In the given circuit, the AC source has angular frequency $ω = 100 r a d / s$ , Considering the inductor and capacitor to be ideal, the correct choice(s) is/are

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The current through the circuit, I is 0.3A

The current through the circuit, I is $0.3 \sqrt{2} A$

The voltage across $100 Ω$ resistor $= 10 \sqrt{2} V$

The voltage across $50 Ω$ resistor $= 10 V$

answer is A, C.

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Detailed Solution

$ε_{0}^{2} = i_{10}^{2} (R_{1}^{2} + X_{c}^{2}) \Rightarrow i_{10} = \frac{ε_{0}}{\sqrt{R_{1}^{2} + X c^{2}}} Q_{1} = \tan^{- 1} (\frac{X c}{R_{1}}) = \frac{π}{4}$

$ε_{0}^{2} = i_{10}^{2} (R_{1}^{2} + X_{L}^{2}) \Rightarrow i_{20} = \frac{ε_{0}}{\sqrt{R_{1}^{2} + X_{L}^{2}}} Q = \tan^{- 1} (\frac{X_{L}}{R_{2}}) = \frac{π}{4}$

$i_{0} = \sqrt{i_{10}^{2} + i_{20}^{2}} x_{L} = w L = 100 \times 0.5 = 50 Ω; X_{c} = \frac{1}{w c} = \frac{1}{10^{2} \times 100 \times 10^{- 6}} = 100 Ω ∴ i_{c} = \frac{20}{100 \sqrt{2}} = \frac{\sqrt{2}}{10} A; i_{L} = \frac{20}{50 \sqrt{2}} = \frac{\sqrt{2}}{5} = \frac{2 \sqrt{2}}{10} A i = \sqrt{i_{c}^{2} + i_{L}^{2}} = \frac{\sqrt{2}}{10} \sqrt{5} = 0.316 A$

Voltage across $R_{1} (100 Ω) = i_{0} R_{1} = \frac{\sqrt{2}}{10} \times 100 = 10 \sqrt{2} Ω$
Voltage across $R_{1} (50 Ω) = i_{20} R_{2} = \frac{2 \sqrt{2}}{10} \times 50 = 10 \sqrt{2} Ω$