Question

You have a 150-? resistor, a 0.400-H inductor, a 8.00-µF capacitor, and a variable-frequency ac source with an amplitude of 3.00 V. You connect all four elements together to form a series circuit.

(a) At what frequency will the current in the circuit be greatest?
  Hz

What will be the current amplitude at this frequency?
  mA

(b) What will be the current amplitude at an angular frequency of 370 rad/s?
mA

At this frequency, will the source voltage lead or lag the current?

leadlag    neither

Answer 1

A.

Imax = Vmax/Z

Frim above expression we can see that current will be maximum when impedance is minimum

Z = sqrt (R^2 + (XL - Xc)^2)

Z will be minimum when

XL - Xc = 0

XL = Xc

2*pi*f*L = 1/(2*pi*f*C)

f = 1/(2*pi*sqrt (LC))

f = 1/(2*pi*sqrt (0.4*8*10^-6))

f = 88.97 Hz

Part B

Imax = Vmax/Z

Z = R at f = 88.97, So

Z = 150 ohm

Imax = 3/150 = 0.02 Amp = 20 mA

Part B

when w = 370 rad/sec

XL = w*L = 370*0.4 = 148 ohm

Xc = 1/(wC) = 1/(370*8*10^-6) = 337.84 ohm

Z = sqrt (150^2 + (148 - 337.84)^2)

Z = 241.95 ohm

Imax = Vmax/Z = 3/241.95 = 12.4*10^-3 Amp

Imax = 12.4 mA

When w = 370 rad/sec, phase angle will be

phi = arctan ((XL - Xc)/R)

phi = arctan ((148 - 337.84)/150)

phi = -51.7 deg

Since phase angle is negative, voltage will lag the current.
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