The maximum capacitance of a variable capacitor is $33\; pF$. What should be the self-inductance to be connected to this capacitor for the natural frequency of the LC circuit to be $810 \; KHz$. Corresponding to A.M. broadcast band of Radio Nepal?

Given, Capacitance $(C) = 33\; pF = 33 * 10^{-12}\;F$

Let, $L$ be the self-inductance connected with the capacitance, so that the natural frequency of this LC circuit is $810\; KHz$,

$i.e\; \; \; \;  f = 810 \; * 10^3 \; Hz $

Now, we have 

$f = \frac{1}{2 \pi \sqrt{LC}}$

$L =$ $ \frac{1}{4 \; \pi^2 \; (810 \; * \;10^3)^2 \; * \; 33 \; * \; 10^{-12}}$  $= 1.17 * 10^{-3}\; H$

Thus, Inductance connected $(L) = 1.17 * 10^{-3}\;H$

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6.8 Wattless and Wattful Current:

The average power over a cycle of AC is given by,

$P_{av} = V_{rms}. I_{rms} . Cos\phi$ .......... (i)

Here $Cos \phi$ is the power factor and $\phi$ is the phase factor.

In pure inductor or an ideal capacitor, $\theta = 90^0$. So average power consumed in a  pure inductor or ideal capacitor is,

$P_{av} = V_{rms}. I_{rms} . Cos 90^0 = 0$

Therefore, Current through pure 'L' or pure 'C' which consumes no power for its maintenance in the circuit is called Wattless Current.

At resonance $X_L = X_C$ and $\theta = 0$

∴ Cos $\theta =$ Cos $0^0 = 1$

Therefore Maximum power is dissipated in a circuit at resonance.

The Current through resistance (R), which consumes power for its maintenance in the circuit is called Wattful Current.

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