In the above circuit, C = √(3)/2 μ F, R2 = 20 Ω, L = (√(3))/(10) H… | JEE Main 2019 (12 Jan, Shift 2)

Q11JEE Main 2019 · 12 JAN · SHIFT 2Difficult

Parallel AC Circuits

In the above circuit, C = $\frac{\sqrt{3}}{2}\ \mu$ F, R2 = 20 $\Omega$ , L = $\frac{\sqrt{3}}{10}$ H and R1 = 10 $\Omega$ . Current in L-R1 path is I1 and in C- R2 path it is I2. The voltage of A.C. source is given by, V = 200 $\sqrt{2}$ sin (100 t) volts. The phase difference between I1 and I2 is:

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Answer: (c)

$\omega = 100$ : $X_L = 100\times\frac{\sqrt{3}}{10} = 10\sqrt{3}\ \Omega$ , $\tan\phi_1 = \frac{10\sqrt{3}}{10} = \sqrt{3}$ , so $I_1$ lags $V$ by $60^\circ$ .
$X_C = \frac{1}{100\times\frac{\sqrt{3}}{2}\times10^{-6}} \approx 1.15\times10^4\ \Omega \gg R_2$ , so $I_2$ leads $V$ by about $90^\circ$ .
Phase difference $\approx 60^\circ + 90^\circ = 150^\circ$ .
With the data as printed this gives (a) $150^\circ$ ; the key gives (c) $30^\circ$ .

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Parallel AC Circuits question from JEE Main 2019 (12 Jan, Shift 2)

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