Various RLC Circuit Problems (Resonance Frequency, Phase, Current)

In summary, the resonance frequency of the RLC circuit is 0.447 Hz, the phase angle is 0.0229°, and the circuit impedance is 20000.0001Ω.
  • #1
Spaghetti
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Homework Statement


An RLC circuit has L = 20mH, R = 20kΩ, C = 250μF.

1) Calculate the resonance frequency for this circuit.

2) For a frequency ω=400 rad/s, determine the phase angle and the circuit impedance.

3) Using ε = 30[V]sin(ωt), calculate the current in the circuit. This will be a function of time.

4) Draw the phasor diagram for t=0 seconds.

Homework Equations


ω[itex]_{o} = \frac{1}{\sqrt{LC}}[/itex]

ϕ = tan[itex]^{-1} \left[\frac{X_{L}-X_{C}}{R}\right][/itex]

[itex]X_{L}[/itex] = ωL

[itex]X_{C} = \frac{1}{ωC}[/itex]

Z = [itex]\sqrt{R^{2} + (X_{L} - X_{C})^{2}}[/itex]

I(t) = [itex]\frac{ε}{Z}[/itex]

The Attempt at a Solution


So I wasn't sure if I was doing any of this correctly; with all of the formulas/equations, it looked like the questions were mostly some simple plug-in questions, but I feel like my numbers just weren't coming out quite right.

1) Calculate the resonance frequency for this circuit.

ω[itex]_{o} = \frac{1}{\sqrt{LC}}[/itex]
L = 20mH = 0.02H; C = 250μF
ω[itex]_{o} = \frac{1}{\sqrt{(0.02H)(250μF)}} = \frac{1}{2.24 sec}[/itex] = 0.447 Hz

2) For a frequency ω=400 rad/s, determine the phase angle and the circuit impedance.
Phase Angle ϕ = tan[itex]^{-1} \left[\frac{X_{L}-X_{C}}{R}\right][/itex]

[itex]X_{L} = ωL = (400)(0.02) =[/itex] 8
[itex]X_{C} = \frac{1}{(400)(250)} = \frac{1}{100,000} =[/itex] 0.00001
R = 20kΩ

ϕ = tan[itex]^{-1} \left[\frac{X_{L}-X_{C}}{R}\right] = tan^{-1} \left[\frac{8-0.00001}{20,000}\right] = tan^{-1} \left[\frac{7.99999}{20,000}\right] = tan^{-1} \left[3.99x10^{-4}\right] =[/itex] 0.0229°

Impedance Z = [itex]\sqrt{R^{2} + (X_{L} - X_{C})^{2}} = \sqrt{20,000^{2} + (8 - 0.00001)^{2}} = \sqrt{4.0x10^{8} + 63.99} =[/itex] 20000.0016Ω

3) Using ε = 30[V]sin(ωt), calculate the current in the circuit. This will be a function of time.

I(t) = [itex]\frac{ε}{Z} = \frac{30[V]sin(400[rad/s]t)}{20000.0016[Ω]} = \frac{30[V]sin(400[rad/s]t)}{20000.0016[Ω]} = 0.00149sin(400t)[/itex]

So I(t) = 0.00149sin(400t)

4) Draw the phasor diagram for t=0 seconds.

I wasn't sure at all how to go about doing this one. As far as I can tell, this one ends up being a graph with voltage and current functions (current found in #3) drawn at t=0, showing how out of phase they are at that time, although I'm not sure how to find the voltage/what to use to find it. Of course, I don't want anyone to draw the graph for me, but some advice on phasor diagrams in general/a push in the right direction would be very nice.

Any help would be greatly appreciated. Thank you in advance!
 
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  • #2
The capacitance is given in microfarads: C=250 *10-6 F. Repeat the calculations.

ehild
 
  • #3
Alrighty. I guess my mind totally drew a blank around that capacitance. :blushing:

I feel a lot better about the frequency now; I'm hoping the others are okay, too.

1) Calculate the resonance frequency for this circuit.
ω[itex]_{o} = \frac{1}{\sqrt{LC}}[/itex]
L = 20mH = 0.02H; C = 250x10^{-6}F
ω[itex]_{o} = \frac{1}{\sqrt{(0.02H)(250x10^{-6}F)}} = \frac{1}{.00223 sec}[/itex] = 447.21 Hz

2) For a frequency ω=400 rad/s, determine the phase angle and the circuit impedance.
Phase Angle ϕ = tan[itex]^{-1} \left[\frac{X_{L}-X_{C}}{R}\right][/itex]

[itex]X_{L} = ωL = (400)(0.02) =[/itex] 8
[itex]X_{C} = \frac{1}{(400)(250x10^{-6})} = \frac{1}{.1} =[/itex] 10
R = 20kΩ

ϕ = tan[itex]^{-1} \left[\frac{X_{L}-X_{C}}{R}\right] = tan^{-1} \left[\frac{8-10}{20,000}\right] = tan^{-1} \left[\frac{-2}{20,000}\right] = tan^{-1} \left[-1.0x10^{-4}\right] =[/itex] -0.0057°

Impedance Z = [itex]\sqrt{R^{2} + (X_{L} - X_{C})^{2}} = \sqrt{20,000^{2} + (8 - 10)^{2}} = \sqrt{4.0x10^{8} + 4} =[/itex] 20000.0001Ω

3) Using ε = 30[V]sin(ωt), calculate the current in the circuit. This will be a function of time.

I(t) = [itex]\frac{ε}{Z} = \frac{30[V]sin(400[rad/s]t)}{20000.0001[Ω]} = \frac{30[V]sin(400[rad/s]t)}{20000.0001[Ω]} = 0.0015sin(400t)[/itex]

So I(t) = 0.0015sin(400t)

4) Draw the phasor diagram for t=0 seconds.
I'm still a little confused about this phasor diagram, but I think I can get by. Again, if anyone could offer a little advice on this, it would be great.

And thank you for catching my mistake, ehild.
 

FAQ: Various RLC Circuit Problems (Resonance Frequency, Phase, Current)

What is resonance frequency in an RLC circuit?

Resonance frequency in an RLC (resistor, inductor, and capacitor) circuit is the frequency at which the circuit exhibits maximum energy transfer. At resonance frequency, the capacitive reactance and inductive reactance cancel out, leaving only the resistance. This results in a maximum current flow through the circuit.

How is resonance frequency calculated in an RLC circuit?

The resonance frequency in an RLC circuit can be calculated using the formula: fr = 1 / (2π√(LC)), where fr is the resonance frequency, L is the inductance in Henrys, and C is the capacitance in Farads.

What is the phase relationship between voltage and current in a resonant RLC circuit?

In a resonant RLC circuit, the voltage and current are in phase with each other. This means that they reach their maximum and minimum values at the same time. This is because at resonance frequency, the capacitive and inductive reactances cancel out, leaving only the resistance, which is responsible for the phase relationship.

How does a change in capacitance or inductance affect the resonance frequency in an RLC circuit?

A change in capacitance or inductance will affect the resonance frequency in an RLC circuit. An increase in capacitance will decrease the resonance frequency, while an increase in inductance will increase the resonance frequency. This is because the resonance frequency is inversely proportional to the square root of the product of inductance and capacitance.

What is the difference between series and parallel resonance in an RLC circuit?

In series resonance, the inductor and capacitor are connected in series with each other, while in parallel resonance, they are connected in parallel. In series resonance, the current through the inductor and capacitor is the same, while in parallel resonance, the voltage across the inductor and capacitor is the same. Additionally, in series resonance, the impedance of the circuit is at its minimum, while in parallel resonance, the impedance is at its maximum.

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