Ask a New Question

Question

consider a series R-C circuit consisting of a 30 ohm resistor and 20*10^-6F capacitor connected to a 100V,50Hz power supply.find the magnitude of the current and the angle by which the current leads the voltage.
9 years ago

Answers

Henry
Xc = 1/WC = -1/(314.2*20*10^-6) = -159.2 Ohms.

Z(Ohms) = 30 -j159.2 = 162[-79.3o].

I = E/Z = 100[0o]/162[-79.3] = 0.617A[79.3o].

9 years ago

Related Questions

In a certain series RLC circuit, Irms = 9.00 A, ΔVrms = 150 V, and the current leads the voltage by... a series RL circuit contains two resistors and two inductors. the resistors are 33 ohms and 47 ohms.... A series RL circuit has two resistors and two inductors. The resistors dissipate 7 W and 12 W. The i... In a series R-L-C circuit at resonance, impedance is In a series R-L-C circuit at resonance, impedance is A.MAXIMUM B.MINIMUM C.CAPACITIVE D.IMDUCTIV... A series RLC circuit is used to tune a radio set to receive NOUN RADIO broadcasting at 105.9 MHz in... In a series circuit, if one bulb burns out, what happens? A.All the other bulbs cannot light up B... In the circuit of series and parallel arrangement,c1=4microfarad,c2=3microfarad and c3=1microfarad,g... 4. In a series circuit there are 17 light bulbs, each providing 1 ohm of resistance. What is the t... You have a series circuit with a current of 6 amps and three resistors on it, with resistances of 10...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use