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A choke coil of resistance $5 Ω$ and inductance 0.6H is in series with a capacitance of $10 μF$ . If a voltage of 200 V is applied and the frequency is adjusted to resonance, the current and voltage across the inductance and capacitor are I₀, V₀, I₁ and V₁ respectively. We have

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$V_{1} = 9 .8 kV$

$V_{1} = 19 .6 kV$

$V_{0} = 9 .8 kV$

$I_{0} = 40 A$

answer is A, B, C.

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Detailed Solution

Given voltage is 200V;

$L = 0 .6 H; C = 10^{- 5} F; R = 5 Ω$

in resonance condition, $Lω = \frac{1}{ωC} \Rightarrow X_{L} = X_{C}$

and the net impedance of the circuit is Z = R

$\Rightarrow$ the net current $i_{0} = \frac{V}{R} \Rightarrow i_{0} = 40 A$

The Potential drop across C and L is same.

$\begin{array}{l} V = {iX}_{L} = 40 A (Lω) = 40 A (\sqrt{6 \times 10^{4}}) Ω; \\ V = 9 .8 KV \end{array}$

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