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.
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
www.google.com/search?q=www.webassign.net/katzpse1/33-p-071.png&tbm=isch&source=iu&ictx=1&fir=DZ2zukEkxA46DM%253A%252CZYncnlTBgSNt9M%252C_&vet=1&usg=AI4_-kRfYBH5nY9Bw1ChMeJ5g2KiokZxWw&sa=X&ved=2ahUKEwi3wvSXnvnoAhXklXIEHaaLAG0Q9QEwAXoECAQQBw#imgrc=DZ2zukEkxA46DM:
4 answers
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.
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.
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.
3 answers