A resistor ‘R’ of resistance $2000 Ω$ and two similar capacitor ‘A’ and ‘B’ of capacitance $1 μF$ each are connected with an A.C. source of 200 volt and 500 rad/s. Now switch is closed at t=0, choose the correct option(s).

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

The phase difference between applied voltage and voltage across the capacitor A is $\tan^{- 1} (\frac{1}{3})$

Power dissipated through resistor is 4 watt.

The phase difference between current supplied by source and current in the capacitor ‘B’ is zero.

The current in the capacitor ‘A’ is $\frac{1}{5 \sqrt{20}}$ ampere.

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$x_{c} = \frac{1}{ωC} = 2000 Ω . . . (i)$

$i^{2} = {(i_{C})}_{A}^{2} + {(i_{R})}^{2}; z' = 1000 \sqrt{2} . . . (ii)$

$V^{2} = {(V_{C})}_{B}^{2} + {(V')}^{2} + 2 V_{C} . V' \cos 45 °; z = \sqrt{20} . {(10)}^{3}; i = \frac{200}{\sqrt{20} \times 10^{3}} = \frac{1}{5 \sqrt{20}} . . . (iii)$

$i_{R} = icos 45 ° {(i_{C})}_{A} = icos 45 ° tanα = \frac{V' \sin 45 °}{{(V_{C})}_{B} + V' \cos 45 °} = \frac{1}{3} . . . (iv) Power = i^{2} R = {(\frac{1}{5 \sqrt{20}})}^{2} . 2000 = 4 W$