Question

A coil of inductance 2mH and resistance 15Ω is connected with in parallel ith capacitor of 0.01μF. The peak oscillator current at resonance is?

Answer 1

As per phaser diagram, For circuit to be resonance, Impedance (resistance) of circuit must be minimum.

So,

$I_C = I_Lsin\theta$

simply putting value of $I_C , I_L$ and $sin\theta$

$\frac{V}{X_C} = \frac{V}{Z_L}(\frac{X_L}{Z_L})$

$X_CX_L = Z_L^2$ . . . . . . (i)

since, $X_C = \frac{1}{\omega C}$ , $X_L = {\omega L}$ and $Z_L^2 = R^2 + X_L^2$

then equation (i) will be,

$\frac{\omega L}{\omega C} = R^2 + (\omega L)^2$

$(\omega)^2 = (2\pi f_r)^2= [\frac{1}{LC} - {\frac{R}{L^2}}]$

$f_r =(\frac{1}{2\pi}) \sqrt{\frac{1}{LC} - {\frac{R}{L^2}}}$ is the resonance frequency.

Again,

$\frac {V}{Z_f} = \frac{V}{Z_L}(\frac{R}{Z_L})$ where $Z_r$ is the total effective impedance of the circuit at resonance.

$Z_f = {\frac{Z_L^2}{R}}$

using (i) again here,

$Z_f = \frac{X_CX_L}{R} = \frac{\omega L}{R \omega C} = \frac{L}{RC}$

Now using these formulas,

Impedance of the circuit will be,

$Z_f = \frac{L}{LC} = \frac{2*10^-3}{15*0.01*10^-6} = 13.33*10^3 \Omega$

For current, use ohm's law,

$I = \frac{V}{Z_f} = \frac{VCR}{L}$ . . . . . . . (ii)

If V = 1V then current will be, using equation(ii) putting value of $Z_f$

$I = 0.075*10^{-3} A$