The figure below shows a series RLC circuit with a 20.0-Ω resistor, a 390.0-mH inductor, and a 23.0-µF capacitor connected to an AC source with
Vmax = 60.0 V operating at 60.0 Hz. What is the maximum voltage across the following in the circuit?
(a) resistor
(b) inductor
(c) capacitor
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Calculate Xl and Xc. Now calcule total series impedance (R +Xl + Xc, using using vectors or the Pythorigean (the two impedances are opposite, so you are dealing with the net reactance).
Now with total impedance, you can calculate current. Then voltage. "Max Voltage" is a bit confusing, is the question asking for rms, or peak? I have no idea.
Given: R = 20 ohms, L = 390 mH(0.39h), C = 23 uF(23*10^-6Farads).
E = 60V.rms? @ 60 Hz.
Xl = 2pi*F*L = 6.28*60*0.39 = j147 ohms.
Xc = 1/(2pi*60*23*10^-6) = -j115.3 ohms.
Z = R + j(Xl-Xc) = 20 + j(147-115 ) = 20 + j32 = 37.7 ohms[58o].
I = E/Z = 60[0o]/37.7[58o] = 1.59A[-58o]. 0-58 = -58.
a. Vr = I*R = 1.59[-58] * 20 = 31.8A[-58o]. Vr is in phase with the current.
b. Vl = I*Xl = 1.59[-58] * 147[90] = 233.7V.[32o]. Vl leads current by 90o.
c. Vc = I*Xc = 1.59[-58] * 115[-90] = 182.9V.[-148o]. Vc lags current by 90o.
Correction: Vr = 31.8V. NOT 31.8A.
Check: E = 31.8[-58] + 233.7[32] + 182.9[-148]
E = (31.8*cos(-58)+233.7*cos32+182.9*cos(-148)) +
j(31.8*sin(-58)+233.7*sin32+182.9*sin(-148)
E = 59.93 - j0.048 = 59.9300192V[0.046o].
The voltage across L and C is greater than 60 volts, but the vector sum is
approximate = to the supply voltage.
To find the maximum voltage across each component in the circuit, we need to use the concept of impedance and the properties of series RLC circuits.
(a) The maximum voltage across the resistor can be found by using Ohm's law. Since the resistor is connected in series with the AC source, the maximum voltage across it will be the same as the maximum voltage of the AC source, which is Vmax = 60.0 V.
(b) The maximum voltage across the inductor can be calculated using the equation V_L = I * XL, where I is the current and XL is the inductive reactance. The inductive reactance can be calculated using the formula XL = 2πfL, where f is the frequency and L is the inductance. In this case, the frequency is 60.0 Hz and the inductance is 390.0 mH (or 0.390 H). So, XL = 2π * 60.0 * 0.390 = 46.55 Ω.
The current can be found using the formula I = Vmax / Z, where Z is the total impedance of the circuit. In a series RLC circuit, the total impedance is given by Z = sqrt(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The capacitive reactance can be calculated using the formula XC = 1 / (2πfC), where C is the capacitance. In this case, the capacitance is 23.0 µF (or 23.0 x 10^-6 F). So, XC = 1 / (2π * 60.0 * 23.0 x 10^-6) = 122.83 Ω.
Plugging in these values, we have Z = sqrt(20.0^2 + (46.55 - 122.83)^2) = 134.97 Ω.
Now, we can calculate the current: I = 60.0 / 134.97 = 0.444 A.
Finally, we can find the maximum voltage across the inductor: V_L = 0.444 * 46.55 = 20.66 V.
(c) Similarly, the maximum voltage across the capacitor can be calculated using the equation V_C = I * XC, where I is the current and XC is the capacitive reactance. We have already calculated XC in part (b) as 122.83 Ω. Using the same current value of 0.444 A, we can find the maximum voltage across the capacitor: V_C = 0.444 * 122.83 = 54.54 V.
Therefore, the maximum voltage across each component in the circuit is:
(a) resistor: 60.0 V
(b) inductor: 20.66 V
(c) capacitor: 54.54 V.