JEE PYQ: Capacitance Question 39
Question 39 - 2019 (12 Jan Shift 1)
In the above circuit, $C = \sqrt{3},\mu$F, $R_2 = 20,\Omega$, $L = \frac{\sqrt{3}}{10}$ H and $R_1 = 10,\Omega$. Current in L-$R_1$ path is $I_1$ and in C-$R_2$ path it is $I_2$. The voltage of AC source is given by, $V = 200\sqrt{2}\sin(100t)$ volts. The phase difference between $I_1$ and $I_2$ is:
(1) 60$^\circ$
(2) 30$^\circ$
(3) 90$^\circ$
(4) None of the above
Show Answer
Answer: (4)
Solution
Capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{2 \times 10^4}{\sqrt{3}}$. $\tan\theta_1 \approx$ close to $90^\circ$. For L-R circuit: $X_L = \omega L = 100 \times \frac{\sqrt{3}}{10} = 10\sqrt{3}$. $\tan\theta_2 = \frac{X_L}{R_1} = \sqrt{3}$, $\theta_2 = 60^\circ$. Phase difference $= 90^\circ + 60^\circ = 150^\circ$. If $R_2$ is 20 k$\Omega$, then phase difference comes to $60 + 30 = 90^\circ$.
BETA
