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A series LCR circuit with resistance R = 120 ohm has an angular resonance frequency of 4 x 10⁵ rad s⁻¹. At resonance the voltage across the resistor is 60 V and across the inductor is 40 V. (i) Find L and C. (ii) At what angular frequency does the current lag the applied voltage by 45 degrees? Construct four plausible option sets and select the computed one.

1. L = 0.2 mH, C = 1/32 microF (31.25 nF), frequency = 8 x 10⁵ rad/s
2. L = 0.1 mH, C = 1/16 microF, frequency = 2 x 10⁵ rad/s
3. L = 0.4 mH, C = 1/64 microF, frequency = 6 x 10⁵ rad/s
4. L = 0.2 mH, C = 1/16 microF, frequency = 4 x 10⁵ rad/s

Correct answer: L = 0.2 mH, C = 1/32 microF (31.25 nF), frequency = 8 x 10⁵ rad/s

Solution

At resonance I = V_R/R = 60/120 = 0.5 A. V_L = I*omega0*L -> 40 = 0.5 * (4e5) * L -> L = 40/(2e5) = 2e-4 H = 0.2 mH. From omega0 = 1/sqrt(LC): C = 1/(omega0² L) = 1/((4e5)² * 2e-4) = 1/(1.6e11 * 2e-4) = 1/(3.2e7) = 3.125e-8 F = 1/32 microF. For a 45 deg lag, X_L - X_C = R: omega L - 1/(omega C) = R. Substituting values and solving the quadratic gives omega = 8 x 10⁵ rad/s. Hence the first option.

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