In the given circuit, the magnitude of V _ L and V _ C are twice...

Since the voltage across the inductor and capacitor is same, the circuit is in resonance. Therefore, the current in the circuit will be,

$i = \dfrac{V}{R} = \dfrac{220}{5} = 44A$.

The voltage across the inductor is double that across the resistor. Therefore, ${V}_{L} = 2{V}_{R} = 440V$.

We know that ${V}_{L} = i{X}_{L} = i \times 2\pifL ⇒ L = \dfrac{440}{44 \times 100\pi} = \dfrac{1}{10\pi}H = \dfrac{100}{\pi}mH$

Therefore, $K = \dfrac{1}{100} = 0 . 01 \approx 0$.