summaryrefslogtreecommitdiff
path: root/608/CH28/EX28.08/28_08.sce
diff options
context:
space:
mode:
Diffstat (limited to '608/CH28/EX28.08/28_08.sce')
-rwxr-xr-x608/CH28/EX28.08/28_08.sce33
1 files changed, 33 insertions, 0 deletions
diff --git a/608/CH28/EX28.08/28_08.sce b/608/CH28/EX28.08/28_08.sce
new file mode 100755
index 000000000..fc8407a4b
--- /dev/null
+++ b/608/CH28/EX28.08/28_08.sce
@@ -0,0 +1,33 @@
+//Problem 28.08: An R–L–C series circuit has a resonant frequency of 1.2 kHz and a Q-factor at resonance of 30. If the impedance of the circuit at resonance is 50 ohm determine the values of (a) the inductance, and (b) the capacitance. Find also (c) the bandwidth, (d) the lower and upper half-power frequencies and (e) the value of the circuit impedance at the half-power frequencies.
+
+//initializing the variables:
+Zr = 50; // in ohms
+fr = 1200; // in Hz
+Qr = 30; // Q-factor
+
+//calculation:
+//At resonance the circuit impedance, Z
+R = Zr
+wr = 2*%pi*fr
+//Q-factor at resonance is given by Qr = wr*L/R, then L is
+L = Qr*R/wr
+//At resonance r*L = 1/(wr*C)
+//capacitance, C
+C = 1/(L*wr*wr)
+//bandwidth,.(f2 − f1)
+bw = fr/Qr
+//upper half-power frequency, f2
+f2 = (bw + ((bw^2) + 4*(fr^2))^0.5)/2
+//lower half-power frequency, f1
+f1 = f2 - bw
+//At the half-power frequencies, current I
+//I = 0.707*Ir
+//Hence impedance
+Z = (2^0.5)*R
+
+printf("\n\n Result \n\n")
+printf("\n (a)inductance, L is %.3f H ",L)
+printf("\n (b)capacitance, C is %.2E F ",C)
+printf("\n (c)bandwidth is %.0f Hz ",bw)
+printf("\n (d)the upper half-power frequency, f2 is %.0f Hz and the lower half-power frequency, f1 is %.0f Hz ",f2,f1)
+printf("\n (e)impedance at the half-power frequencies is %.2f ohm ",Z) \ No newline at end of file