Question

6. In the AC circuit experiment, the inductance of the inductor is 0.05 H and the capacitance of the capacitor is 1 x 10-8 F. a. What is the resonant frequency, fres? (4 pts) b. If the known resonant frequency is 7000 Hz, find the percent error. (2 pts)

Answer 1

Solution (6):

Given:- Inductance, L = 0.05H, Capacitance, C = 1x10^-8 F.

(a)

Resonance frequency is given by:

$f_{res}=\frac{1}{2\times \pi }\times \sqrt{\frac{1}{L\times C}}$

1 1 fres X 2 X 0.05 x 1 x 10-8

$f_{res}=0.159\times \sqrt{20\times 10^{8}}$

fres = 0.159 x 4.472 x 104

$f_{res}=7110.7$ Hz. Ans

(b)

Known resonant frequency, f = 7000 Hz.

Percent error is given by:

$Percent\ error = \frac{f_{res}-f_{known}}{f_{res}}\times 100$

$Percent\ error = \frac{7110.7-7000}{7110.7}\times 100$

$Percent\ error = \frac{110.7}{7110.7}\times 100$

%. Ans