The circuit opposite can be used as a \"Q meter\". With a current generator supp

Question

The circuit opposite can be used as a "Q meter". With a current generator supplying a current i = sin 4000t milliamps, C is adjusted until a minimum voltage, V, appears across the entire RLC circuit. Under these conditions, it is found that V = 0.3 V and the voltage across the capacitor, Vc = 12 V. Calculate the Q of the circuit. Calculate C, L, and r. Hints: Which quantities are peak and which rms? How does compare with Vl at resonance? Take care whether you are using omega or f. Design a series RLC resonant circuit for a resonant frequency of 60 kHz and a bandwidth of 1.2 kHz. The available coil has inductance of 3.183 mH and a Q of 60.

Explanation / Answer

a) as the setup is done the value of C is adjusted the circuit goes in resonance

so w = 4000 cause i = I*sin(wt)

so V = IR = 0.3

==> R = 0.3/6*sqrt(2) = 0.04 ohms

at resonance VL = VC = 12 V

==> VL = I*wL

==> 12 = 6*sqrt(2)*w*L

w = 4000

==> L = 12/(6*sqrt(2)*4000)

so Q = w0*L/R = 4000*(12/(6*sqrt(2)*4000))/(0.3/(6*sqrt(2))) = 40

b) r = 0.04 ohms

L = 0.4 milli Henry

C = 1/w^2L = 1/(4000*(12/(6*sqrt(2)*4000))) = 0.2 milli Farads

2) Q = w0*L/r

==> r= w0*L/Q = 60*10^3*3.183*10^-3/60 = 3.183 ohms

Q = 1/w0RC

==> C = 1/(wo*R*Q) = 1/(60*10^3*3.183*60) = 0.087 micro Farads

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.