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RLC Circuit Consider a series RLC circuit with L = 0.5H, C = 3.0 μF, R = 10.02 is connected to an AC voltage source V(t) = Vo sin(wt) which has a maximum voltage amplitude Vo = 10V. (a) What is the resonance frequency wo? (b) Find the rms current(Irms) at resonance. (c) Let apply a drive at frequency ofw = 3000 rad/s. Assume the current response is given by I(t) = Io sin(wt-o) Calculate the amplitude of the current(Io) and the phase shift(o) between the current and the driving voltage. .

RLC Circuit Consider a series RLC circuit with L = 0.5H, C = 3.0 μF, R = 10.02 is connected to an AC voltage source V(t) = Vo sin(wt) which has a maximum voltage amplitude Vo = 10V. (a) What is the resonance frequency wo? (b) Find the rms current(Irms) at resonance. (c) Let apply a drive at frequency ofw = 3000 rad/s. Assume the current response is given by I(t) = Io sin(wt-o) Calculate the amplitude of the current(Io) and the phase shift(o) between the current and the driving voltage. .

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RLC Circuit = Consider a series RLC circuit with L = 0.5 H, C = 3.0 μF, R V(t) = Vo sin(wt) which has a maximum voltage amplitude Vo = 10V. = 10.02 is connected to an AC voltage source (a) What is the resonance frequency wo? (b) Find the rms current(Irms) at resonance. = (c) Let apply a drive at frequency of w 3000 rad/s. Assume the current response is given by I(t) = I₁ sin(wt — 6) Calculate the amplitude of the current(I) and the phase shift() between the current and the driving voltage.
Transcribed Image Text:RLC Circuit = Consider a series RLC circuit with L = 0.5 H, C = 3.0 μF, R V(t) = Vo sin(wt) which has a maximum voltage amplitude Vo = 10V. = 10.02 is connected to an AC voltage source (a) What is the resonance frequency wo? (b) Find the rms current(Irms) at resonance. = (c) Let apply a drive at frequency of w 3000 rad/s. Assume the current response is given by I(t) = I₁ sin(wt — 6) Calculate the amplitude of the current(I) and the phase shift() between the current and the driving voltage.
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Hello, this part is a bit vague. What is the content of this part? I can't see clearly. . . Thank you so much

:: Impedance (2)=√√√√₁² + (x₂-xc)² 2 = √10² + (1500-111-11) ² = 1388.9262 Vo 40° I => 10 20⁰ R+J (xxc) 10+3 (1500-111-11) MA I = 7•2 × 10³ 2-89-58° = √7.2 sin (wt - 89.58) I₂ = 17-2 MA | AM² 40 = 189.58"] ANG WAS THIS HELPFUL?
Transcribed Image Text::: Impedance (2)=√√√√₁² + (x₂-xc)² 2 = √10² + (1500-111-11) ² = 1388.9262 Vo 40° I => 10 20⁰ R+J (xxc) 10+3 (1500-111-11) MA I = 7•2 × 10³ 2-89-58° = √7.2 sin (wt - 89.58) I₂ = 17-2 MA | AM² 40 = 189.58"] ANG WAS THIS HELPFUL?
Step 3 в) быу? Урсон =vo=5 XV yim's .: Vrmis Voimis 5 707 vot 10V қ
Transcribed Image Text:Step 3 в) быу? Урсон =vo=5 XV yim's .: Vrmis Voimis 5 707 vot 10V қ
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