In a series circuit, C=2 μ F , L=1 mH and R=10 Ω. When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

Question

In a series circuit, C=2 μ F , L=1 mH and R=10 Ω. When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be (A) 1: 1 (B) 1: 2 (C) 2: 1 (D) 1: 5

Tardigrade Admin · Accepted Answer

Current will be maximum in the condition of resonance, so i max =(E/R)=(E/10) A Energy stored in the coil, WL=(1/2) L i max 2 =(1/2) L((E/10))2 =(1/2) × 10-3((E2/100)) =(1/2) × 10-5 E2 joule Energy stored in the capacitor, WC=(1/2) C E2 =(1/2) × 2 × 10-6 E2 =10-6 E2 joule ∴ (WC/WL)=(1/5)

Q. In a series circuit, $C = 2 μ F, L = 1 m H$ and $R = 10 Ω$ . When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

1 : 1

1 : 2

2 : 1

1 : 5

Q. In a series circuit, C=2μF,L=1mH and R=10Ω. When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

1 : 1

1 : 2

2 : 1

1 : 5

Current will be maximum in the condition of resonance, so imax​=RE​=10E​A Energy stored in the coil, WL​=21​Limax2​ =21​L(10E​)2 =21​×10−3(100E2​) =21​×10−5E2 joule Energy stored in the capacitor, WC​=21​CE2 =21​×2×10−6E2 =10−6E2 joule ∴WL​WC​​=51​

Q. In a series circuit, $C = 2 μ F, L = 1 m H$ and $R = 10 Ω$ . When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be