Question

A series *RLC* circuit with a resistance of 122.0 Ω has a
resonance angular frequency of 3.4 ✕ 10^{5} rad/s. At
resonance, the voltages across the resistor and inductor are 60.0 V
and 40.0 V, respectively.

(a) Determine the values of *L* and *C*.

L |
= | ............. H |

C |
= | ..................F |

(b) At what frequency does the current in the circuit lag the
voltage by 45°?

......................Hz

Answer #1

(a)

Impedance of the series RLC circuit,

Z = (R^2 + (wL - 1 / wC)^2)

At resonance,

Z = R

Voltage across resistor, V = 60 V

Voltage across the inductor = V*wL / Z = 40

(V / R)*wL = 40

(60 / 122) * 3.4*10^{5}*L = 40

**L = 0.232 mH**

At resonance, wL = 1 / wC

C = 1 / w^2*L

C = 1 /
[(3.4*10^{5})^{2}*0.232*10^{-3}]

**C = 3.72*10 ^{-8} F**

(b)

the circuit lag the voltage by 45°,

tan(phi) = (1 / wC - wL) / R = -1

(1 / w*3.72*10^{-8} - w*0.232*10^{-3})/ 122 =
-1

0.268*10^8 / w - w*0.232*10^{-3} = -122

By solving,

w = 5.26*10^5 rad/s

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