An AC voltage of the form Δv = (100 V) sin(1000t) is applied to a series RLC circuit. Assume the resistance is 400 , the capacitance is 4.60 µF, and the inductance is 0.500 H. Find the average power delivered to the circuit.
W
v = Vmax*sinWt
v = 100*sin(1000t)
Vrms = 0.707 * 100 = 70.7 Volts.
W = 2pi*F = 1000
6.28*F = 1000
F = 159 Hz.
Xl = W*L = 1000 * 0.5 = 500 Ohms.
Xc=1/W*C=1/(1000*4.6*10^-6)=-217.4 Ohms
tan A = (Xl+Xc)/R
tanA = (500+(-217.4))/400 = 0.7065
A = 35.2o
Z = R/cos A = 400/cos35.2 = 489.5 Ohms
[35.2o].
Irms = Vrms/Z =
70.7/489.5[35.2]= 0.1444A[-35.2o]
P=Irms^2 * R = (0.1444)2 * 400= 8.34 W.
NOTE: All of the power is consumed in the resistor.
Well, to find the average power delivered to the circuit, we can use the formula P_avg = 0.5 * V_0 * I_0 * cos(ϕ), where V_0 is the maximum voltage, I_0 is the maximum current, and ϕ is the phase difference between voltage and current.
First, let's find the maximum voltage V_0. From the given equation, we have V(t) = (100 V) sin(1000t). Since the maximum value of sin is 1, V_0 = 100 V.
Now, let's find the maximum current I_0. We can calculate the impedance Z of the circuit using the formula Z = √(R^2 + (X_L - X_C)^2), where R is the resistance, X_L is the inductive reactance, and X_C is the capacitive reactance.
The inductive reactance X_L is given by X_L = ωL, where ω is the angular frequency and L is the inductance. Plugging in the values, we get X_L = (1000 rad/s) * (0.500 H) = 500 Ω.
The capacitive reactance X_C is given by X_C = 1 / (ωC), where C is the capacitance. Plugging in the values, we get X_C = 1 / ((1000 rad/s) * (4.60 µF)) = 1 / (4.60 × 10^-4 Ω) = 2174 Ω.
Now, let's calculate the impedance Z: Z = √(400^2 + (500 Ω - 2174 Ω)^2) ≈ 2495 Ω.
Since the voltage and impedance are in phase in a purely resistive circuit, the phase difference ϕ is 0 degrees. Therefore, cos(ϕ) = cos(0) = 1.
Finally, using the formula P_avg = 0.5 * V_0 * I_0 * cos(ϕ), we get P_avg ≈ 0.5 * (100 V) * (V_0 / Z) * (1) ≈ 2 W.
So, the average power delivered to the circuit is approximately 2 watts (W).
To find the average power delivered to the circuit, we can use the formula:
Average power = (1/2) * Vm * Im * cos(ϕ)
Where:
- Vm is the maximum voltage of the source
- Im is the maximum current in the circuit
- ϕ is the phase angle between the voltage and current waveforms
First, let's find the values of Vm, Im, and ϕ.
Given:
- AC voltage, Δv = (100 V) sin(1000t)
- Resistance, R = 400 Ω
- Capacitance, C = 4.60 µF (microfarads)
- Inductance, L = 0.500 H (henries)
The RMS voltage and current can be found using:
Vrms = Vm / sqrt(2)
Irms = Im / sqrt(2)
For a sinewave, the RMS value is the peak value divided by the square root of 2.
Vrms = 100 V / sqrt(2)
Vrms ≈ 70.7107 V
Now, let's find the impedance (Z) of the RLC circuit, which combines the resistance (R), capacitance (C), and inductance (L).
Impedance, Z = sqrt(R^2 + (Xl - Xc)^2)
Where:
- Xl = 2πfL (inductive reactance)
- Xc = 1 / (2πfC) (capacitive reactance)
- f = frequency (in Hz)
Given:
- Frequency, f = 1000 Hz
Xl = 2 * π * 1000 Hz * 0.500 H
Xl ≈ 3142.74 Ω
Xc = 1 / (2 * π * 1000 Hz * 4.60 x 10^-6 F)
Xc ≈ 6.876 Ω
Z = sqrt(400 Ω^2 + (3142.74 Ω - 6.876 Ω)^2)
Z ≈ 3143.32 Ω
Now, we can find the current (Im) using Ohm's Law:
Im = Vrms / Z
Im ≈ (70.7107 V) / 3143.32 Ω
Im ≈ 0.0225 A
To find the phase angle (ϕ) between the voltage and current, we can use:
ϕ = arctan((Xl - Xc) / R)
ϕ = arctan((3142.74 Ω - 6.876 Ω) / 400 Ω)
ϕ ≈ 1.555 rad
Finally, we can calculate the average power delivered to the circuit:
Average power = (1/2) * Vm * Im * cos(ϕ)
Average power ≈ (1/2) * (100 V) * (0.0225 A) * cos(1.555 rad)
Average power ≈ 0.793 W
Therefore, the average power delivered to the circuit is approximately 0.793 watts.
To find the average power delivered to a circuit, you need to calculate the average of the instantaneous power over a complete cycle. The instantaneous power in an AC circuit can be expressed as P(t) = V(t) * I(t), where V(t) is the voltage across the circuit and I(t) is the current flowing through the circuit at time t.
In a series RLC circuit, the current can be found using the voltage across the circuit and the total impedance (Z) of the circuit, which is the combination of the resistance (R), inductive reactance (XL), and capacitive reactance (XC).
The voltage across the circuit is given as Δv = (100 V) sin(1000t).
The total impedance (Z) can be calculated using the following formula:
Z = sqrt(R^2 + (XL - XC)^2)
where XL = ωL and XC = 1/ωC, and ω is the angular frequency of the AC voltage signal, given by ω = 2πf, where f is the frequency.
In this case, the angular frequency ω is 2π times the frequency, which is 1000 Hz. Given the values of R, L, and C, you can calculate the impedance.
Once you have the impedance, you can calculate the current by dividing the voltage by the impedance, I(t) = V(t) / Z.
Now, you can calculate the instantaneous power at any given time t by multiplying the voltage and current, P(t) = V(t) * I(t).
To find the average power, you need to integrate the instantaneous power over a complete cycle and divide it by the period of the signal. Since the signal is a sine wave with a frequency of 1000 Hz, the period T is given by T = 1/f, where f is the frequency.
After finding the average power P_avg, you can substitute the given values into the equations and calculate the result.