# Average power, current, and capacitance of RLC circuit?

a) A series RLC circuit is connected across a 4.73×10^1 kHz source. The 1.00 H inductor and the 1.10×10^3 kohm resistor are fixed whereas the capacitor is variable. At what capacitance will the current in the resistor be a maximum?

b) What will be the current in the resistor if the source voltage is 4.00×10^1 V (rms) and the capacitance is 2.83 pF?

c) What is the average power dissipated in this circuit?

@texas: those answers are wrong :( did you convert the units in the question to SI units?

### 3 Answers

- oubaasLv 78 years agoFavorite Answer
a)

current will be max at resonance , i.e. :

Xc = Xl ....ωL = 1/ωC....ω^2*L*C = 1

C = 10^12/(47,300*6.2832)^2 = 11.32 pF

b)

xl = ωL = 6.2832*47,300 = 297.195 kohm

xc = 1/ωC = 10^12/(6.2832*47300*2.83) = 1,188.972 kohm

x = xc-xl = 891.776 kohm

r = 1,100 kohm

z = √r^2+x^2 = 1,416.074 kohm

I = 40*10^6/1.416 = 28.25 μA

P = r*i^2 = 1.1*10^6*28.25^2*10^-12 = 878 μwatt

- Anonymous4 years ago
your question is at a loss for words. The impedance at 159 ohms is 159 ohms. perchance you advise 159 Hz? use accurate the following: Xc = a million/(2?fC) XL = 2?fL X = XL – Xc Impedance Z = ?(R² + X²) .

- RogerLv 78 years ago
a) when XL = XC

XL = 2π(47.3)(1.0) = 297.2 ohms

XC = 1/(2π(47.3)C) = 297.2 ohms

C = 11.3 microfarads

b) f = 47.3 hertz, XC = 1/(2πfC) = 1.19 x10^9 ohms

XL = 297.2 ohms

Z = 1.1 x10^3 - j1.19 x 10^9 ohms

|Z| = 1.19 x 10^9 ohms

I = 40/1.19 x 10^9 = 3.36 x 10^-8 amps

P = (i^2)(R) = 3.96 x10^-12 watts