A series RLC circuit with a resistance of 116.0 Ω has a resonance angular frequency of 5.1 ✕ 105 rad/s. At resonance, the voltages across the resistor and inductor are 60.0 V and 40.0 V, respectively.
(b) At what frequency does the current in the circuit lag the voltage by 45°?
find current at resonance: 60/116
find Xl at resonance (which also is Xc). From those, find L, and C.
so if we want current lagging, the Xl-Xc has to be the net reactance, and you want 45 deg or Xl-Xc to be equal to R
now, knowing L and C (and R) what w will give that Xl-Xc?
At resonance:
I = V/R = 60/116 = 0.52A.
Xl = Vl/I = 40/0.52 = 77.3 ohms.
Xl = 2pi*F*L = 77.3
6.28*5.1*10^5L = 77.3
L = 2.41*10^-5 h.
At -45 degrees:
Xl = R = 116 ohms.
Xl = 2pi*F*L = 116
6.28*F*2.41*10^-5 = 116
F = 7.65*10^5 rad/s.
To find the frequency at which the current in the circuit lags the voltage by 45°, we first need to calculate the impedance of the circuit at resonance.
The impedance of an RLC circuit is given by the formula:
Z = √(R^2 + (XL - XC)^2)
Where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.
Given that the resistance (R) of the circuit is 116.0 Ω and the resonance angular frequency (ω) is 5.1 ✕ 10^5 rad/s, we can calculate the inductive reactance (XL) and capacitive reactance (XC) as follows:
XL = ωL
= (5.1 ✕ 10^5 rad/s)(40.0 V)
= 2.04 ✕ 10^7 Ω
Where L is the inductance.
To calculate XC, we can use the formula:
XC = 1 / (ωC)
Let's assume the capacitance (C) is C Farads. Rearranging the formula, we get:
C = 1 / (XCω)
Substituting the values we know, we get:
C = 1 / ((40.0 V) (5.1 ✕ 10^5 rad/s))
= 4.90 ✕ 10^(-9) F
Now that we have the values for R, XL, and XC, we can calculate the impedance (Z) at resonance:
Z = √(116.0 Ω)^2 + (2.04 ✕ 10^7 Ω - 1 / (5.1 ✕ 10^5 rad/s)(4.90 ✕ 10^(-9) F))^2)
Calculating this equation gives us the value of Z at resonance.
Once we have the impedance at resonance, we can find the angular frequency (ω0) at which the current lags the voltage by 45°. The formula relating the impedance and the lag angle is:
tan(θ) = (XL - XC) / R
Where θ is the phase angle.
Solving for θ, we find:
θ = arctan((XL - XC) / R)
= arctan((2.04 ✕ 10^7 Ω - 1 / (5.1 ✕ 10^5 rad/s)(4.90 ✕ 10^(-9) F)) / 116.0 Ω)
Once we have the phase angle, we can calculate the angular frequency (ω) at which the current lags the voltage by 45° using the formula:
ω = ω0 - θ
Where ω0 is the resonance angular frequency.
Calculating this equation gives us the value of ω at which the current lags the voltage by 45°. Finally, we can convert this angular frequency to frequency by dividing it by 2π:
f = ω / (2π)
Calculating this equation gives us the frequency at which the current lags the voltage by 45° in Hz.