливальному контурі дорівнює 100кГц. Якою стане частота коливань якщо зменшити електроємність конденсатора у 8 раз, а індуктивність котушки збільшити у 2 рази? срочно!!! даю 100 балів

Ответы

Автор ответа: 555ai555

Ответ:

Объяснение:

f1 = 100кГц

C2 = C1/8

L2 = L1*2

f2 = ?

f2 = 1/ (2*п*√(L2*C2)) = 1 / (2*п*√(L1*C1/4))

f1 = 1/ (2*п*√(L1*C1))

f2/f1 = (1 / (2*п*√(L1*C1/4))) / (1 / (2*п*√(L1*C1))) = √4 = 2

f2 = 2*f1 = 2*100 = 200 кГц

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