Explanation
$$V_R = 2sin (1000t)V$$ So, $$V_{max}(R) = 2V$$
So, $$w = 1000$$
$$2\pi f = 1000$$
$$f = \dfrac{500}{\pi}Hz$$
$$X_L = wL$$
$$= 1000\times 4$$
$$= 4000 \Omega$$
$$R = 100 \Omega$$
$$i_R = \dfrac{V_R}{R} = \dfrac{2}{100} = 2\times 10^{-2}A = i_L$$
($$\because$$current is same in series)
$$V_{max}(L) = i_L\times X_L$$
$$= 4000\times 2\times 10^{-2}$$
$$= 80V$$
So, $$V_L = 80 sin (1000t + \dfrac{\pi}{2})$$
Since in inductor voltage lead by $$\dfrac{\pi}{2}$$
Impedance = $$X_L - X_C = \omega L-\dfrac{1}{\omega C}$$
$$= 2\pi f L -\dfrac{1}{2\pi f C}$$
$$= 2\pi \times 50\times 1-\dfrac{1}{2\pi \times 50\times 10\times 10^{-6}}$$
$$= \pi \left (100 - \dfrac{10^6}{2{\pi}^2\times 50\times 10} \right )$$
$$= \pi \left(100-\dfrac{10^6}{2\times 10 \times 50 \times10} \right)$$
$$(\because {\pi}^2 = 10)$$
$$= 0$$
So, impedance $$= 0$$
Natural frequency $$= 125000 Hz$$
New frequency $$= 125000 - 25000$$
$$ = 100000 Hz$$
Frequency of L-C circuit
$$2\pi f = \dfrac{1}{\sqrt{LC}}$$
So, $$2\pi f_1 = \dfrac{1}{\sqrt{LC_1}}$$ ---(1)
$$2\pi f_2 = \dfrac{1}{\sqrt{LC_2}}$$ ---(2)
Dividing (1) by (2)
$$\dfrac{f_1}{f_2} = \sqrt{\dfrac{C_2}{C_1}}$$
$$\sqrt{\dfrac{C_2}{C_1}} = \dfrac{125000}{100000} = \dfrac{5}{4}$$
$$\dfrac{C_2}{C_1} = \dfrac{25}{16}$$
$$k = \dfrac{C_2}{C_1} = \dfrac{25}{16} = 1.56$$
In the problem $$X_C = 4\Omega$$and $$X_L = 4\Omega$$So, V across $$X_C$$ and $$X_L$$ will be same and in opposite direction, So net voltage will be zero. Since voltmeter is connected parallel to Capacitor and inductor so, it will read 0 volts.
Current $$=$$ $$\dfrac{V}{impedance}$$
$$Z = R $$ as $$X_L = X_C$$
Current$$= \dfrac{90} {45} = 2\ A$$
Since current lead and lag are same So, circuit is in resonance, so, circuit is purely resistive circuit.
So, $$i = \dfrac{V}{R}$$
$$= \dfrac{200}{100}$$
$$= 2A$$
A
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