Answer: a) 4623 rad/sec b) 26.5 A c) 60 ohms. d) 3.2 ohms. e) 0º
Explanation:
a) At resonance, the circuit behaves like it were only resistive, so the reactive part of the impedance, is equal to 0.
So, the capacitive reactance and the inductive reactance are equal each other , as follows:
Xc = Xl ⇒ ω₀L = 1/ω₀C ⇒ ω₀ = 1/√LC
Replacing by the values of L and C, we get the value for ω₀, the angular frequency at resonance, as follows:
ω₀ = 4623 rad/sec
b) At resonance, the current can be obtained applying the Ohm´s Law as for any resistive-only circuit, as follows:
Irms = Vrms / R = (√2/2) Vmax / R = 85 V / 3.2 Ohms = 26.5 A
c) Xc = 1/ω₀C = 60 Ohms.
Xl = ω₀L = 60 Ohms.
d) At resonance, as the total reactance becomes 0, the only component of the impedance that remains is the resistive part, so we can write the following : Z= R = 3.2 Ohms.
e) As the circuit behaves like if it were resistive only, voltage and current are in phase at any point of the circuit, so θ = 0º.
