In a series circuit, a generator (1400 Hz, 12.0 V) is connected to a 15.0- resistor, a 3.80-ìF capacitor, and a 5.70-mH inductor. Find the voltages across (a) the resistor, (b) the capacitor, and (c) the inductor.

Answer 1

Xl = 2pi*F*L = 6.28*1400*0.0057 = j50 ohms.
Xc = 1/(2pi*F*C) = 1/(6.28*1400*3.8*10^-6) = -j30 ohms.
Z = R + j(Xl-Xc) = 15 + j(50-30) = 15 + j20 = 25 ohms[53o].
I = E/Z = 12/25[53o] = 0.48A[-53o]. The current lags the voltage by 53o.

a. Vr = I*R = 0.48[-53] * 15 = 7.2V.[-53o].
b. Vc = I*Xc = 0.48[-53o] * 30V[-90o] = 14.4V.[-143o]. Lags current by 90o.
c. Vl = I*Xl = 0.48[-53o] * 50[90o] = 24V.[37o]. Leads current by 90o.

E = Vr+Vc+Vl = 7.2[-53o] + 14.4[-143o] + 24[37o]

Answer 2

Check: E = 12 + 0.027i = 12.0000304V[0.128o].
The magnitude of Vc and Vl is greater than the supply voltage, but the
vector sum = the 12V.-supply as it should.