12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L = 400 mH, and C = 10 μF. Find the bandwidth of the network, the half-power frequencies, and the voltage across the network at the half-power frequencies.

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V (jwlow) = 1414.21 < 45°V

V (jwhigh) = 1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1 / 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1 / (SQRT LC)

wo = 1/SQRT 400*10¯³ * 10*10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q + SQRT (1/2Q) ² + 1]

wlow = 500 [-1/2*5 + SQRT (1/2*5) ² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q + SQRT (1/2Q) ² + 1]

whigh = 500 [1/2*5 + SQRT (1/2*5) ² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y (jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y (jwlow) = (1/1000) + (1/j*452.49*400*10¯³) + (j*452.49*10*10^¯6)

Y (jwlow) = 0.001 - j5.525*10¯³ + j4.525*10¯³

Y (jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V (jwlow) = I/Y (jwlow)

V (jwlow) = 2 / (1-j).10¯³

V (jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y (jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y (jwhigh) = (1/1000) + (1/j*552.49*400*10¯³) + (j*552.49*10*10^¯6)

Y (jwhigh) = 0.001 - j4.525*10¯³ + j5.525*10¯³

Y (jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V (jwhigh) = I/Y (jwhigh)

V (jwhigh) = 2 / (1+j).10¯³

V (jwhigh) = 1414.2 < - 45°V