Answer to Question #136597 in Electricity and Magnetism for Caylin

Question

An audio amplifier drives a series RLC circuit consisting of an 8ohm loudspeaker, a 160µF capacitor and a 1,5mH inductor. The amplifier output is 15V rms at 500 Hz.

a) calculate power delivered to loudspeaker
b) calculate maximum power that amplifier could deliver
c) how capacitor would change for resonance to happen?

Please assist.

Answer 1

R = 8 ohm

$C = 160\; µF = 160*10^{-6}\; F$

$L = 1.5\; mH = 1.5\times10^{-3}\; H$

V = 15 V

f = 500 Hz

Peak voltage $V_0 = \sqrt{2}V$

$V_0 = 15\times\sqrt{2} = 21.21 \;V$

Circuit impedance:

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

$X_L = 2ΠfL = 2Π\times500\times1.5\times10^{-3}= 4.71\; ohm$

$X_C = \frac{1}{2ΠfC}= \frac{1}{2Π\times500\times160\times10^{-6}} = 1.99\; ohm$

$Z = \sqrt{8^2 + (4.71 – 1.99)^2}=8.45\; ohm$

Peak current through circuit:

$I_0 = \frac{V_0}{Z} = \frac{21.21}{8.45} = 2.51 \;A$

a) Power delivered to loudspeaker

$P_R = I_0^2R=2.51^2\times8=50.4\; W$

b) Maximum power that amplifier could deliver

$P = I_0^2Z = 2.51^2\times 8.45 = 53.23\; W$

Voltage across capacitor

$V_c = I_0X_C = 2.51 \times 1.99 = 4.99\; V$

c) Charge on capacitor

$Q = CV_c = 160\times 10^{-6} \times 4.99 = 0.798\; mC$