In an LC circuit shown in figure C=2μF and L=2μH. At t=0 charge on capacitor is 43 μC and at same instant is increasing at a rate of 6 μC/s.Choose the correct statement(s):

q=q0sin(ωt+π6) i=q0ωcosπ6 i=q0232⇒q0=243=83 Alternatively q022C=1(43)22+12×2×(6)2 q0=83 12Li02=q022C⇒i0=43

In an LC circuit shown in figure C=2μF and L=2μH. At t=0 charge on capacitor is 43 μC and at same instant is increasing at a rate of 6 μC/s.Choose the correct statement(s):

q=q0sin(ωt+π6) i=q0ωcosπ6 i=q0232⇒q0=243=83 Alternatively q022C=1(43)22+12×2×(6)2 q0=83 12Li02=q022C⇒i0=43

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In an LC circuit shown in figure $C = 2 μ F a n d L = 2 μ H .$ At t=0 charge on capacitor is $4 \sqrt{3} μ C$ and at same instant is increasing at a rate of $6 μ C / s .$
Choose the correct statement(s):

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

Total energy stored in circuit is $24 μ J$

Total energy stored in circuit $48 μ J$

Maximum charge on capacitor is $8 \sqrt{3} μ C$

Maximum current through inductor is $4 \sqrt{3} μ A$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$q = q_{0} \sin (ω t + \frac{π}{6}) i = q_{0} ω \cos \frac{π}{6} i = \frac{q_{0}}{2} \frac{\sqrt{3}}{2} \Rightarrow q_{0} = \frac{24}{\sqrt{3}} = 8 \sqrt{3} A l t e r n a t i v e l y \frac{q_{0}^{2}}{2 C} = \frac{1 {(4 \sqrt{3})}^{2}}{2} + \frac{1}{2} \times 2 \times {(6)}^{2} q_{0} = 8 \sqrt{3} \frac{1}{2} L i_{0}^{2} = \frac{q_{0}^{2}}{2 C} \Rightarrow i_{0} = 4 \sqrt{3}$