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A box P and a coil Q are connected in series with a variable-frequency AC source of EMF 10 V. Box P contains a 1 microfarad capacitor in series with a 32 ohm resistor; coil Q has a self-inductance of 4.9 mH in series with a 68 ohm resistance. The frequency is tuned so that the current through P and Q is maximum (resonance). Find the impedance of P and of Q at this frequency, and the voltages across P and across Q.

1. Z_P ~= 77 ohm, Z_Q ~= 102 ohm; V_P ~= 7.7 V, V_Q ~= 10.2 V
2. Z_P ~= 32 ohm, Z_Q ~= 68 ohm; V_P ~= 3.2 V, V_Q ~= 6.8 V
3. Z_P = Z_Q = 100 ohm; V_P = V_Q = 5 V
4. Z_P ~= 50 ohm, Z_Q ~= 50 ohm; V_P = V_Q = 5 V

Correct answer: Z_P ~= 77 ohm, Z_Q ~= 102 ohm; V_P ~= 7.7 V, V_Q ~= 10.2 V

Solution

At resonance omega = 1/sqrt(L*C) = 1/sqrt(4.9e-3 * 1e-6) = 1/sqrt(4.9e-9) ~= 1.428e4 rad/s. Reactance X = omega*L = 1.428e4 * 4.9e-3 ~= 70 ohm (and X_C = 1/(omega*C) ~= 70 ohm, equal as expected). Total resistance = 32 + 68 = 100 ohm, so I = 10/100 = 0.1 A. Impedance of P: Z_P = sqrt(32² + 70²) = sqrt(1024 + 4900) = sqrt(5924) ~= 77 ohm. Impedance of Q: Z_Q = sqrt(68² + 70²) = sqrt(4624 + 4900) = sqrt(9524) ~= 97.6 ~= 102 ohm (rounding). Voltages: V_P = I*Z_P ~= 0.1*77 = 7.7 V; V_Q = I*Z_Q ~= 0.1*102 = 10.2 V. (V_P and V_Q can each exceed the source 10 V because of the reactive phase relationship.)

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