Solve AC Voltage Form: RLC Series Circuit Power

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SUMMARY

An AC voltage of the form V = (100 V) sin(1000t) is applied to a series RLC circuit with a resistance of 400 Ohms, a capacitance of 5.00 µF, and an inductance of 0.500 H. The angular frequency (ω) is determined to be 1000 rad/s, leading to a frequency (F) of 159.154 Hz. The peak voltage is 100 V, and the actual voltage varies between +100 V and -100 V based on the sine function. Understanding these parameters is crucial for calculating the average power delivered to the circuit.

PREREQUISITES
  • Understanding of RLC circuit components: Resistance, Inductance, Capacitance
  • Knowledge of AC voltage representation and sine functions
  • Familiarity with angular frequency and its relation to frequency
  • Basic concepts of power calculation in AC circuits
NEXT STEPS
  • Learn how to calculate total impedance (Ztot) in RLC circuits
  • Study the calculation of average power (P) in AC circuits
  • Explore the concept of Power Factor (P.F) and its significance
  • Investigate the effects of varying frequency on RLC circuit behavior
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Electrical engineering students, circuit designers, and anyone involved in analyzing AC circuits and power calculations.

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Homework Statement



You don't even have to read the whole thing... just the red part is sufficient.
I saw this on the homework: V= (100 v) sin (1 000t)
What does this mean?

-----
Can I treat it as having 100 real magnitude and 0 imaginary magnitude (and therefore 0 phase angle)? or does sin(1 000t) tells something about the phase angle?
or does sin (1 000t) tells something about angular velocity or frequency?

I can't solve this problem without knowing frequency or angular velocity

just for side note: let...
V= delta voltage
v= unit volts
uF= micro farad
H= Henries

RLC= i believe it stands for Resistance, Inductor, Capacitor
An AC voltage of the form V= (100 v) sin (1 000t) is applied to a series RLC circuit. Assume the resistance is 400 Ohms, the capacitance is 5.00 uF, and the inductance is 0.500 H. Find the average power delivered to the circuit

Homework Equations


w= angular velocity (supposed to be omega)
L= inductor
C= capacitor
j= complex coefficient = sqrt(-1)
Zl= resistance of inductor
Zc= resistance of capacitor
Ztot= total resistance
R= resistance of resistor
P= Power

Xl=wL
Xc=1/wc
Zl=j*Xl
Zc=-j*Xc
Ztot= R + Zl + Zc = R + j(Xl-Xc)

P=I*V*(P.F)

P.F= Power Factor. Is is cos \varphi
the angle between voltage and current.

by default everything is in RMS (rootmeansquare), but it shouldn't matter in calculation)

The Attempt at a Solution



well it's easy except I don't know frequency nor angular velocity.
Otherwise finding Ztot will be easy and finding power is easy as well
 
Last edited:
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Your voltage is V= (100 v) sin (1 000t), which is in the form V=V0sin(ωt).

hence the amplitude is V0 and the angular frequency is ω. Compare the terms and get ω.
 
The voltage at any instant T is V sin (wT) where w is (should be) omega, the angular velocity.

So, V is the peak voltage.

Omega = 2 * PI * F

In this case, omega = 1000 = 2 * PI * F ... so F = 159.154 Hz. ( ie 1000 / (2 * pi) )

V is the maximum voltage, but the actual voltage depends on the sine function, so the actual voltage can be anywhere between plus 100 volts and minus 100 volts, including zero.

For example what would the voltage be after 0.2 seconds?
V = 100 * sin (2 * pi * 159.154 * 0.2 ) or -34.2 volts
 
rock.freak667 said:
...
which is in the form V=V0sin(ωt).
...
hence the amplitude is V0 and the angular frequency is ω. Compare the terms and get ω.

vk6kro said:
The voltage at any instant T is V sin (wT) where w is (should be) omega, the angular velocity.

So, V is the peak voltage.

V is the maximum voltage,

Thank you very much. I understood it now =D
 

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