A circuit containing resistance R ₁, inductance L ₁ and capacitance C ₁ connected in series resonates at the same frequency 'f ₀ ' as another circuit containing R ₂, L ₂ and C ₂ in series. If two circuits are connected in series then the new frequency at resonance is (a) 3/4fr (b) 3/2fr (c) 2fr (d) fr

Answer: (d). Same resonance: L₁C₁ = L₂C₂ = (1)/(ω₀²), so 1/C₁ = ω₀²L₁ and 1/C₂ = ω₀²L₂. In series: L = L₁ + L₂ and 1/C = 1/C₁ + 1/C₂ = ω₀²(L₁ + L₂). ω² = 1/LC = (ω₀²(L₁+L₂))/(L₁+L₂) = ω₀² So the answer is (d): the resonant frequency is unchanged ( fr = f₀).

A circuit containing resistance R ₁, inductance L ₁ and capacitance C… | MHT CET 2021 (23 Sep, Shift 2)

Q91MHT CET 2021 · 23 SEP · SHIFT 2Medium

Resonance, Bandwidth & Q Factor

A circuit containing resistance R ${}_{1}$ , inductance L ${}_{1}$ and capacitance C ${}_{1}$ connected in series resonates at the same frequency 'f ${}_{0}$ ' as another circuit containing R ${}_{2}$ , L ${}_{2}$ and C ${}_{2}$ in series. If two circuits are connected in series then the new frequency at resonance is

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Answer: (d)

Same resonance: $L_1C_1 = L_2C_2 = \frac{1}{\omega_0^2}$ , so $\frac{1}{C_1} = \omega_0^2L_1$ and $\frac{1}{C_2} = \omega_0^2L_2$ .
In series: $L = L_1 + L_2$ and $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \omega_0^2(L_1 + L_2)$ .
$\omega^2 = \frac{1}{LC} = \frac{\omega_0^2(L_1+L_2)}{L_1+L_2} = \omega_0^2$
So the answer is (d): the resonant frequency is unchanged ( $f_r = f_0$ ).

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Resonance, Bandwidth & Q Factor question from MHT CET 2021 (23 Sep, Shift 2)

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